Binomial & Geometric Distributions | Difficult Questions from 2025-2026 Papers | A-level Math S1

Añadido: 2026-05-18

[00:00:00]exact number for this. It's just an estimate. Okay. Let's look at the next one now. Person number four.

[00:00:06]Okay. Now, Suri or whatever that name is has a bag containing four blue discs and three yellow discs.

[00:00:16]She selects one disc from the bag at random, notes its color, and returns to the bag.

[00:00:22]She repeats this process until she obtains a yellow disc.

[00:00:29]Find the probability that she obtains a yellow disc for the first time on her on on her fourth selection. For the first time, does that remind you of something? First success, that's a geometric distribution. You just have to be careful.

[00:00:43]Is the probability remaining constant?

[00:00:47]Is the probability constant? Yes, because she is returning to this to the to the bag.

[00:00:54]It's with replacement.

[00:00:56]Probability is remaining constant.

[00:00:59]So, it'll be a geometric distribution.

[00:01:01]We can say it's a geometric distribution.

[00:01:06]Do we know the probability for success in this? We're talking about yellow disc. There are three yellow discs out of seven. So, probability of success is going to be three out of seven. And then, probability of failure, that will be four out of seven. What we're looking for is probability of getting our first success on the fourth attempt. What's the formula that we've got for a geometric distribution?

[00:01:28]Q raised to the power r minus one times p. Input values in that, see what you get from that. 4 over 7 whole raised to the power Okay, I'm uh raised to the power 4 minus 1, which is 3, into 3 over 7. You can see what that is.

[00:01:49]What's this number?

[00:01:50]Okay, you'll write down the actual number first, the exact number. So, if it's a fraction, you can write down a fraction.

[00:01:55]If it's decimals, you can write down a lot of decimal places first of all, and then you write down the number to three significant figures. In this case, it's not necessary. They're asking for probability. Probabilities are often written in fraction form as well. You can leave it as it is as well. If you do want to write a decimal, it rounds off to 0.080.

[00:02:20]Okay, let me actually show you the actual number. This is the actual number 192 over 2401. This is the decimal value that we're getting. It will round off to 0.080.

[00:02:30]That will be your final answer.

[00:02:33]Okay.

[00:02:35]Makes sense, right? Let's move on.

[00:02:37]Now, they're saying find the probability that she obtains a yellow disc for the second time on her fifth selections.

[00:02:43]On her fifth selection. Again, we've seen this question multiple times in recent past papers.

[00:02:48]We want the second success to come on her fifth selection. Now, what does that mean?

[00:02:54]1 2 3 4 5 The fifth attempt should be her second success.

[00:03:07]So, we can say this fifth attempt is a success. This has to be a success.

[00:03:14]But, for this to be her second success, we want one more success in these first four attempts.

[00:03:29]Okay?

[00:03:34]What we need is one out of four successes.

[00:03:43]So, we can consider this as a binomial.

[00:03:46]That we have one success out of the first four. How do you find that probability? That's a binomial.

[00:03:52]The formula for that is nCr p raised to power r q raised to power n minus r.

[00:04:00]There are four possible attempts here in total.

[00:04:05]So, n is equal to four.

[00:04:08]Probability of success is 3/7.

[00:04:12]It'll be 4C1 into 3/7 power 1 into 4/7 power 4 minus 1, which is 3.

[00:04:22]This gives us the probability of getting one out of the first four attempts as a success.

[00:04:30]And then we want another success on the fifth attempt. What's the probability of success again?

[00:04:37]3/7. We multiply this by 3/7 and that is going to give us the result.

[00:04:42]Now, let me repeat this.

[00:04:45]We want the fifth attempt to be our second success. What that means is in the first four attempts we want one success, so we treat that part as binomial, that we want one success, exactly one success out of the first four attempts.

[00:05:02]That's a binomial distribution. It'll become this.

[00:05:06]And then we want the second success to come on the fifth attempt.

[00:05:10]So, we multiply that number by 3/7 and that is going to give us the result. You can see You can see what that turns out to be. You're saying that's 0.137085.

[00:05:20]Or write down the fraction first. Round it off to three significant figures.

[00:05:23]Now, if you do that, that'll that'll be 0.137 and that's your final answer.

[00:05:33]Yeah, so first success can be anywhere in the first four attempts. That is why we're using a binomial in this, right?

[00:05:42]So, this binomial accounts for all possibilities that the first success could come in any one of those positions.

[00:05:50]You could also do it without binomial.

[00:05:52]You could say probability of getting one success and three failures.

[00:06:02]That's going to be this for one particular order. But then that one success could come in this position or this position or this position or this position. For that reason, you multiply this by four.

[00:06:14]That's exactly what 4C1 is. 4C1 is also equal to four.

[00:06:18]You can just do it logically directly as well, or just use binomial, and that will give you the result as well.

[00:06:24]Okay, that's 0.137. That's your final answer.

[00:06:27]Two second part, we've done questions like these multiple times now, right?

[00:06:35]A six is obtained for the third time on the seventh throw. How would we approach this?

[00:06:40]The second part of this?

[00:06:44]We have this six-sided dice. It's thrown repeatedly until a six is obtained.

