SSB TGT PCM PAPER ANALYSIS || PROBABILITY PYQS SOLUTION (2019 TO 2025) ||

[00:00:00]Hello everyone welcome to the competitive YouTube channel so dekhnatu aaj video me discussion kar probability pyqs to humein set theory relation function quadratic equations ta apna coordinate geometry conic section aur ta apna number system ko already humein discussion kar sarchati jo apna m se pYQ series ra question gu ko dekhti nhi humein channel ro playlist section ko j sethi apna SSB te pYQs paay jave sethi apna tako click kar dekhti pave [nasal sound] so aaj humein probability pyqs ko discussion kar 2019 nahi ki ltr 204 parant j ki bhi sab question pacharachi aur ko apnakar portion pach probability chapter to sahi sab ko bhi humein discussion kar detail session dekh to aap man ko ap PYQ series bhaag tale video ko like kar channel ko j subscribe kar nati tale subscribe kar rakhantu to chalant humein video ko start kar accha aur throat ki session If we start our study with the number system then we have to solve the previous video discussion of the question to solve the problem of probability then we have to discuss the question thila eta up kar what is the solution set mode x - 3 > = 4 then if we get x - 3 then we have to solve x - 3 instead of x + 4 then we have to solve it thik achi to a ko hume x - 3 ne ki solve karba so basically more solution approach than just empty eta ko minus ko plus karba di which p ansarila thik achi to k k nhi eta ko solve kar dekhchi dekhantu then if x - 3 > 4 gibbob and x - 3 < = -4 then we have to solve the mode thik achi to a ko hume the inequality piaba to first inequality Solve this inequality and get x > = 4 + 3 which equals 7, then x is less than 7 and solve the second inequality and get x < = -4 + 3 which equals -1, then x is less than -1. So if we look at the number line, then -1 is less than infinity and -1 is less than -1 to infinity. So, if we look at the number line, then x belongs to -infinity and -1 is less than -1 to infinity. So, if we look at the number line, then x belongs to -infinity and -1 is a closed interval because it is equal to the union of 7 to infinity and 7 to infinity is a closed interval. So, if we look at this equality, then we have option A, option B, or correct answer, then this question answer is probability question, so let us look at the first probability question. Humko kaun diya hela in a binomial distribution the mean is 4 and the variance is three then its mode se humko pala thik to se kah ki binomial distribution a se binomial distribution mean ho hum 4 and variance ho 3 to se humko mode ke hobo pachala thik achhi to pratham jin ho ki hum ko dekh binomial distribution bahut sara question dekhiya pe mile ka ki na binomial distribution ko syllabus re mention ki rah aur binomial distribution ro sab shift re praty to question pachara se insert bha se dekh je bhi binomial distribution question pachar ho max question upkar mean variance e sab ko nahi ki question pachara ho thik achhi to binomial distribution re kaun hue dekhantu humein jo g binomial distribution dekhva se binomial distribution mean ro phula kaun hoye na If we have n * p then the variance of the result is npq. If you see that the result is as per the given result and p is as per the given result, then if we perform an experiment then it tells us that if we toss a coin five times then the value of the result is five. Similarly, if we toss a coin five times then the probability of success is 10. If we toss a coin five times then the probability of heads is 10. If we want heads then the probability of heads is 10. If we want heads then the probability of heads is 10. If we want heads then the probability of success... relation between p and q is 10. If we have p + q then the relation between p and q is 10. If we have p + q then the relation between p and q is 10. If we have p + q then the relation between So basically we have to give you a theoretical idea, we have to solve this question, let's say the mean is four, so mean four means np row value is four and variance is three means npq value is three, so if we calculate the value of this data then obviously we have to calculate q, so if we divide npq by np, then q is ok, then npq / np, then np is 0.5 then Q is 0.5q value, and npq value is 3 and np value is 4, so Q is 4, so which one should we take out p, so look at p, p is 1 - Q, so 1 - 3/4 is 1/4, so we have to take out the angle, now we have to calculate the value of n, so calculate it, sir, see if n equals to np, then divide p by pi, then we have angle pi when N is pi So np value should be 4 and p row value should be 1/4 so the value should be 16 so we have to add n pi to p pi to q [nasal sound] we are asking the question, if we have to add the mode row value in the question then first of all the values ​​should be taken to the side and then on that we have to calculate the mode and calculate the mode so if we calculate the mode then the mode should be written to the side so basically we have to add the mode to where it is inside, okay see, we have to add n + 1 * p - 1 and if we have to add the mode to the right then it should be brought to n + 1 * p inside, okay, that means the value is not inside, we have to add the mode to the value [nasal sound] so basically we have to add the current from n + 1 p - 1 so n [nasal sound] will be 16 + 1 17 and p will be p 1/4 means 17/4 = 4 - 1 < r and for 17 / 4 then if we have to solve 17/4 then we have to solve it 16 = 14 means 4.25 = -1 which is 3.25 so I directly wrote down 3.25 and if 3.25 < < = r < < = 4.25 then if we take the integer value of 3.25 then if we do not take 4.25 then we will see the option and the option which we have to see is the number and option c is four then we will see the option c is four so I hope [nasal sound] that we can solve this type of question on our own so next question is what will happen if two fair dice are rolled then what is the conditional probability that the first die lands on the six If the sum of the numbers on the dice is 1 then we can calculate the conditional probability of the event.

