Mock Discussion Session: Statistics II

[00:00:29]Good evening everyone.

[00:00:32]>> Good evening ma'am.

[00:00:33]>> Good evening. How are you all? All good.

[00:00:35]>> Yes.

[00:00:36]>> Yeah. Okay. So, how's the progression going on?

[00:00:42]It's going on but there are few problems actually ma'am mock there is a multiple uh this question >> choice uh >> no no uh this is uh uh this first question uh in mock no but >> it is not uh asked in pyq's >> so what >> uh ma'am it's Wait a second.

[00:01:16]Moment is it has this uh um multiple discrete random variables when joint PMF of uh X Y and Zed.

[00:01:26]>> H correct. So it's in the content right it's from week 1 to four only right?

[00:01:32]>> Yes ma'am.

[00:01:33]>> No so anything can come up right?

[00:01:37]>> Okay. So it is not based on the previous year question paper. It's based upon the content.

[00:01:42]>> Okay.

[00:01:43]>> Okay. Yes sir.

[00:01:46]>> Okay.

[00:01:46]>> I was having this problem in uh this question. Can we discuss?

[00:01:50]>> Yeah. So I will be discussing uh the whole question like the whole question paper.

[00:01:56]>> Okay. Yeah. My screen is visible to everyone right?

[00:01:59]>> Yes ma'am.

[00:02:00]>> Okay. What about others? Have you guys done the mock paper?

[00:02:06]Yes, ma'am.

[00:02:08]>> Yeah. How was it? Was it easy? Fine. How was it?

[00:02:14]>> Some questions were easy, some were difficult.

[00:02:17]>> Okay.

[00:02:18]>> Yes, ma'am. I found difficulties in the graph question.

[00:02:22]>> Graph question. Okay.

[00:02:25]>> Okay. So, let's do the first question.

[00:02:27]So, I'm just um Okay. So, you guys have opened up your question like the mock discussion, right? Because I'm not going to share that thing. I'm just going to share this the solution this notepad only. Okay.

[00:02:46]>> Okay.

[00:02:46]>> Is the system going to live stream?

[00:02:49]>> Yeah. Yeah.

[00:02:51]>> So in the first question they are saying that my X is following a Bernardi distribution with 1x3 as a parameter.

[00:02:56]Then Y is also following a Baldi with 1x2 as a parameter. Zed is following a Bernardi with 2x3 as a parameter and what all is given f(x,yz) that is the joint distribution is given to be 1x4 fz given y = 1 or = to0 is given to be 1x 3 and all. I need to find out the value of f of x given y =0 = 1 0. Okay, this is what I need to find out. Okay.

[00:03:29]Now first of all if this thing how will I write it? Like the formula the formula will be what? f(x) given y = 0 = 1. This can I write it? F(x,yz) 0a 1 / f of what will it be? Y za 1.

[00:03:56]>> Yes.

[00:03:56]>> Yes.

[00:03:57]>> Okay. Now if you see here this value is actually given to be here which is 1x4 correct but I need this value. So first of all I will find out this value. So what is my f of y 0a 1?

[00:04:11]Can I write it fz given y = 1 0 / sorry multiplied by f of 1. Can I >> everyone agrees with this thing?

[00:04:29]Yes ma'am.

[00:04:30]>> Okay.

[00:04:30]>> No ma'am. Um it should be zed is here.

[00:04:34]It's given zed is equal to 1. You're giving here zed equal to zero.

[00:04:45]Sorry.

[00:04:50]Zero corner.

[00:04:59]f y =0 sorry.

[00:05:08]So this will be um 016 0 comma one that is y is taking zero and zed is taking one correct.

[00:05:42]So f of y set one second.

[00:06:14]Yeah. Okay. Fine. That was fine. Right.

[00:06:17]See the value is f of z given y is equal. See f of y given >> I don't think anything is wrong here.

[00:06:24]>> Yeah. That was fine only. So this is actually what f of z y 0a 1 divided by f of y1.

[00:06:33]So this will be f of z y 0a 1 that is f of z given y = 1 um 0 f of y1 now this will be f of z uh given y = 1 what will be this value f of z given y = 1 this is given to be 1x 3 according to the Yes or no?

[00:07:06]>> Yes. No.

[00:07:07]>> Yeah. Okay. And then f of y1. What is f of y1?

[00:07:16]>> Half.

[00:07:18]>> It will be half.

[00:07:20]Okay.

[00:07:22]So this is my f of zed that will be 1 by 6. Okay. Now [clears throat] so just put this value here. So this will be 1x4 divided by 1x 6. Correct?

[00:07:39]>> No ma'am ma'am it's f 5 z1 and here y is taking value zero and zed is taking value one and here it is taking zed value zero and y is taking value one. It's opposite.

[00:07:55]Yes.

[00:07:59]It's uh yes ma'am. It's as uh f of z f of z is given to be 1 by 3 find the value of 0 0 1 m you should 1 - 1 by Okay.

[00:08:40]>> Now there are like two two conditional probabilities using we are using >> Z given Y = 1 is 0.

[00:08:56]Z.

[00:09:03]So from this we just got the value of 0 comma 1.

[00:09:14]Why is it Okay. So this is my y z value is 1 comma 0 is 1x 6. Okay. So I need the value of 0. Okay. So what we going to do is so we know that my x and y is taking values.

[00:09:34]It's a bol distribution right 1x 2 and 1 uh 2x3. Correct? Okay. So my y is this and my zed is this. Fine. So now y is taking two values 0 and one. And it's taking 0 and one. Fine. So now so it's written my f of z y is 0a 1.

[00:09:54]That means when my z is y z, this should be y z.

[00:10:04]When my y is zero and my zed is one, it's going to be 1x 6. Okay. And then f of just Okay. Then f of y1 is this is 1x 6 set >> ma'am I think it will be z y only should be 1 comma 0.

[00:10:50]So my y is taking this thing. So this will be okay. So then my y is this. My z is 0. That is 1x 6.

[00:10:59]This is 1x 6.

[00:11:02]This thing is f of z. This thing is f of pi. Okay. Now what all it's given?

[00:11:11]Okay. So what is the probability that my when my y is zero it will be taking probability 1x2 and this is also 1x2 correct and f of zed is also this will be 1x 3 this will be 2x3 okay now what will be this value that is when zed is taking one and y is taking one can you guys tell me 2x 6 >> yes ma'am >> yes and this will this will also be 2x 6 H and this will be 1x 6. So we have made the PMF table. Now from here we need to find out this value which is the f of y 0a 1. So what from this graph f of y za 1 what will be this value?

[00:12:02]2x 6.

[00:12:06]Yes or no?

[00:12:10]That is this this value.

[00:12:11]>> Yes ma'am.

[00:12:12]>> Okay.

[00:12:12]>> Yes ma'am. Yes sir.

[00:12:13]>> Yeah. So now we got this value. So just put it here. So this will be 2x 6. Okay.

[00:12:19]So what will be my answer? This will be 3 by 4. Okay. Which is 0.75.

[00:12:28]Question.

[00:12:30]>> Yes ma'am. Please wait. I'm just noting it. Um >> I will approve this like in the supplementary content only.

[00:12:40]Okay.

[00:12:41]>> Yes sir.

[00:12:42]>> I will be uploading it like in the supplementary content. You don't have to >> note it down.

[00:12:52]>> Okay.

[00:12:53]>> Yes ma'am.

[00:12:54]>> Okay. So is the question clear to everyone?

[00:12:57]>> Yes ma'am.

[00:12:58]>> Okay. So now it is clear.

[00:13:00]>> Now it's fine. Okay.

[00:13:01]>> Yes ma'am.

[00:13:03]>> Now what is the second question? A box contain seven milk chocolates and three dark chocolates.

[00:13:10]>> Uh this was done.

[00:13:12]>> This is done. Okay, let I'll just um solve all the questions because if somebody has doubts, we can just go into it. So seven milk chocolates and three dark chocolates. Two chocolates are drawn at random. So from here two chocolates are being drawn without replacement from the box. Let a random variable x denote the number of milk chocolates drawn. So how many milk chocolates he can draw out? He can draw 0 1 and two. Correct? Yes ma'am.

[00:13:35]>> What is the probability that X equals to 0? That means there are no milk chocolates. So 7 3 C2 by 10 C 32 then >> sorry >> 3 C2 >> 3 C2 divided by 10 C >> 10 C2 >> then probability that X = to 1 this will be from the seven we'll take one and from here we can >> C1 3 C >> one is here right so this one 10 C2 >> 10 C2 >> probability X = 2 will be 7 C2 divided by 10 C2 okay I'm not solving other thing this is the this will be The first will be 1.5 1x 15 then second will be 7 by 15 and third will be 7 by 15 we will draw the table and we have the PMF >> 74 >> 01 two yeah it's total one coming so this is fine now what is the expected value so what will be the expected value expected value will be 0 into 1x 15 plus 7 7 by 15 into 1 + 7 x 15 into 2 because what is my expectation it is actually summation of x into probab probability of X correct for the discrete random variable.

[00:14:43]Okay. So solve this and can you tell me the answer? What is the answer coming out to be?

[00:14:49]>> 1.4.

[00:14:50]>> 1.4. Sure. Is that the answer?

[00:14:55]Uh is it coming for everyone?

[00:14:58]>> Yes ma'am.

[00:14:59]>> Yes sir.

[00:15:00]>> Okay. Fine.

[00:15:00]>> Yes ma'am.

[00:15:02]>> So till question three there is no doubt right? Everything is fine.

[00:15:06]>> Yes ma'am.

[00:15:06]>> Okay. Let's jump to question number four. Question number four is a fair six-sided marked one on one face.

[00:15:17]>> So we have six traders here. So at one we have on one phase one then two one is two phase that means two two then three on the remaining three pieces that is 3 three three. Okay. A dice twice. Let a random variable x denote the number of obtained on the first row. So x is the number obtained in first row.

[00:15:43]Y is you know the number obtained on the second row number obtained in the second row. Now define a new random variable zed is equals to x + y. So find the range of zed. So what will be the range of zed?

[00:16:00]First you will find out. Okay. So the thing is you can find out like this. So what what all x values uh that it can take? It can take any values right from 1 2 and three right.

[00:16:11]Similarly the y also can take the same thing. So we can just make a table. So what we can do is so we can have like uh with one I can have two with one I can have sorry with 1 1 then 1 2 then 1 3. Correct?

[00:16:27]I can have a situation like this. Right?

[00:16:32]>> Yes.

[00:16:34]Correct. Okay. Now [clears throat] then next 2 1 2 2 2 3 correct similar 3 1 3 2 and 3 3.

[00:16:52]Okay. So I can have these different situations here. Okay. So what all Zed is what? X + Y. So if I add both of the things if I like 2 3 4 3 4 5 then 4 5 6. So what is the range coming out to be?

[00:17:13]What are the different distinct values that you can take?

[00:17:16]>> It is taking 3 4 5 and 6. Correct? Is this clear? So the range is actually coming out to be 2 3 4 5 6. Is it fine?

[00:17:31]>> Yes ma'am.

[00:17:33]>> Okay. Now for the fifth question that is probability my zed is equals to three.

[00:17:38]How will I count? That means my probability x + y should be equal to three. So at what point I will have three? That means this one.

[00:17:49]>> Yes ma'am.

[00:17:50]>> Okay. this one any other place that is there are two points right so when are the possibilities when my probability that my x = to 1 y = 2 and probability that my y is um x = 2 and y = 1 correct so now can you tell me the um the probabilities you can say 1x 6 into 2x 6.

[00:18:50]Yes.

[00:18:52]+ 2x 6 into 1x 6 I'm not getting yes or no.

[00:19:06]>> Yes ma'am.

[00:19:07]>> Yes ma'am.

[00:19:09]>> 1 by 9 is it?

[00:19:11]>> Yeah. So this will be coming out to be 1 by 9.

[00:19:14]>> Yes ma'am. So this is I guess the answer is 1 by 9 uh option is there. Okay. So is this fine?

[00:19:23]>> Yes ma'am.

[00:19:24]>> Okay.

[00:19:31]>> Then the next question I guess so we have the joint PMF. We need to find out the coariance.

[00:19:40]So what is the formula of coariance?

[00:19:45]X Y >> expectation of XY minus expectation of X into expectation of Y. Correct? Okay. So first of all you will find out expectation of XY. Expectation of XY will be what?

[00:19:59]It will be kind of like X into Y f of XY correct.

[00:20:06]This is what you're going to do right?

[00:20:08]>> Yes ma'am.

[00:20:09]>> So this will be what? 0 into 0 into 1x 12 then plus 0 into 1 into 1x 6 and so on you will write out right so what we're going to do is we're going to just leave the first row and the first column okay so in short I will be having um 1 into 1 into 1x3 plus 2 into 1 into 1x X >> correct?

[00:20:42]>> Yeah ma'am.

[00:20:43]>> Yeah. So this will be what? 1x3 + 1x 3 which is 2x3.

[00:20:50]Okay. Now similarly what is my expectation of x? Can you guys calculate and tell me?

[00:21:05]>> 2x3.

[00:21:07]>> Expectation of x is 2x3. What is expectation of y?

[00:21:20]>> One >> one. Okay. So now just put this value.

[00:21:25]So this will be two. So if you see here my coariance is coming out to be zero right. So 2x3 minus 2x3 into 1. So this will be zero. So coariance is zero. That means they are uncorrelated.

[00:21:39]Everybody is getting that answer.

[00:21:42]>> Yes ma'am.

[00:21:43]>> Yes.

[00:21:43]>> Okay. Now can I move on? So till here no doubts till question six.

[00:21:51]Uh ma'am could you explain this again? I don't understand.

[00:21:56]>> Okay. So basically uh what what I need to explain like from where >> uh from the second step that I understood co is this after that okay you have taken this formula >> m Okay.

[00:22:24]>> Okay. Got it.

[00:22:26]>> Got it.

[00:22:27]>> Yes sir.

[00:22:30]>> Okay. So shall we proceed to the question number seven?

[00:22:34]>> Yes sir.

[00:22:35]>> Okay. So now question number seven is uh consider a function R to R is that FX.

[00:22:40]So FX is given to be CX² CX 0 where one this one is given two where C is a real constant find C such that the function F is a valid density function. Okay. So how will you find out what is the method that you will do?

[00:23:07]Anyone >> take >> integrated?

[00:23:11]>> We will integrate. We will integrate it.

[00:23:14]Integrated from where to where?

[00:23:17]>> How will you do it?

[00:23:20]>> And the first equation from one to >> the second then we add it the first first equation from 1 to two.

[00:23:27]>> Correct.

[00:23:28]>> Then we add it the second equation from 2 to 4. 2.

[00:23:33]>> Yeah, that's it. So this will be equal to one. So now if you solve this so this will be x qx 3 1 by 2 + c x² by 2 2x4 = 1. So now this will be what 8x 3 - 1x 3 + 16 by 2 + 4x2 correct. So now if I solve this so this is coming out to be - by 3 + >> ma'am - 4x - 4 by 2 >> yeah 1 minus so this will be 16 - 4 into 6 so this will be 2 5 by 3 so c will be 3 by 25 Yes.

[00:24:34]Is what is the answer that you guys are getting?

[00:24:38]>> Yes ma'am.

[00:24:39]>> Yes ma'am. Same.

[00:24:40]>> Okay. So this is my answer that is 3x 25. Is that fine?

[00:24:46]>> Uh ma'am please wait. Uh what is okay?

[00:24:49]3x 25.

[00:24:50]>> Uh three. Okay.

[00:24:54]>> Yes ma'am. Okay >> ma'am. For the uh density function always to do integration like they give the int but if we don't have the in >> sorry s one second >> uh for density functions to find out the >> uh value we have to do the integrations right always like this.

[00:25:13]>> Yeah. So the yeah that is the thing for continuous you will always do integration. So for the continuous always remember do integration and for the discrete if you have discrete distribution you will always do sum.

[00:25:24]Okay. So you know that the sum the integr integration of all the f(x) dx will be equal to one and summation of uh px all the probability will be equal to one." Okay.

[00:25:37]>> Got it. Thank you.

[00:25:38]>> Okay.

[00:25:40]Now let's proceed to question number eight.

[00:25:44]What is question eight? So the joint PMF of two discrete random variables. So my f of xy x y is given to be 1 by 27 2x + y where my xy belonging to 0a 1a 2 1. Yes.

[00:26:10]Identify the current joint payment table of x. Yeah.