[00:08:00]If we roll the dice then the conditional probability of the event is 1 then we can calculate the normal probability of the event. If we throw the dice then the sample space has cardinality of 36 then the sample space has cardinality of 6 ro² which is equal to 36 and if we look at the sum of the numbers on the dice then we can calculate the event. If we roll the dice then the event results in 1 then the event is... If we calculate the first dice then we get 6 1 6 2 6 3 and 6 1 6 2 and we get the second event then we get 6 1 6 2 6 3 and we get 6 1 6 2 and we get 6 1 6 2 and we get the conditional probability then if we take the first dice then the intersection will be the same outcome as 6 2 then we get the required probability that if E1 and E2 intersect then we get one of the outcomes as 6 2 then we get the required probability that if E1 and E2 intersect then we get one of the outcomes as 6 2 Divided by total outcome, if the total outcome is 5 then the answer is five then the probability of this is 1/5 then we have option D then option D is the correct answer and the correct answer is 1/5 then we have the next question so next question we have to say that if the probability of A and B is 2/3 and the probability of A and B is 1/6 and the probability of A is 1/3 then we have to say that the option is within the correct option and we have to correct the option. Ok, first we have the option and we have the option A and B which are dependent events and A and B are independent events and we have the option A and B which are disjoint events and we have the option D If we look at the union then we can see the probability of A + B minus the probability of A intersection B. If A is disjoint then A is divided by zero. So what will we get from the union of A + B? Okay [nasal sound] but if the intersection is 1/6 then it will also be disjoint, okay or not. So, let us check whether AB is either dependent or independent. If we check whether it is dependent or independent then we have to look at the independent criteria. If we look at the probability of A into the probability of B, if A is independent then A and B are divided by zero then we will have to look at it.

[00:12:08]If this intersection is good then we have to calculate the probability of A and B by the product of the probability. So we have to calculate the first formula [nasal sound] so if we have to calculate the probability of A then the probability of A is 2/3 and the probability of A is 1/3 and B is 1/6 then we have to write the probability of B and the intersection is 1/6 so we have to solve it so that the probability of B is equal to 2/3 + 1/6 - 1/3 so 2/3 = 1/3 = 1/3 and 1/3 = 1/6. If we add 1/3 = 1/6 then the LCM is six times 2 If the product of a and b is independent then option A is cut off and the probability of a is dependent is 1/2 and the probability of b is 1/2 then the product of a and b is 1/6 and the probability of a is 1/3 and the probability of b is 1/2 then the product of a and b is 1/6 and the probability of a and b is 1/3 and the probability of b is 1/2 then the product of a and b is 1/6 and the probability of a and b is independent then option A is cut off and the probability of a and b is dependent then the option B is the correct answer so the next question is to answer your question and if the previous information is given then you can join our SSB TGT PCM batch by clicking on the number you see on the screen and you can WhatsApp it to 7873430697 and you can download the PDF by clicking on the number you see on the screen. then video description re telegram group link a se click kar join jaantu se apan pdf pave [nasal sound] aur baaki rahila katha ki ki hum classes etc. code hue dekhantu classes e application re hue classes to aap download kar na hoti play store ko ja ke download kar pave to directle se to bhi batch ko join kar pave otherwise a number whatsapp kar le join sethi up discount pave thik acchi ab jawa next question ko so dekhantu next question ko diyaala humko when a hand of seven cards is drawn from a well successful deck of 52 cards to se kahla ki 52 cards a se 52 cards humko kaun kar kar le seven card ko humko pick kar le to what is the probability that it contains three kings to sethere teenta king as probability kate to prathm humko bahar kar sample space to sample space jo calculate karbo to sample space cardinality kbo 52C7 eta hi jo apan ko sample space cardinality From here we calculate and press it, okay and next time we have to calculate the 3rd card, then we have to calculate the total of the 7th card and from here we have to calculate the 3rd king, then the total king is what it is.