[00:26:16]Okay. So this is actually so this let's suppose this is my x this is my y. So what are the values that x is taking?

[00:26:29]>> 012 >> 012 >> 0 1 and 2. And what are the values that y are taking?

[00:26:34]>> Same 012.

[00:26:36]>> Okay. Now what is the value when x is 0 and y is 0?

[00:26:44]Zero >> zero >> zero.

[00:26:48]Okay.

[00:26:50]When and what is this next value?

[00:26:56]>> 2x 27.

[00:26:58]>> 2x 27.

[00:27:00]>> 2x 27. So how did you guys do it in this equation? That is 1x 27. 2x + y. Just put the value of x as 1 and y =0. Right?

[00:27:11]So this will be 1 by 27. 2 into 1 + 0.

[00:27:16]So this will be 2x 27.

[00:27:20]Is this fine to everyone?

[00:27:23]Okay. Can I get this into the next one?

[00:27:28]>> Yes ma'am.

[00:27:29]>> What is this value when x is 2 by 0?

[00:27:40]4 by 27.

[00:27:41]>> 4 by 27.

[00:27:43]>> Yeah. Next. X0 by 1.

[00:27:46]>> 1 by 27.

[00:27:48]>> 1 by 27.

[00:27:50]>> Okay. Next.

[00:27:53]>> 3x 27.

[00:27:56]Next.

[00:27:58]>> 5 by 27.

[00:28:01]>> Next. 2 by 27. 2 by 27. 2x 27. Okay. 4 by 4 by 2 by 27. Hm.

[00:28:13]67.

[00:28:16]>> Yeah. So this is my joint amount of X and Y. So is this fine to everyone?

[00:28:24]So I guess everybody's getting this answer. Yes or no?

[00:28:27]>> Yes.

[00:28:28]>> Yes ma'am.

[00:28:29]>> Okay. Yes.

[00:28:29]>> So then next question is question number nine.

[00:28:39]So what is my 9? I need to find the probability that my x + y is less than = 2 given that my x is greater than zero.

[00:28:47]Okay. So the same thing the conditional thing right.

[00:28:52]So I can write this as probability of x + y less than = 2 intersection this and probability that x is greater than zero. This one correct. So now I need to find out like the thing is my x is taking values it should be greater than zero and the values after adding x + y it should be less than equal to two right it can take value 0 1 and two that's it okay so what all uh the probabilities I can take can you guys tell me one and two only x= 1 and two baba that's my x = 1 comma y = 0 >> 0. Yes, ma'am.

[00:29:41]>> No ma'am. One >> by one.

[00:29:44]>> Ma'am, because um x should be greater than zero. So x is one and y should also be one.

[00:29:52]>> No, it's just saying that my x is greater than zero. It's not talking about anything about y, right?

[00:29:59]>> Uh yes, ma'am. Okay. Yeah. Yeah.

[00:30:02]So now you tell me like what all values >> x = 1 and y = 1.

[00:30:11]>> x = 1, y = 1. Okay, fine. Then >> x2 y= >> x = >> 2 y = >> two. Any other?

[00:30:30]Then y >> okay one second. If I take this thing >> y should be zero.

[00:30:38]>> Y should be zero ma'am.

[00:30:40]>> Correct.

[00:30:43]Any other?

[00:30:45]>> No ma'am.

[00:30:47]>> That's it.

[00:30:49]Okay.

[00:30:50]>> In first case you are having x= to 1 and y is equals to 0. So it should be it shouldn't be >> it should be >> how it is how how you have taken that >> see what is the condition that they are stating is they they are stating that my x + y the first condition is my first condition is if I add both of these terms it should be less than equal to two that means what all values I can have after adding it should be either zero it should be either one or it should be two correct >> right and the second condition that they have stated is my x should be greater than zero. So what all values that x can take? It can take only 1 comma 2.

[00:31:30]Correct?

[00:31:31]>> Okay. Okay. Okay.

[00:31:32]>> Right. Yeah. So that is the conditions that we have to use. So according to this condition, these are the probabilities that I can only take.

[00:31:39]>> Yes. Got it. Got it.

[00:31:40]>> Yeah.

[00:31:42]>> Okay. So now, >> so what is the probability that x= to 1 and y is equals to 0?

[00:31:57]X= 2 by 2 by 27 >> 2x 27 Then next value when X is 1 and Y is 1 by 3 by 27 >> 3 by 27 and then when X is 2 and Y is 0 >> 4 by 2 >> 47 Okay now what is probability that my X is greater than zero can you Tell me this value >> one.

[00:32:26]>> One. Sure.

[00:32:29]>> X = 1 and 2 X= to 1 and 2.

[00:32:35]>> Yeah. Probability that my X is 1 and plus probability that my X is 2.

[00:32:40]Correct.

[00:32:41]>> Probability X = to 1 and Y = 2.

[00:32:45]>> No, this whole term >> 9 by 27 and 15 by 27. Yes.

[00:32:50]>> Correct. So this is 9 by 27 and 15 by 27. Now you guys solve this and tell me what is the answer like in fraction only tell me what is the answer that you're getting 0.375 is it >> 0.375 okay uh what about the others like everybody's getting this answer or like different answers Yes ma'am.

[00:33:29]>> Yes ma'am.

[00:33:30]>> Answer.

[00:33:31]>> No it's 0.375 becomes that.

[00:33:36]>> Okay fine. So this is so yeah that is the only question.

[00:33:43]>> The denominator can be also written as 1 minus probability of x less than equal to 0.

[00:33:50]>> 1 - x less than huh that is will be very easy. Right? Then you need to find out only the probability of x= to 0. That's it.

[00:33:59]Is the question clear to everyone? This one?

[00:34:02]>> Yes.

[00:34:02]>> This one?

[00:34:03]>> Yes.

[00:34:04]>> Now, next question.

[00:34:08]Question number 10.

[00:34:13]Use the following information. An unbiased coin is tossed three times independently. Let x represent the x is actually the number of heads and y is number of tails in the first two toes.

[00:34:37]Okay. Define another random variable u is equal to xy. Find the pmm of u. Okay.

[00:34:44]So how will you guys find out this thing?

[00:34:48]Ma'am, we'll first write down like uh the all the possible combinations.

[00:34:53]>> Correct. So, so if we are tossing out the coin three times, right? So, what are the possible uh outcomes? The first can be HH, then HHT, then TH, then we can have uh then we can have HTH, right? then tth then THT then HTT and TTT correct these are the possible values that I could have now >> sir >> now if you look into here like I'm just talking about for each of the outcomes so for this case what will be X what will the value of X here >> it will be three what will be the value of Y here >> zero >> it will be zero now similarly For this thing, what will be the value of x here?

[00:35:46]>> Two.

[00:35:48]>> Y >> zero.

[00:35:50]>> Correct. Now here x >> two >> y >> one >> one >> and y on the second.

[00:36:00]>> Sorry. Sorry.

[00:36:02]>> On the head head tail y should be one.

[00:36:06]>> No, just read the question. Y specifically what the number of tails in the first two. That means we will be taking consider only tails which are in the first two tosses. Okay.

[00:36:16]>> Now for the fourth one this x= to 2 and y = to >> one >> one. This one x = to >> 1 1 >> y >> 1 2 for this one x = to >> 1 1 >> y1 >> y1 x = 1 1 >> y1 >> y1 >> y1 >> x= 0 >> 0 >> y >> correct. Now here what will be the value of u >> zero >> zero then this thing here >> zero for the second part could you explain for tst why it is x= to 1 and y equ= to one >> which one which one >> this th >> t okay so in this number of head is what x is number of head right so this will be only one head and y is what number of tails in the first two tosses so that means We considering only these two tosses, right? So these two twos only one tail is there.

[00:37:20]>> Yes ma'am. Yes.

[00:37:20]>> Okay. Now here is two. This one is two.

[00:37:23]This one is 2 one one.

[00:37:28]>> H. So now what all values that you can take?

[00:37:32]012 >> 0 1 0 1 >> Then the probabilities >> for zero >> 3 by 8 by 8 >> 1 by 8 >> 3 by 8 >> 3 by 8 >> 1 by 3 by 8 >> 2 by 8 >> 2 by 3 by 8 >> 3 by 8 by 8 >> 3.

[00:37:54]>> So this is my PMF of you. Is this fine to everyone?

[00:38:02]Yes ma'am.