[00:15:16]First of all we have to calculate the total king, that is four times, then pick the 7th card, if there is a 7th card then the 4th card is king, then the 4th card is also three times king, then the remaining 48 cards are 58 cards, the 4th king card is 58 cards, and the remaining 48 cards are 48 cards, so we have to pick the 3rd king randomly, so it does not matter, then write 48 and 4, and multiply it by 3 because of counting principle, you need to separate the 3rd card If we separate and complete the multiplication then [nasal sound] let us divide the first king by black and the rest by black, let us divide the work by this, let us multiply, okay, so what should we do, basically calculate the probability of the value, by pressing this, we will get just eta, okay, simplify and write it down, if you have given me the combination form, or if you have given me the number form, then solve the value, then what should we do, let me write 52 52 - 7, so 52 - 7 means 45 * 7, okay, solve it, it will be solved automatically and look at 4C3, 4 / 3 = 4 [nasal sound] so write eta 4C3 4, okay, so four is into 484 8, 48 / 48 - 4, add 44, okay 44 and 4cor and if we divide the data then who will look at it, then 52cor * 45cor * 7 and if we divide it then who will divide it, numerator will read the denominator upar ko padebo so now 44cor * 4 / 4 * 48cor so if we cut it then obviously we will cut it, we will write 52cor * 51 on 50, write 49 and then write 48cor * 8 on 49 and then we will directly cancel it out [deep breathing sound] [sudden heavy breathing sound] so just this part we cut the throat, it will be 49 parant rahila and then look 44 and 45 well 44 45 45 If you cut it then it will remain 45 and [nasal sound] it is the 45th year if you cut it then it will remain 50th year if you cut it then it will remain 50th year if you cut it then it will remain 50th year if you cut it then it will remain 50th year if you cut it then it will remain 50th year if you cut it directly... 13 * 17 * 7 / 9 so 13 * 17 * 7 / 9 sorry 17 ok so what should I do if I put the sample space above and if I do n't agree then the probability baab reciprocal hobo ta means the probability number is 9 / 13 * 17 * 7 am thibo ok akchi ko karo chha na sample space ko bottom place the space above ok akchi so seethi [nasal sound] no inconvenience just empty reciprocal parchi so above then nine so nine we have the data space and empty us see the unit digit so 7 * 7 = 49 9 3 27 so last we have seven then we have to assume 1 547 eta, we have to give the correct answer eta to option B then option B is the correct answer thik achhi so ab aap apan mane kaun kar pub question ko solve kar pub jo ki question pachar ho a type question dekhi par asantu next question kaun diyala if a fair cubic die is rolled then the probability that the perfect square number comes up is then say if a fair cubic dice is rolled then the sample space is keta cardinalitybo chhatabob ka ki na g dice ko throw karachi aur se pala ki ki apan perfect square number asib probability to perfect number ke achi g hi achi set ho one and g four de ok then the cardinality is up event row to so if we have to calculate the probability then say 2/6 that is = 1/3 then 1/3 is option B then option B is correct Answer: Read the question slowly so that you can easily understand it.

[00:21:28]Then look at the next question. If two unbiased dice are thrown, then the probability of getting a sum less than six is ​​equal to 6π times the probability. So first, if we throw two unbiased dice, then the probability of getting a sum less than six is ​​equal to 6π times the probability. So, first, if we throw two unbiased dice, then the sample space of the data dice is equal to 36. So, if we say that the sum term is less than six, then what is pi pva, then we need one pi pva, two pi pva, three pi pva, four pi pva, five pi pva, right now, if we say that the minimum value of the dice is one and the minimum value of the sum is one and the minimum value is 1 + 1. So, the minimum value is 1 and the minimum value is 5. So, if we say that we need 5. So instead of two, we have got 1 + 1. Okay, so let's say we have 1 + 2 cases. So, if we have three cases, we have three cases.

[00:22:30]1 + 2 + 1. So, if we have four cases, we have 1 + 2 + 3 + 3 + 6 + 4 + 10. So, if we have three cases and five cases, we have 1 +... 18 given option A given option A then option A is your correct answer [nasal sound] on as next question so next question what is the probability of A given probability of B given A intersection B now [nasal sound] we have to look at A intersection B complement then A intersection B complement means we have to look at A - B so the probability of A - B is not A set is not B set so A part which is good is A set is A - B part so A - B is Kmti bahar kara ja A set to us intersection part ko hte dawa ok intersection part ko hte dawa to us directly get basically a - b get which is A intersection B complement so [nasal sound] a ke achi number dekho 0.7 intersection ke achi 0.3 so 0.7 - 0.3 If 0.4 is 0.4 then 0.4 is the correct answer.

[00:24:24]Option D is there. Then look at the next question. If three dice are thrown together then the sample space of cardinality is 6 to the power 3. Okay, which is 216 [nasal sound]. Then look at the next question. The probability of getting a total of at least six is ​​there. If the total is at least six then the probability of getting a total of at least six is ​​there. If the total is at least six then the minimum six is ​​there. Look at the minimum six is ​​there. If we get six then we get six pi pi pi 8 pi 10 pi 11 pi 12 pi 13 14 15 16 17 18 then the maximum is 18. But sir, don't look at any dice. The maximum number of dice is six three. If dice means 6 + 6 + 6 is 18 then we get maximum pi so [nasal sound] let us check the set so we calculate the even number of 18 if there is no six ro then it is better to check the six ro minimum number and press six ro minimum number which is s on s, see three s on s, four s on s, five s on s or s, calculate and press and then subtract one ro, which is s, is it s, right? Answer is three or not, is 1 + 1 + 1 let us get three pi so minimum three s on s, three ro is the only choice, but do not add the number and give us four pi, so if we see the three ro number then 1 2 1 will be the same [nasal sound] let us write 1 2 also and write 2 1 also and then 1 2 1 then 1 1 2 then 2 1 then ok ah no number p ki teenta number apan ko sum four need pathwa katha ok ah so we have two two nine and three nine then ok three j to katha hi so basically this is only our choice then one to one so three pair pi pwa and same sum term ko five row me pair se dekhna kar dekhnatu so first we have one nine, second if we have one one one three hi pub ok ah this case re hum threeta case banabo ok jati ithi banla so ya threeta case banabo and par one nati two nati one two three and two so [nasal sound] our case bana ok ah and then who is up 1 2 1 3 1 ok ah you have this number bana so total we have to look at this pair pale three and six So total we have 10 pairs, okay so the probability we get is 10/216 or 345, but we need not need 18, so if we subtract one and then we get 216 - 10, that is equal to 206/216, or if we cancel out 3/100, then 3/8 is option A, so option A is correct answer, okay so it [nasal sound] we have to search for the numbers,... If we toss a coin then what is the probability of heads?

[00:28:36]If the probability of getting a head is 1/2 and the probability of getting a tail is 1/2 then it is very simple question. Next, if the probability of getting a spade card is 1/4 then if the probability of getting a spade card is 1/4 then if the probability of not getting a spade card is 1/4 then if the odds are higher then odds then odds are lower. If the odds are higher then odds then odds are higher and odds are higher then odds. If the sum of one and 1/4 is minus one then 1 - 1/4 x 1 equals 3/4 then 3/4 is option C then option C is the correct answer [nasal sound] and if you want to join the batch then Number WhatsApp kar pave PDF palle hi telegram group ko join karntu thik acche next question ko so dekhontu next question ko diyala a card is taken out at random from a pack of 52 cards number from one to 52 so se kahchi 52 card rakhaichi pratyak re number likhi dechi ek ro nahi ki 52 paryantha the probability that the number on the card is a prime number number less than 20 so [nasal sound] se card ko utavata se card prime number probability ke se prime number less than 20 to ko chota prime number ke a prathama set humko dekh pab dekhontu humko dekh three a prime number five a seven a par up 11 a 13 a 17 a 19 a to ek gula prime number ya chha prime number nahi to ket prime number hagala up ko chhata prime number hagala aur total apan ko number ke 52 so probability ke probability haibo 6 / 52 and ya ko humein ya ko cancel kar kar kar kate up to re bhi katibo ok to three asibo and two re kat le 26bo so 3 / 26 ok sorry thi mu g number chhadi di se ho dui ok to thi up number kela look at eight number sorry I wrote six I wrote it sorry it was my mistake ok to do 8 / 52 [nasal voice] 8 / 52 ko probability humein ya ko eta four re cancel out hobo so four re kati a two asibo eta hijab 13 to 2 / 13 hobo yaar probability ki a humein option d re to option dbo correct answer ok acchi to tick calculate kola [nasal voice] bh se count karbe ke ho nale bh means silly mistake ho chance rah th ta par next question ho ki in a simultaneous throw of two dice if [nasal voice] [nasal voice] humein deta dice ko papad chhati den probability of getting a total even at even at even probability kate to hi karthale type question sample space ho up ke 36 ko samta at da to kono number he pub dekhantu gajab 26 re hijbo aur [nasal voice] g 62 re hijbo 35 re hijbo 53 re hijbo aur g 4jbo thik a kya nahi to total ke payale paanchta payale to 5/36 5/36 kothi option si re to option si ho right answer to kita question to apne bahut hi mmth easy thaye jo ki apne dekhno dekhhu ki solve kar pave aur kita question tic difficult tha tricky thi jo thik ki apne ko bhabiya ko pabbo next question dala ki what is the probability of getting an odd number on the top when an unbiased cubic die is thrown so if we throw a die then our top odd number probability of odd number is one good sir look then one good three good 5 a [nasal sound] so basically three odd number is a sample space so what is 3/6 as probability so its value is 1/2 so 1/2 is a good option a so option [ __ ] correct answer next question the set of all the outcomes of an experiment is called a jo humaan experiment perform kar chha probability re jo humaan kam karti like tossing a coin or throwing a die or ko hi experiment kaha jaaye se experiment jit all the outcomes outside of that outcome is called a collection of sample space ok achhi ta par [nasal sound] then next question so next question see kaun diyaachi a box containing seven If the first box contains five white balls and five black balls and the other box contains 10 white balls and six black balls then one ball is drawn at random from each box. So each box contains two balls.