[00:38:04]>> Yes or no?

[00:38:05]>> Yes sir.

[00:38:06]>> Yeah. Okay.

[00:38:09]Now next question.

[00:38:15]probability that my u is a given that my u is greater than 1 is equal to 2x 5. I need to find out the value of a. So how will I solve it?

[00:38:39]So this is what? So I can write this as probability of u = to a intersection of u greater than= 1 divided by probability u greater than equal to 1 equals to 2x 5. Correct? Now first what is my probability of u greater than equal to 1? This will be what? Probability of u = to what plus probability of u = to 2.

[00:39:00]Correct?

[00:39:03]>> Yes ma'am.

[00:39:04]>> Yeah. So what is probability of u equals to what? It is 2x 8. 2x 8 >> + 3x 8 which will be 5 by 8. Right? Now what is probability that u is = a u is greater than equal to 1.

[00:39:22]See if you see here like from this equation the first equation can I write it 2x5 into probability of u greater than equal to 1?

[00:39:35]>> Yes.

[00:39:35]>> Yes sir.

[00:39:36]>> Yes. Yes. So now I can write this 2x 5 into 5 by 8. Correct?

[00:39:44]>> Which is actually coming out to be 2x 8.

[00:39:46]Yes.

[00:39:48]>> Yes ma'am.

[00:39:50]>> So now so the probability is coming out to be 2x8. Now if you look into here this thing is actually telling me when probability of u is equals to a and my u is greater than equal to 1. At what point of this I will be having 2x8 as a probability. You guys tell me.

[00:40:07]at first at first. So this is actually the probability that my u is equals to 1. So now if you look into this condition if my probability of u is equals to 1 and u is greater than 1 this will be actually equal to probability of u = to 1 only right and this is the probability of 2x8. So my value of a is at what equals to 1.

[00:40:33]Is this clear to everyone?

[00:40:36]Yes ma'am.

[00:40:37]>> Yes ma'am.

[00:40:38]>> Ma'am can explain again that uh uh two bit how how you say p= one.

[00:40:46]>> Oh you okay. So first of all this is my equation right?

[00:40:50]>> Yes.

[00:40:51]>> Okay. So from this I can find out this value by just multiplying it here. So I will be getting this thing. Correct.

[00:40:58]>> That's >> now I have found out this value which is 5 by 8. Correct.

[00:41:03]>> Correct. Okay. So this will be 2x 5 into 5 by 8 correct.

[00:41:08]>> Okay. Now this is 2x 8. Now if you look into here. So what is this actually saying? This is actually saying that the probability of u = to a comma u greater than equal to 1 is coming out to be 2x8.

[00:41:19]This is my probability. Now if you look into here my probability when u is equals to 1 is also 2x8. Yes.

[00:41:27]>> So can I write it equal to probability of u = to 1?

[00:41:30]>> Okay.

[00:41:32]>> Right. Now if you Yeah. If you just check if my a in place of a if I just put one also I'll be getting this equation yes or no >> yes >> probability u equals to 1 and u greater than equal to 1 the common part between them is basically u equals to 1 only right so this should be not the comma this should be in that this thing >> okay so then I got a equals to 1 is this fine >> yes m >> okay now can I move to the next question then.

[00:42:05]>> Now could you show the PMF table of this question?

[00:42:08]>> PMF table this one.

[00:42:10]>> Yes ma'am. 3x 8 2x 8 and 3x.

[00:42:13]>> One thing not ma'am it is not matching with the option. It's uh 3x 8 1x4 and uh 2x 2xal 1x4.

[00:42:25]>> Thank you.

[00:42:26]>> Okay. Yes sir. Ma'am, can you like once more say that what did you do do with A?

[00:42:33]I didn't understand that.

[00:42:36]>> Yeah, this part that U= to A and U greater than equal to 1.

[00:42:40]>> H. So, see my probability U equals to 1 intersection of U greater than equal to 1. So from this common if I take out common U is equals to 1 will be my common right.

[00:42:53]>> Yes ma'am.

[00:42:54]>> Yes. So that is what so so what I did in place of a I just put the value as one correct.

[00:43:01]>> So then only I'm getting this uh this uh equation you can say right so from that thing only I got to know that my a is equals to one.

[00:43:12]>> Okay ma'am got it.

[00:43:14]>> Okay.

[00:43:17]Yeah, I got you >> for the mock paper. The uh deadline is till midnight.

[00:43:30]>> Yes, ma'am.

[00:43:31]>> Yes, ma'am.

[00:43:34]>> Very nice.

[00:43:36][laughter] >> Very nice.

[00:43:42]>> Okay, everyone will get 100 by 100.

[00:43:44]>> Yeah, for sure.

[00:43:47]That's why uh you were asking me the answer, right?

[00:43:52]Okay. [laughter] Okay, that's fine. See, it's your thing.

[00:43:56]You are going to give the exam. Fine.

[00:44:00]Okay. But I don't care about the answers. I just need to know like you are getting the concept like how to solve the question. That is more important because in the exam you won't be getting uh the answers and all. So I need to know whether you're getting the concepts and all. Are you getting the concept?

[00:44:15]>> Yes ma'am. Yes ma'am.

[00:44:16]>> Yeah.

[00:44:16]>> Yes ma'am.

[00:44:18]>> Okay. So I need the results.

[00:44:22]>> Okay. Then >> ma'am how how you have said the paper is it difficult?

[00:44:30]>> I said it's fine only if you know um like in the sense if you have gone through the lectures very nicely. Okay.

[00:44:39]Uh then I guess you'll be able to score it. Okay. So mostly it's from the lectures and all the graded assignment activity questions.

[00:44:49]Okay. So if you have gone through the lectures very nicely the topics and all then you'll be able to solve it.

[00:44:56]Okay.

[00:44:58]>> I would say that.

[00:44:59]>> Yeah.

[00:45:05]Okay. So till 11th we done. Let's do question number 12. Shall we? Shall we proceed now further?

[00:45:12]>> Yes sir. Yes.

[00:45:14]>> Okay.

[00:45:20]Question number 12. Right. Okay. So, where what is there? What are they saying?

[00:45:26]A fed is thrown two times independently.

[00:45:28]Let X1 represent the number obtained in the first row. So, X1 is number obtained in first row.

[00:45:43]Then x2 is number obtained in the second row. Okay.

[00:45:50]Define a new random variable x is that x is max of x1 x2. Okay. How you will guys solve it?

[00:46:02]Find the cdf. So first we need to find out the cdf. Yeah. So first for uh finding out the cdf first you find out the pmf. How will you guys find out the pmf?

[00:46:24]X1 >> X1 can be 1 2 3 4 5 or 6 and X2 can also be 1 2 3 4 5 6. So like we can plot the values and uh calculate the probability in the first and the second respectively.

[00:46:38]H >> we can do that.

[00:46:41]that will give the PMF initially.

[00:46:50]>> Okay. So you are saying that um first you will find out the values that is um >> yes ma'am that plot x1 and then x2 and then give 1 2 3 4 5 6 and then like we can calculate x1 comma x2 the pmf.

[00:47:07]>> Okay. So without the PMF also like if I don't uh write the PMF like see like directly can I find out the CDF >> ma'am I think yes >> yes how >> yeah because uh max of uh x1 comma x2 is given now so what we can do is we can uh like take the cases like 1a 1 uh 1a 2 1 3 1 4 1 5 1 6 and then like calculate like that See what we can do is see directly if we just calculate see f of x k what is this? That is actually probability that my x is less than equal to k. Correct?

[00:47:53]Yes or no? Now what is my x? x is probability that maximum of x1 x2 is less than equal to k. Correct?

[00:48:06]Yes.

[00:48:07]>> Yes. Now can I write it in this way? Probability that my x1 is less than equal to k, x2 is less than equal to k.

[00:48:18]>> Yes ma'am.

[00:48:20]>> And both of them x1 and x2 are independent right because they are independently thrown. So can I write it x1 is less than k into probability that my x2 is less than equal to k?

[00:48:33]>> Yes. Yeah. Now if you see here my x1 and x2 they are both independent. So I can say probability that my x1 is less than equal to k² both will be equal right.

[00:48:50]>> Yes ma'am.

[00:48:51]>> X1 or because yeah because they are independent. So these value and this value will be equal only right.

[00:48:57]>> Yeah.