[00:33:28]What is the probability that both the balls are white? If the first ball is drawn with probability 7/1 then the first box contains white balls and the second box contains white balls. If the second ball is drawn with probability 10/16 then the total number of balls drawn is 10/16 and if the first box contains five white balls and the second box contains six white balls then the total number of balls drawn is 10/16 then the first box contains five white balls and the second box contains six white balls and the second box contains five black balls and the third box contains six black balls and the third box contains six black balls and the fourth box contains six black balls and the fourth box contains six black balls and the fourth box contains six black balls and the fourth box contains six black balls and the fourth box contains six black balls and the fourth box contains six black balls and the sixth... / 96 Right Answer [nasal sound] But next question is, who gave the right answer? Two cards are drawn at random from a pack of 52 cards. If we draw 52 cards then what is the probability that one is spade and the other is heart. So [nasal sound] Spade is one and heart is one with probability. So we see that the spade card is 13. The heart card is 13. So we see that the 13 cards are needed and the 13 cards are needed and the total cards are 52 to 52.

[00:34:54]So the value we see is 13B and the value we see is also 13B and 52 / means 52C2 or see 52real 52 - 2 50real 2real s up to 52 * 51 / 2 a kati le 26 hochi to 26 * 51 to [nasal sound] 13 kati le to haibo to humko answer mil na 13 / 102 to 13 / 102 koti de option c dachi to option c yaar correct answer to ta par next question ko so dekhantu next question diya ko in a well travelled pack of 52 playing cards picking of two cards from it can be done in to humein 52 cards through gta card ko humein pick karti to humein extra vare pick kar p to seta ho ko 52c2 to eta kole 52dial by 50dial * 2 ya ko cancel out kar ke apne ko 52 * 51 / 2 directly bhi ko humein bahut tha kar leni to 26 / 51 unit digit ho six code unit digit six ho g jagat to 1 326 ho yaar right Answer is Next An event containing only a single sample point is called an event which has only one outcome or a simple event.

[00:36:25]So option D is correct answer. So [nasal sound] Equally Likely (possibly impossible) (possibly impossible) (possibly okay) So correct answer is Option D is correct answer.

[00:36:36]Next calculate the probability of getting exactly three heads in a simultaneous toss of four coins. If we toss four coins simultaneously then the probability of getting exactly three heads is pi times. If we toss four coins exactly then the probability of getting three heads is px3. If we toss four coins exactly then the probability of getting three heads is px3. So write the binomial distribution formula.

[00:37:05]Which one is equal to [nasal sound]?

[00:37:10]To write n number of outcomes within us, we have to write ncr p r q to the power 1 - r or n - r ok if we write this then in this place if we see that the total toss is four then n 4 then 4c 3 and p is the head probability to 1/2 or to the power of three and [nasal sound] second q is our failure to mean tail probability also we have 1/2 now n - 4 means 1 pπ upo ok so if we calculate 4c3 to the power then our 4 pi is the next jig that is 1/2 to the power 3 1/2 power 1 that is 1/2 power 4 only so if we write 1/2 to the power 4 then four asila 1/16 asila cancel out then how much do we have If we get the answer of 1/4 then 1/4 option C is correct answer [nasal sound] then next next question is ko so next question is dekho diya the probability of getting the sum as a prime number when the two dice are thrown together then we get an even prime number so we need to get a prime number as 1 and 2 then the rest of the prime numbers are 8, 9, 10, 11 and 12 so if we get a prime number then we will get a prime number as 1 and 2 then we will get a prime number as 1 and 2 then we will get a prime number as 1 and 2 then we will get a prime number as 1 and 2 then we will get a prime number as 1 and 2 [nasal sound] then three is akhi so three is akhi If we get 1 2 1 then we will get the total data. If we get 1 4 1 2 3 and 32 then we will get the fourth number. If we see the path from 11 then we will see 61. If there is any inconvenience then write it down. 25 52 34 43 and if we see only 11 and 11 then we will see 11 and 5 65 and if we see only 11 and then we will see 5 / 36. So basically we will get the data. If we see the total then it will be 10 5 15 15. So, the probability [nasal sound] 15 / 36 is one to three. If we send 5 to 12 then 5/12 means option B is correct. Answer is ok.

[00:40:03]But know 143.

[00:40:05]Who is called if a and b are mutually exclusive event mutually If A and B are mutually exclusive then the probability of A and B being mutually exclusive is 0.5 and the probability of B being 0.6 is 0.6 then A and B being mutually exclusive is 0.5 and the probability of A being mutually exclusive is 0.6 so A is not B and the probability of A being mutually exclusive is 0.5 and the probability of B being equal to B is 0.6 so A is not B and the probability of A being mutually exclusive is 0.6 so [nasal sound] the probability of A being equal to B is 0.5 and the probability of B being equal to B is 0.6 That = 0.4 So 0.5 + 0.4 Which = 0.9 So 0.9 is the correct answer option A [nasal sound] But next question is the letter of the word balancer are placed in a line at random So the word balancer letter Gu is placed in a line randomly What is the probability that all the vowels together are placed in a line randomly So first think of the word, if a vowel is placed in a line at random So the vowel is one, two, three, three.

[00:41:46]Vowel A is one, two, three, three and total letters are one, three, four, five, six, seven, eight So total letter eight, consonant, consonant, five, five, five, eight, thik, thik, thik, so basically if our letter Gu to letter Gu arrangement is arranged in a line at random then we have to do the permutation and see the correct answer.