[00:48:58]>> Now the probabilities. So what will be this value? Probability of X 1 less than equal to K. If I need to find out see what will be the values of X1. X1 will be like 1 2 3 4 5 and 6. Right? And what will be the probabilities? It will be 1x 6 1x 6 1x 6 and so on 1x 6.

[00:49:21]Correct.

[00:49:22]>> Uniform.

[00:49:24]>> Uniform. Correct. Now if I need to find out probability that my x1 is less than equal to 2. So how will I find out? It will be probability that my x = 1 + probability x = 2. Correct? H. So this will be 1x 6 + 1x 6 which is 2x 6.

[00:49:40]Correct? Now similarly if you see here probability if my x1 is less than equal to 4 this will be what?

[00:49:47]4x 6.

[00:49:48]>> This will be yeah ma'am >> 4x 6. Now so if you see here this is basically what? So um 2 >> 4 * 1 by 6 >> yeah so whatever it is so I can just generalize it as x1 less than equal to k is basically k by 6 can I generalize it because if you look into here if my x1 is less than equal to 2 it's coming out to be 2x 6 x1 less than equal to 4 it's coming out to be 4x 6 similarly if I go with probability that my x1 is less than equal to 5 it will be 5x 6 Yes ma'am. Yes ma'am.

[00:50:25]>> Yeah. Yeah. So my this value is equal to this value. So can I generalize this into this thing?

[00:50:30]>> Yes ma'am.

[00:50:31]>> Right. So now if you see here this value can I write it? K by 6².

[00:50:40]>> Yes ma'am.

[00:50:40]>> Yes ma'am.

[00:50:41]>> Right. So what is my f of x k by 6² >> square?

[00:50:53]Is this fine?

[00:50:56]>> Yeah ma'am.

[00:50:58]>> Yes.

[00:51:00]>> Okay.

[00:51:01]>> Okay. [laughter] Ma'am uh in this question uh instead of max if they ask for minimum ma'am uh I just thought that I don't understand why first we take x greater than um k by uh k by sec like x greater than in cumulative we will take x greater um I don't understand that cumulator x you're talking about this thing probability that x is greater than something this No ma'am if it was instead of max if it was minimum um in the question and we have to find the cumulative uh SEMF then how would we proceed in that?

[00:51:47]Yeah. So it will be what minimum of x y same same thing is greater than equal to some x. So this will be what x is less than equal to x probability this into probability that my y is less than equal to x then you will multiply and that's it like if both are independent >> ma'am in minimum uh we have to take less than I think >> so um I'm not uh clear on that [snorts] >> clear on like >> and basically she is saying that she gets is confused in case of right when you write maximum greater than equal to k less than equal to k and minimum greater than equal to k like that.

[00:52:34]>> No, I mean uh if if in the question instead of max it was minimum the function was minimum and we have to calculate the cdf of it. So uh we know that cumulative is always x is less than equal to k but we cannot do that when it comes to minimum.

[00:52:51]>> You can minus it right? You already know that when I have a minimum thing right I have to write it in this form correct.

[00:52:59]So if I have to convert it to this form can I write it 1 minus probability of x less than this thing can I like if I need to like in like in um general case if I need to find out probability that my x is less than equal to some value a can I write it 1 minus probability of x is greater than a >> yeah yes so just convert it in this form and just solve it so you'll be solving it this part and just subtracting it with one >> yes ma'am correct but why do we take x greater than a like in minimum like why do we take this part um why is it easier like I don't get the concept behind it >> concept in the sense like you're asking when uh like in the minimum why do we take less than equal to some value right >> uh no we we first calculate x greater than a and then we minus it right so why do we take x greater than a like uh why does it not work in like we don't do that in maximum but in minimum we do that so yeah okay so let's suppose if my uh let's suppose if my x is actually maximum of some value x1 comma x2 correct and my y is minimum of y1 comma y2.

[00:54:34]Right? This this this is the thing. Um okay. So now the probability that my x is uh like if I need to find out the probability that x is less than some value a.

[00:54:52]So this is actually what probability that it is maximum of x1 x2 less than equal to a and we will just go with it right this is how we use uh like how we solve the maximum thing correct >> correct >> now if I say y is greater than equal to a this thing >> okay >> yes now how will I write it will be what minimum of y y1 comma y2 to greater than equal to 8. Right?

[00:55:25]>> Right.

[00:55:28]>> Then next step uh then it will be um P of X1 greater than equal to A and P of X2 greater than equal to A. P of >> Y1 greater than equal to A into P of Y2 greater than equal to A. Correct.

[00:55:55]>> Correct.

[00:55:56]>> H. Uh why why did we do this?

[00:56:02]Um because um that's what we did in the max one which >> in the maximum they do the same thing because they're independent. So we can um multiply it individually.

[00:56:19]>> H correct. So they are independent that's why we did this thing right.

[00:56:24]>> Yes. Correct.

[00:56:26]>> H okay. So now so now this will be what my f of that is probability of y uh 1 greater than equal to a whole square. Right? I can write it this way. Yes ma'am.

[00:56:45]>> Right. So uh if you see here when my minimum is there I am writing it in this way. Correct.

[00:56:53]>> Yes.

[00:56:54]>> Yes. Similarly when I have maximum I'm writing it in this way. This you don't have a doubt right? Now >> um like why do we write in actually in minimum y greater than equal to a like that I don't get. See we are talking about the minimum value right?

[00:57:15]>> Yes.

[00:57:16]>> So from this minimum value what will be my lower limit? It will be some value a right >> right? Yeah, that is what so that's why we are huh >> okay so now in the question if let's suppose if my uh you talking if my x was minimum right so we are talking about probability that my x is less than equal to some a so if I need to convert this into this form correct so what will I do it I'll just one minus probability of that something x is greater than a correct >> correct yeah >> okay ma'am thank you ma'am Okay. Is the question I guess the question is clear to everyone. Yes or no?

[00:58:01]>> Yes.

[00:58:03]>> Okay.

[00:58:05]>> So now which question was it?

[00:58:07]>> Ma'am please show the last line.

[00:58:10]you understand?

[00:58:11]>> Yes.

[00:58:12]>> Ma'am, just a general question I had ma'am. Whenever we we are given such questions of max and minimum like should we always like um we should always try first to get the CDF right like from there we can get many data like whatever the question is asked.

[00:58:27]>> Correct. Correct. Correct.

[00:58:29]>> Okay. So that will be much easier for you because if you go with like you can go with the you know this question can also be solved in some other way you know that the the thing that you are telling uh telling me so this question how you will solve it you'll just take out uh so first of all what all um chances you can have the first thing is so uh like how one person told me like x1 x2 that is 1 2 >> yeah I only told >> six huh so this will be like this correct so now if you see Here this will be what 1x 36 1x 36 1x 36 and so on.

[00:59:04]Similarly this will be also be 1x 36 1x 36 1x 36 and so on. Correct?

[00:59:12]>> Yes.

[00:59:13]>> I'm not writing whole thing. So this is what? So now if you look into here my x is what? X is basically maximum of x1 x2.

[00:59:23]Correct? So now x can take values what 1 2 3 4 5 and 6 correct or not? I'm just talking about not x1 x2 I'm talking about x. So x can take values 1 to 6.

[00:59:36]>> Yes ma'am.

[00:59:37]>> Okay. Now what will be the probabilities of x when x= to 1. So x= to 1 that means my maximum from both of these is coming out to be 1. So the at what point it is 1x 36 only.

[00:59:50]>> Yes ma'am.

[00:59:52]>> Right. So this will be 1x 36. Now similarly for two it will be what? This one, this one and this one.

[01:00:01]>> Yeah ma'am.

[01:00:02]>> 3x 36. Yes or no?

[01:00:05]>> Yes ma'am.

[01:00:06]>> Okay. Now similarly for the 31 it will be what?

[01:00:10]This thing. Yeah.

[01:00:13]>> Right. So in total I have five >> 5 by 36 5 by 36. Similarly you can calculate. So for four it will be >> 7 by 36.

[01:00:25]>> 7 by 36 and for five 9 by 36 >> 6 >> and 11 by 36. Correct?

[01:00:31]>> Now this is what my PMF is actually right. I have calculated the PMF from this I can calculate CDF. CDF will be what?