[00:42:11]If we arrange n number of letters then we arrange them three times, it is also a repetition, right? So, at this place you should look at this vowel, keep the vowel Gu, the letter means nine is a consonant, there are five G vowels, that is, the total is the sixth letter, so we should arrange the sixth letter six times, so we should keep the vowel Gu in G place, but if the third letter is also arranged within itself, then three vowels, three letters are arranged within itself, then we should arrange them three times, and the total number of letters is eight, then we should arrange the eight times, okay, so if we calculate this, then we should look at the three times, six times and we should write 8 * 7, the remaining six times will be cancelled out, or we will have to divide it by two, three, four, six times, then we will get 3/28, so 3/28 Koti option C so option C is the right answer [nasal sound] but 133 number 133 number kaun diya to create a code for each employee in a company se kochi g company a seti pratyaksh employee r humko kaun kar got g code ko creat karab to so the code is up to five letters a to e and 10 digits a to 0 to can be used ek jas ko humein use kar code banewa thik achhi a code consist of three letters teenta letter rahbo follow by three digit teenta digit rahbo ta par apan ko teenta letter rahbo ok so what is the probability that all the three letters are same in the code so se [nasal sound] kahachi ki tale teenta jako jo letter humein use teenta jako letter up same as probability kate e se question pahchala to pratham hume ko dekhnu pab ki apni The sample space is made of 10 digits, okay, total arrangement is 100, if we see the arrangement, basically let us make a box, first we see the 3rd letter, then where will be the 3rd digit, so if you see the 3rd letter, 3rd letter means how many choices in the first place, 5th choice in the second place, 5th choice in the third place, 5th choice in the third place, 8th letter, okay, so from 125 [nasal sound] if you see in the digit, keep 10 digits, keep 10 digits, keep 10 digits, keep 10 digits [nasal sound] 0 to inside, okay, if there is no such thing then it is 000, so the total up results means 100 times the sample space, means 125000 in the sample space, okay now you know that all the letters need to be equal. If we have to use the same pattern then the letter a is our five letters a a ro e parant to which is the favourable outcome va to s pbo a s pbo nale bb b bb cc dd ee ee to panchta hi eti apan ko outcome bani pabo j hum ko samansaman dekh to a ke 5 aur ta par 1000 to ke payale hume 5000 payale so yaar probability kela 5000 / 125000 zero 0 katila 5 125 ko katva to 1/2 to 1/2 ko option a re to option az bha bhai yaar correct answer next question is the mean and the variance of a random variable x having the binomial distribution r4 and to respectively to kahthali a type question bahut sara bachra ho to se bacha ki px1 value keete to prathm to mean aur If we take random variable variance then np = 4 and np = 4 then if we subtract Q from p then p = 2/4 then p = 1/2 so p = 1/2 then p = 1/2 so we need to subtract n from p so we can see that n = 1/2 which is = 4/1/2 which is equal to 8 then we need px x 1 so if 8 = 1 then p = 1/2 to the power of 1/2 then 1/2 to the power of 1/2 is 1 and q = 1/2 to the power of 8 - 1 that is = 7 so if we solve p = 1 then we can see that 1/2 to the power of 8 so we get 8 = 2 to the power of 3 and 2 to the power of 8 so solve it If 1/2 to the power of 5 means 1/32 then 1/32 koti diyaachi option si lechi to option si ho correct answer ta par asantu next question kaun diya dekhantu the probability that A speaks the truth is 3/5 the probability that B speaks the truth is 3/4 what is the probability that they contradict each other when they are asked to speak a fact then [nasal sound] say that A is the truth what is the probability 3/5 B is the hundredth what is the probability 3/4 what is the probability that they contradict each other jane aju ko contradict kar probability ke hobo so a question jo humein answer calculate kar ko total probability re ba to humein yaar probability dekhva to probability ko dekhantu a If the probability of a term is 3/5 then the probability of a term is 3/4 then the probability of a term is 3/4 is 1/4 then the probability of a term is 1 - 3/5 is equal to 2/5 * b then the probability of a term is 3/4 or it is 3/20 or 6/20 then how many times will we get 9/20 then option D is the right answer then option D is the right answer then next question consider the following relations for the two events E and F. If SSB question asks you to know which event is given by E and B then correlation is correct then consider karntu to If we look at the probability of e from the union f then the probability of e plus the probability of f minus the probability of e from the intersection f then [nasal sound] we have to minus the square root of... If we also ask him, he should correct it. If we look at the relationship, which one is correct, then the third one is correct. Look, already we have which one, which one, which one, which one, which one, which one, which one, which one, which one, which one, which one, which one, which one, which one, which one, which one, which one, which one, which one, which one, which one, which one, which one, which one, which one, which one [suddenly loud breathing sound] [deep breathing sound] In an experiment, positive and negative values ​​are equally likely to occur, the probability of obtaining at most one negative value in five trials is said to be that in an experiment, positive and negative values ​​which are equally likely to occur, so equally likely means that we have many positive and negative values ​​with probability and probability and also with probability and [nasal sound] The probability of obtaining at most one negative value in five trials is the maximum negative value. If we get five trials then the maximum negative value is zero. If we get one then the maximum negative value is zero. So, if we get one negative value then we get zero. Second, if we get one negative value then we get five trials. Okay, so [nasal sound] first, 5C0.

[00:51:41]Second, we have to raise the probability of a positive number that is = 1/2 to the power 0 and the probability of a negative number that is = 1/2 to the power 5 - 0. So, 5C0 plus P1 is 5C1 to the power 1 to the power 4. So, we have to solve 5C0.