[01:00:41]Probability that my X is less than equal to X. It will be what? First it will be 1x 36 then 4x 36 then 9x 36 by 36 and so on right 25 x 36 and then >> correct now if you look into this pattern so if you see here can I write it in this form 2² by 6 square 3x 6 square 4x 6 square 5x 6 square and this is 1x 6 whole square can I write it? So in short can I say it's like k by 6 whole square?

[01:01:21]>> Yes ma'am.

[01:01:22]>> Yes.

[01:01:23]So in this way also you can solve it. So in this I didn't uh go with that fx maximum thing. So but this thing is actually lengy for you like if you go >> examination hall these things will not strike actually >> actually. Yeah correct. So this is like to understand. Okay.

[01:01:43]>> Yes ma'am.

[01:01:44]>> Fine.

[01:01:45]Is this fine to everyone?

[01:01:46]>> Yeah. Yes ma'am. Yes.

[01:01:49]>> Okay. Now what is the next question?

[01:01:53]We need to find out probability of x = to question number 13. Probability that x = to 4. What is probability of x= to 4?

[01:02:10]>> 7 by 36.

[01:02:11]>> 7 by 36.

[01:02:13]7 by 36.

[01:02:18]>> Okay. Or like this is so you just got it from here, right? If I didn't prepare this PMF, how will you calculate it then?

[01:02:28]>> Then f(x) of 4 minus f_sub_x of 3."

[01:02:31]>> Correct? So I can just calculate it by f of x of 4 minus f_sub_x of 3. That's okay. So this will be what? What is f(x) of 4?

[01:02:47]>> 2x3 square >> 2x3 whole square.

[01:02:52]>> Okay. Okay. So it will be 4x uh 4x 6 whole squareus 1 square correct. So if you solve this for 16 16 - 9 so I hope this question is fine to everyone.

[01:03:18]>> Yes ma'am.

[01:03:19]>> Okay let's proceed to the next question.

[01:03:21]Question number 14. It is saying that my x is following a poa distribution with lambda as a parameter. I need to find out the value of lambda which 3 into probability of x = 3 = 2 into probability of x = 2 + 4 into probability of x = 1 is given. Okay. So now you guys tell me first if my X is following a poa distribution what will be the PMF of X E to the power minus lambda E to the power minus lambda >> lambda X by X factorial >> by X factorial correct so now if it's saying probability that X = 3 it will be e ra to the power minus lambda lambda raised to the power 3 upon factorial 3 correct >> yes ma'am >> so just put this value so it will be 3 e to the power minus lambda lambda to the power 3 upon factorial 3 >> the equation will give the lambda >> huh so 2 into e the power minus lambda lambda the power 2 upon factorial 2 + 4 into e the power minus lambda lambda the power 1 upon factorial 1 correct So now if you look into here this part, this part this part will get cancel. So I will be left out with what?

[01:04:45]So this 3 factorial can be written into 3 into 2 into 1. So this three will also get cancel right. This two will also get cancel and from here so I will be left out with what? Lambda cq / 2 = lambda² + 4 lambda. Correct?

[01:05:06]Now here also if I take lambda common from the right hand side that is lambda equals to lambda + 4.

[01:05:16]>> Okay ma'am could you scroll it a bit uh scroll it down.

[01:05:20]>> Okay.

[01:05:23]>> Okay.

[01:05:25]Okay.

[01:05:27]>> Okay. Yes.

[01:05:34]>> So this value will be what? Lambda square is equals to 2 lambda + 8 + 8.

[01:05:40]>> So this will be now what will be the value of lambda here?

[01:05:46]What are the two values that lambda? We have >> four andus 2.

[01:05:50]>> 4 andus.

[01:05:52]>> So what will be the value that lambda can take only?

[01:05:56]Y - 2 >> Y - 2 >> lambda is greater than zero.

[01:06:05]>> Very good. So my lambda is greater my lambda will always take greater than zero. That's why the negative value we cannot consider it. Okay.

[01:06:13]>> Yes ma'am.

[01:06:14]>> Okay.

[01:06:18]>> Okay. So can we proceed to the next question then?

[01:06:24]Yes ma'am.

[01:06:25]>> Okay. Now what is Chabisha's inequality?

[01:06:28]So the next question is about Chabisha's inequality. So they are saying I need to find out probability the lower bound of this thing.

[01:06:42]where mu and sigma square are where mean and variance of x is mu and sigma square. Enter the answer correct to two decimal places.

[01:07:03]What is the ticious inequality?

[01:07:17]Okay.

[01:07:19]>> Mod of x - mu is less than or equal to k sigma.

[01:07:25]>> x - mu is less than equal to k sigma.

[01:07:29]Okay.

[01:07:30]>> Is greater the probability is greater than or equal to >> 1x k². Sure.

[01:07:43]>> I think it will be less than or equal to 1 by K.

[01:07:49]>> Sure.

[01:07:56]>> Ma'am, I think modus mual >> should be greater than.

[01:08:02]>> Yes ma'am. greater than equal to >> k sigma >> yes >> yes yes ma'am greater than equal than less than equal to yes ma >> right and the next one I can write it probability of x - mu less than equal to k sigma greater than = 1 - 1x k² >> k square >> yes okay now but I need in this form right so what is the other form do you guys remember that form Do you remember probability of mu minus k sigma less than x less than mu + k sigma is actually greater than equal to 1 - 1x k². Do you remember this form?

[01:08:48]>> Yes ma'am.

[01:08:50]>> Right. So just compare this form with this thing. Okay. So now if you look into here my question is given to be - 2 sigma x - mu correct this is given and we need to find out this one it should be 1 - 1x k² now if you look into here my mu - k sigma is actually equal to minus 2 sigma >> yes ma'am >> right and my mu plus k sigma is actually equal to 2 sigma right?

[01:09:27]>> Yes ma'am.

[01:09:29]>> Okay.

[01:09:31]>> K= 2 >> k = >> two.

[01:09:37]>> K= >> ma'am put the new both side of equality.

[01:09:44]No, I mean blue blue in blue equation there is x is subtracted with >> ah in the blue one.

[01:09:58]>> I'm sorry I didn't get it. Could you explain this uh third line uh this blue and how you're basically comparing the two equations?

[01:10:05]>> H >> okay >> now it's fine.

[01:10:13]>> Yes ma'am.

[01:10:14]So now if you look into here my mu - 2 sigma is actually equal to - 2 sigma and mu + 2 sigma is actually equal to 2 sigma correct sorry one second this is actually equal to mu minus k sigma and this is actually mu + k sigma yes or no >> yes ma'am right right >> now if I just compare this thing I get k is equals to two here yes ma'am >> right now can I put this k= 2 in this value so it will be what greater than equal to 1 - 1x 2² >> 3x4 h >> so this will be greater than equal to 1 by 1x 4 which is greater than 3x4 which is 05 so this is the probability that - 2 sigma - 2 sigma Okay. So now the thing is do you guys know like when to use this formula and when to use this formula.

[01:11:24]>> Then the second one is for lower bound.

[01:11:26]The first one.

[01:11:27]>> Yeah correct. So actually this thing this actually is telling me about the lower bound.

[01:11:33]Okay. And this thing is actually telling me about the upper bound.

[01:11:38]Fine.

[01:11:41]So whenever the question asks about the upper bound just use this formula.

[01:11:45]Whenever the question is asking about the lower bound just use this formula okay or this part. Mostly uh the question will be asked in this form only. So remember this part.

[01:11:58]Okay.

[01:12:00]This is fine.

[01:12:02]>> Yes ma'am.

[01:12:03]>> Yes.

[01:12:04]>> Yes ma'am.

[01:12:06]Now next question that is question number 16.

[01:12:11]This question we won't be asking actually.

[01:12:19]Okay. So my diagram is actually This is alpha comma 0.

[01:12:46]This is 0 comma beta.

[01:13:05]Okay. Now, which of the following is true?

[01:13:09]So, how will you guys solve this question? Do you guys know?

[01:13:13]So, let's name it first. So, can I name it as let's suppose this is zero. This is A.

[01:13:22]This is B, C and D. Okay, let's name it like this.

[01:13:33]Now you need to find out the equation.

[01:13:37]Any ideas?

[01:13:40]Um, area is equal to 1.

[01:13:43]>> Total area is equals to 1. Okay, fine.

[01:13:46]So how will you find out the total area?

[01:13:51]Um ma'am we will find out the right hand side first and since it is symmetrical >> so we'll find the right side first and find the area through of trapezoid.