[00:52:07]So, obviously we have to raise 1 to the 32. So, we have to solve 5C0 and 1/32. Let us look at 1/32 5/32 means 6/32 now or if we cancel out the sum of the numbers to get three times 16 then 3/16 is option D then option D is the correct answer then next question is what is the probability of getting four sixes and another number in five random rolls in a balanced die if we throw the dice then we get four sixes as the probability then basically we look at the five throws if we get five within the five we need six we need four sixes so 5 6 5C4 is the total event cardinality divided by the sum of the sample space to the power of 5 6 to the power of 5 or to solve it then five as the denominator we have to tick the six [nasal sound] we have to multiply five fives then it is 216 as you have to multiply it by 36 so 216 / 216 * 36 just multiply it by 216 so almost 7776 okay so 7776 means 5/7776 correct answer which is given option B then the question is difficult nathila just empty calculative tick thela now who put the pata ko bahar kar naan of these bhi diya ok blank o digit ko dekhi humein kar pani nahi k non of these bhi answer hi pare so seth p a ko tick check kar de ok je bhi non of these diya hi de so ta on next question kaun diya what is the probability of getting five heads and seven tails in the 12 flips of a balanced coin a very good question a dekhantu se apan ko confuse kare If we toss a coin and get five heads and seven tails, we get 12 outcomes. If we toss a coin and get five heads, then the remaining seven are obviously tails. If we flip a coin, the outcome will be either heads or tails. If we flip a coin and get five heads, then the remaining seven are obviously tails.

[00:54:56]So basically, here you have to know the probability of five heads and seven tails. If we calculate the tails, then the total will be 12. If we toss a coin and get 12, then the total will be 12. If we toss a coin and get five heads, then the probability of five tails will be 12. And [nasal sound] the probability of heads will be 1/2. So the probability of tails will be 1/2. So the solution will be 12. So, option B is correct. Answer is fine, good, different seven tails, no need for it, either our five heads, or our seven tails, okay, whatever happens, yes, but the next question is, the coin is flipped three times, who will flip the coin three times [nasal sound] ah, the event is represented by getting at least one head, so the event is represented by getting at least one head. 2 to the power of 3 which is equal to 8 So A is the total sample space of eight outcomes. If we calculate the outcome of an event then A is the minimum number of outcomes... If we write a head then we get an even number of tails and if we toss a coin... If we take the intersection of the sample space of the outcomes, then the total number of outcomes is equal to this, so if we take the intersection of the sample space of the outcomes, then the total number of outcomes is equal to this, so if we take the intersection of the sample space of the outcomes, then the total number of outcomes is equal to this, so [nasal sound] we should be very careful in answering, we should neglect the zero part, we should leave it, that is, the three-point option is also given, and if we take the answer, then the next question is, if the mean of a binomial distribution is 8, then the probability of success P is 0.4, then what is the value of n, then the nth value is 0.4, okay, okay, so then we should say that the mean of the binomial distribution is 8, that is, the NP value is 8, and the Pth value is 0.4, okay, sir, so 0.4 means that we Write 4 / 10 and eta cancel out ki cancel out ki kar kar thik a rakhi danthu so we have n to n k n p / p that is equal to 8 / 4/10 so write 8 * 10 / 4 which is equal to 20 so n value ke pale 20 pele mane option d so option d is correct answer so par dekhantu next question aaj last question se kahi ki, two dice are rolled and the outcome is valid only if the sum of the numbers rolled is either an even number or a prime number so if the sum of the numbers rolled is an even number or a prime number then the outcome is valid only if the sum of the numbers rolled is an even number or a prime number then the outcome is considered valid so what is the total number of outcomes in the sample space so what are the outcomes from the sample space mane paba ki thik achi so [nasal sound] dekhantu If we roll the dice then either we get an even number or a prime number so if we roll the dice then we get two pi pi pi and then next up we get a prime number so if we roll the dice then we get two pi pi and then we get three four pi pi and then we get four because it is an even number so if we roll the dice then we get five pi pi and then six pi pi and then we get eight pi pi and then nine is not even because nine is not even number or prime number so if we roll the dice then we get one and then we get one and then we get one and then we get two 2 3 4 5 6 7 8 and then we get nine and then we get two 3 4 5 6 7 8 and then we get nine and then we get two 2... So if you throw all the dice only then you will get nine so you can have a sample space of total 36 dice so if you throw the dice then you will get 36 numbers from 36 then you will get nine so you will get nine just by emptying the teeth so rest you will get all these numbers [nasal sound] then you will get nine from 36 and 63 and 45 and 54 and if you do n't like any other number then basically 36 number is 36 number then you will get 32 ​​number so 32 number is your outcome in the sample space so if you have any question or probability then you can discuss the video to Bhagwa video and subscribe to the channel or if you want to join the batch then please do so You can join by WhatsApping the number and if you need a PDF then you can join the Telegram group and download our application by searching for Jam Class on Play Store and downloading it and publishing it.

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[01:03:18]Okay [nasal sound] so you saw the next video, but Jai Jagannath, thank you.