[01:14:03]>> Okay. So basically so if you look into here we actually need to find out which area we need to find out this area.

[01:14:11]Yes or no?

[01:14:12]>> Yes or no? Yes. That means I need to find out the area of O A B D B D B D B D B D B D B D B D B D B D B D B D B D B D B D B D B D B D B D B D B Correct.

[01:14:25]And if I find out this area, I can just double it to find out this this area, the whole area.

[01:14:34]Yes or no?

[01:14:36]This will be equal to one.

[01:14:37]>> Yes, ma'am.

[01:14:38]Now this O A B can I just divide it into one triangle and one rectangle?

[01:14:47]>> Yes.

[01:14:48]>> Like it will be easy for me to calculate the area. Correct.

[01:14:50]>> Yes ma'am.

[01:14:50]>> Right.

[01:14:51]>> Yes ma'am. So this will be 2 into area of triangle which one that is ABC plus area of the rectangle O A CD. Correct?

[01:15:11]Now what is the area of uh triangle?

[01:15:16]height into >> half into base into >> half into base into height. So tell me what is my half into what is my base?

[01:15:33]>> Uh theta by 2 base is beta by 2.

[01:15:37]>> This is beta by 2 is alpha >> 0 to alpha >> base is alpha >> base is alpha. Okay. And what is my height?

[01:15:47]>> Beta by 2 beta by >> beta beta by >> beta by 2 >> beta by two.

[01:15:52]>> Is this fine for everyone?

[01:15:55]>> Yes ma'am.

[01:15:56]>> Yes ma'am. Okay. Now what is the area of triang rectangle OCD?

[01:16:01]>> Length into bread.

[01:16:03]>> Uh length into bread. So what is my length?

[01:16:06]>> Alpha >> alpha >> beta by 2 >> beta by 2 >> beta by >> beta by 2.

[01:16:12]>> So this is equals to 1. Now we will solve this equation. So this will be 2 into alpha beta by 4 + alpha beta by 2 = 1. If I take out 2 so this will be 3x 4 right? So 2 into 3x 4 into alpha into beta = to 1. So alpha beta will be 2x3 >> 2x3 >> 2x3 >> So this will be the equation correct.

[01:16:52]>> Yes ma'am.

[01:16:55]>> One second. Give me a moment.

[01:17:05]Okay. Now next question.

[01:17:08]So [clears throat] uh you guys don't have any issue in this problem right? Can we move forward?

[01:17:16]>> Yes ma'am.

[01:17:17]>> Okay. Now question number 17.

[01:17:20]What is question number 17 is asking?

[01:17:23]It's saying that we need to find out probability of mod of x is less than alpha by 2. Okay.

[01:17:31]Now can I write it in this way?

[01:17:35]probability that my x is between alpha by 2 and alpha by 2.

[01:17:44]>> Yes ma'am.

[01:17:46]>> Yes.

[01:17:46]>> That means we are actually talking about let's suppose >> uh let's suppose this is my alpha by 2.

[01:17:58]So this will be minus alpha by 2. So we are actually going to check for which area? This part right?

[01:18:11]Yes or no?

[01:18:13]>> Yes.

[01:18:13]>> So let's suppose this is my um this is like G, this is like E and this is F.

[01:18:27]Fine.

[01:18:31]Now if I need to find out this thing, how will I find out?

[01:18:41]See, I need to find out this um if I need to find out this whole thing. If I just find out this part, I'll just double it and it will be the total area.

[01:18:51]Correct?

[01:18:52]>> Yes ma'am.

[01:18:53]>> Okay.

[01:18:53]>> Yes ma'am. Now same thing we will do it.

[01:18:56]So my probability this will be actually what? Double of area of which one? O A GF.

[01:19:09]Just check O A GF.

[01:19:15]>> Okay. Now the same thing we will do it.

[01:19:18]It will be 2 into area of first the triangle, right? that is a G E.

[01:19:29]Okay. Plus the rectangle O A E F.

[01:19:39]Correct?

[01:19:42]>> Yes or no? Right. Now what will be 2 into area of triangle [snorts] A G E?

[01:19:49]Can you guys tell me?

[01:19:52]half into alpha by 2 into beta by what?

[01:19:57]>> Beta by 4 magnitude by 2 altitude.

[01:20:05]Okay. So my triangle will be half into what is my base? Base is alpha by 2 into first of all if you look into here for E the point will be what? Alpha by 2 comma beta by 2. Correct?

[01:20:18]I know this point. So I I need to find out this length actually. How will I find out this length? If I have this point and this point, if I subtract it, then I will get this point. Correct?

[01:20:30]>> Yes ma'am.

[01:20:31]>> But I need to find out this value, right?

[01:20:34]>> Yes ma'am.

[01:20:35]>> How will I find out?

[01:20:37]>> Any any idea?

[01:20:39]>> Now it's a midpoint. So um >> but I don't know this point at uh beta is there. So this is my alpha comma beta cuz this is 0 comma beta by 2. So beta by 2 plus beta / 2 >> ma'am it's beta by 2 by 2 to beta by 4 >> beta by 4. So this point will be alpha by 2 by beta by 4.

[01:21:20]Yes or no >> everyone?

[01:21:23]>> Yes.

[01:21:27]U ma'am could you explain how did we got beta by 4 how did it okay so okay like uh so one point uh one way is like this is the midpoint okay because this is like half of it right so this is my midpoint so if this is my midpoint then these two points are already given so can I find out the y point what is the formula of midpoint Right.

[01:22:01]What was the midpoint formula?

[01:22:04]>> y2 - y1 by 2 >> y2 - y1 by 2, x2 - x1 by2. Correct?

[01:22:14]This is the formula. So this is my x1.

[01:22:16]This is my y1. This is my x2. This is comma y2. And this is what my x and y is. Right?

[01:22:24]So I need to find out this thing. So in this case what we're going to do is so the for the first one see formula is if you need to find out the xaxis x and y it will be x2 - x1 / 2a y2 - y1 / 2. Okay. Now this will be what? Uh x2 is alpha alpha -0 by 2 comma y2 is beta beta - 2. So it will be beta minus beta by 2 / 2.

[01:23:17]Okay. So this will be what? alpha by 2 comma beta by 4. So this is what my x comma that is the point axis point is is this fine?

[01:23:34]>> Yes ma'am. Got it.

[01:23:36]>> Got it. Okay. So now if you find out this value so this will be what? This will be alpha x2 beta x 4 plus what will be the uh rectangle o e f. So this will be alpha x2 into beta x 4 or beta by 2 m beta by 2 I think >> beta by 2.

[01:24:04]>> So this will be alpha by 2 into beta by 2. So now this will be what? 2 into by 2 alpha beta by 16 plus alpha beta by 2 2's are 4 correct. So now this will be 2 into uh 4 right 5 by 16 into alpha beta.

[01:24:39]Now what is alpha beta from the above equation?

[01:24:42]>> 2x3 >> 2x3. So this will be 2 into 5x 16 into 2x3.

[01:24:49]Okay. So this will be what? 5 by 12.

[01:24:58]Yes or no?

[01:25:00]Yes. Okay. Then solve it and you will get some answer. Is this question clear to everyone?

[01:25:07]>> Yes ma'am.

[01:25:08]>> Yes ma'am.

[01:25:09]>> Okay.

[01:25:13]Okay. Then the last question that means like 18th question.

[01:25:21]Okay. Question 18th.

[01:25:23]So two friends Ammon and Deepa play a game in which each of them repeatedly and independently tosses a fair coin.

[01:25:28]The first to get head once outright. If both get heads together for the first time, it is a damp of that. A man wins outright. Okay. X1 and X2 be the number of tosses. Ammon and Deepak need respectively to get heads. X1 and X2 are dash and follow a dash distribution.

[01:25:43]Tell me what will be the answer for the first one.

[01:25:52]Geometric.

[01:25:54]The >> first one X1 and X2 are >> independent.

[01:25:58]>> Independent >> are dependent.

[01:26:02]>> Okay. Then and follows which distribution?

[01:26:06]>> Geometric.

[01:26:07]>> Geometric.

[01:26:08]>> Geometric. Why geometric?

[01:26:12]First H.

[01:26:17]Curry why geometric? Why it is geometric?

[01:26:26]>> You don't know how many times we have to >> whether success will come after they get ahead.

[01:26:32]>> Yeah. So basically what is the situation? The situation is we have to toss until we get ahead. Right. And what is the geometric? Geometric is we uh the first success after the uh the failure, right? failure right how many tosses we uh like how many we have to toss so that we can get one first success right that is what geometric is so it is the same case the game is going on right so that's why it's a geometric distribution now probability that my ammon wins it will be equal to what probability that >> x1 is less than x2 >> x1 is less than Next, Russia.

[01:27:18]>> Mom, because if the other person gets more trials, >> sorry, >> because the first player is going to be by if he he will win if he get less number of trials.

[01:27:34]>> Okay. So, the situation here is X1 and X2 is basically what they are the number of tosses, right? And who will win the uh game? He will uh the first person who will get the first head. Right? So if x1 x1 is ammon is tossing the coin and x2 is deep is tossing the coin. And what we need to find out we need to find out that ammon wins the game. So when will ammon wins the game when he gets the first uh head right and if he wants to get the first head the number of tosses that he should have should be less than the number of tosses that deep will be having. Yes or no?

[01:28:12]Is this part clear?

[01:28:13]>> Yes. Yes.

[01:28:14]>> Okay. Now, probability that X1 is less than equal to X2. Now, next thing, can I write it in this? Uh, this will be summation of K is going from 1 to infinite. Probability that my X1 is equals to K into probability of X2 less than K.

[01:28:42]X2 less than K. Okay. Then next summation of K = 1 to infinite 1 by 2 to the power K. Then next one 1 by 2 per K. How?

[01:29:05]Because there have equal number of chances now. So 1 by 2 per k equal number of chances.

[01:29:15]Can anyone explain with this part how we got 1x 2k raised to the power k?

[01:29:32]What is x2?

[01:29:39]Deeper getting uh tails.

[01:29:42]>> Yeah. X2 is basically the number of tosses that Deep is having, right? And what is K like like he is this what is the situation that he should get head.

[01:29:51]So this thing is actually saying that like till K he didn't get any head. So after K only after K tosso he will be having head. Yes or no?

[01:30:03]>> Yes. Right.

[01:30:06]So, so that is so what is this probability? It will be 1x2 raised to the power k only.

[01:30:13]>> Yes.

[01:30:14]>> Yes. Okay. Yeah. So, this is so now this is what basically now you guys solve this thing. How will you solve?

[01:30:24]Any idea?

[01:30:27]So summation of k = 1 to infinite 1 by 4 raised to the power k this is what a geometric expression >> uh ma'am if it is ma'am it was a if it was a unfair coin then 1 by two upon k this ratio would change now >> yeah of course it will change and they will give the parameter right according to that >> okay >> okay so now you tell me Um yeah this is geometric expression then this for the infinite series it will be what 1x 4 upon 1 - 1x 4 correct >> yes ma'am >> right so this will be 1x 4 3x 4 which is 1 by 3 which is 0.33 Is this fine?

[01:31:31]>> Yes ma'am.

[01:31:32]>> Everyone.

[01:31:34]>> Yes ma'am.

[01:31:35]>> Yes ma'am.

[01:31:35]>> Ma'am. If Aman wins outright then like in the first throw only Ammon should be getting ahead right.

[01:31:43]>> Yeah. Maybe in the first row in the second throw as well you like there can be a situation right >> because the number of throws are not given here. Yeah, ma'am. I got it. But they are saying that outright. That means I think in the first throw, right?

[01:32:06]Then why will they use that word outright?

[01:32:22]Yeah. So they are saying that if like if uh let's suppose we can have a situation in which for at first row both of them got tail in the second row also both of them got tail in the third one got head and the second one got uh tail so in the first like um yeah so basically what so they're saying that Right.

[01:32:53]>> Okay. So it's like tail tail. So this is the like um Aman is playing and Deepak is playing right? Tail tail then he got then he got head and he got tail right.

[01:33:02]So now he that Ammon uh wins the game because he got the first head. So whoever gots the first head at outright like at first then he will win the game.

[01:33:13]That's what >> Okay. Got it.

[01:33:20]>> Okay.

[01:33:22]>> Yeah. Yeah. Got it.

[01:33:26]>> Ma'am, this mean that for few tosses there will be time now.

[01:33:30]>> Sorry.

[01:33:32]>> Uh if the first one getting is going to mean to for few uh tosses it will it should be tie for first few tosses >> for few. Yeah. If they get both head like if both of them get head it will be tie situation.

[01:33:51]>> Yes ma'am.

[01:33:52]>> Like one of them should get ahead like that.

[01:33:57]>> Okay ma'am.

[01:33:59]>> Okay. So like we the situation is kind of like this.

[01:34:04]So this is like Aman Deepak. So let's suppose they get both tail.

[01:34:10]Okay. So then so the situation like according to the question they're saying if both get heads then only it's a tie right? So again they will uh toss a coin. Let's suppose they got both got head. So this will be a tie situation.

[01:34:21]Again they will throw the coin. So this will be like tail head. So here who wins? Deep wins.

[01:34:29]Correct. This is what so now he um he wins it. Now similarly if I change the situation here. So this will let's suppose this is h and now ammon wins here.

[01:34:43]What the point?

[01:34:45]Yes ma'am.

[01:34:46]>> That is what this is the situation here.

[01:34:49]So who gets the first head? That is the winner. Okay. So here my so let's suppose my deep he did not get the head.

[01:34:57]Right. But Ammon got head here. So he is the winner. So that is what outright for who got the first head when both of them toss the coin. Okay. If both of them get head then it will be a tie situation.

[01:35:11]Then again they will play and they will check whoever got the head first. Okay, that is the game. Got it.

[01:35:21]>> Yes ma'am.

[01:35:25]>> Okay. So that was all in quiz one. So is the are the questions clear to everyone like is it fine?

[01:35:38]>> Yes ma'am. as the solution and everything. I know everybody has checked the solution and all but okay it is not my concern. The main thing is you should know how to solve the question and the concept wise. Okay. So just [clears throat] be careful with the question what all questions they have asked. Go through the concept very well.

[01:35:58]Okay. The lectures what all the professor has told all the PPTs. Okay.

[01:36:06]go through the examples and all go like just go through the concept very well.

[01:36:14]Fine.

[01:36:16]>> Yes ma'am.

[01:36:17]>> Is that clear?

[01:36:18]>> Yes ma'am. Ma'am and the graded and the practice assignments also right?

[01:36:23]>> Huh? Practice assignment also you can go through graded assignment.

[01:36:26]>> Ma'am could you uh explain the 13th question once again? Uh show the 13th question.

[01:36:32]>> Show the 13th question.

[01:36:44]This one.

[01:36:45]>> Yeah. Yes, ma'am.

[01:36:48]Okay.

[01:36:50]Ma'am, could you scroll it down?

[01:36:53]Uh, no. Above.

[01:36:56]Yeah.

[01:37:00]Ma'am, how do you got uh this uh could you explain a bit?

[01:37:04]>> Which one?

[01:37:05]>> This uh PMF and CDF this is like okay you have first.

[01:37:19]>> Okay. Thank you.

[01:37:20]>> Got it.

[01:37:21]>> Yes ma'am.

[01:37:27]>> Okay. So yeah that was all otherwise everything is fine.

[01:37:35]Okay practice very well okay prepare very well just give your 100%.

[01:37:41]Okay.

[01:37:44]>> Okay then ma'am ma'am will there questions from week zero?

[01:37:49]>> No no no only from week 1 to four.

[01:37:52]>> Okay.

[01:37:54]>> And we should focus more on week 1 to four. you said the professor lecture uh slides right the PPTs.

[01:38:02]>> Yeah, like the activity questions and the practice. So, basically the more like uh if you know the uh if you have gone through the lectures very well and the activity questions and all then yeah >> okay ma'am.

[01:38:16]>> Okay.

[01:38:18]>> Yeah. Thank you ma'am.

[01:38:20]>> Okay.

[01:38:20]>> Thank you ma'am.

[01:38:21]>> Oh thank you ma'am.

[01:38:23]>> Thank you ma'am.

[01:38:24]>> Good night ma'am.

[01:38:24]>> Okay. Good night. Just all the best everyone. Yeah. Bye. Bye.

[01:38:28]>> Thank you, ma'am.

[01:38:29]>> Yeah. Bye. Bye.

[01:38:30]>> Thank you.