[00:06:50]Now they're saying a six is obtained for the third time on the seventh throw.

[00:06:54]Probability of obtaining a six, we can say probability of success, that is one out of six. Probability of failure is five out of six.

[00:07:05]The first part would be straightforward, right? The first will just be a geometric distribution. That will be Q raised to power seven times P.

[00:07:12]Probability of X equals eight. That will be equal to this, which will be five over six whole raised to power seven into one over six, and that will give you the result.

[00:07:22]Second part is slightly tricky, but we've done multiple questions like these before in in earlier papers. If the if six is obtained for the third time on the third seventh throw, the way we understand this is we say this is the first throw, second throw, third, fourth, fifth, sixth, seventh.

[00:07:42]The seventh attempt is a success. But it is the third success.

[00:07:49]For this to be the third success, we want two successes out of the first six.

[00:08:06]And then we could say the seventh success uh uh uh success on the seventh attempt that would be the third one.

[00:08:18]So, these first six, we could think of these as a binomial distribution with n equal to six. Out of six, we want two successes.

[00:08:30]That will be 6C2 into p, which is 1/6 raised to the power 2 into 5/6 raised to the power 6 - 2.

[00:08:40]That's the probability of getting two successes in the first six attempts.

[00:08:46]And then you get another success on the seventh attempt. You multiply this number by 1/6, and that gives you the result.

[00:08:55]We've done similar questions like these before as well. This is how you you would approach this.

[00:09:01]Make sense, right? You can do this.

[00:09:09]Right, 0.0335 that's the result that we're getting. Okay, 0.0335.

[00:09:17]And that's the result that you would get from this.

[00:09:20]Okay, this is getting repeated in recent papers again and again. You need to make sure that you understand this well. The words used to that next.

[00:09:30]A fair six-sided dice with faces labeled 1 2 3 4 5 6 is thrown repeatedly until a three is obtained. Does that remind you of something?

[00:09:42]You keep doing something until you get your first success.

[00:09:47]That's geometric distribution.

[00:09:49]The number of throws taken is denoted by the by the by the random variable X.

[00:09:53]Find the probability of X equals 8.

[00:09:55]That's a geometric distribution.

[00:09:59]And the probability of success is 1 over 3 1 over 6. Three is one out of those six numbers.

[00:10:08]Q is going to be 5 over 6. Probability of X equals 8 therefore is going to be 5 over 6 raised to power 8 minus 1.

[00:10:19]Remember the formula?

[00:10:20]Probability of X equals R for a geometric distribution that is Q raised to power R minus 1 times P. So 5 over 6 into 5 over 6 raised to power 8 minus 1 into 1 over 6 and that gives you the result. You can see what that is.

[00:10:35]What's this number?

[00:10:43]What is this number anyone?

[00:10:46].0465. Thank you. 0.0465.

[00:10:50]Now again, it's a geometric distribution still. They're saying find the probability of X less than 9. How do you find this probability? We have a formula for inequalities in geometric distribution as well and that looks like this. Probability of X greater than R is given by Q raised to power R.

[00:11:06]Can we convert it to that form somehow?

[00:11:10]We can think of it like this.

[00:11:12]These are all the numbers that we want all the way from 1 2 3 up to 9. What are the numbers that we do not want? 10 11 and so on. The total probability for all of these numbers is going to be 1.

[00:11:24]Subtract from 1 the numbers from 10 onwards. How do you describe these numbers in greater than form? These numbers are greater than nine. So, 1 minus probability of X greater than nine. And that will give you 1 minus Q raised to the power nine. Input values in this, 1 minus 5 over 6 raised to the power nine. And that will give you the result. You can see what that is.

[00:11:48]What do you get from this?

[00:11:53]What's this number?

[00:12:01].80619 to three significant figures.

[00:12:05]And that will become 0.806.

[00:12:08]And that's the final answer.

[00:12:10]Okay, that was part B of this.

[00:12:14]Yeah, 1 minus probability of X greater than nine, right?

[00:12:19]Okay, I'm sorry about that. I'm sorry about that. You're You're right. I read that wrong.

[00:12:24]You're absolutely right.

[00:12:27]It's less than nine, right? So, nine is not included there.

[00:12:34]It should have been up to eight here.

[00:12:37]And then 9 10 11 and so on with so on, these are not included. Sorry about that.

[00:12:44]So, they're starting from nine here.

[00:12:46]Total is one. These numbers are described as greater than eight. Yes, so it will be 1 minus probability of X greater than eight. 1 minus Q raised to power eight. That means 1 minus 5 over 6 raised to power eight. Yes, that's correct. This is what is going to give you the result.

[00:13:02].767 four. Okay, that's the probability that we end up getting.

[00:13:11].7674, that rounds off to 0.767.

[00:13:17]Okay, that's the final answer that we get.

[00:13:20]Okay, acha. Now it says, "Find the probability that a three is obtained for the second time before the sixth row.

[00:13:30]What about this now?

[00:13:36]This is quite complicated now.

[00:13:38]Before the sixth row, what is that?

[00:14:03]Okay.

[00:14:04]Find the probability that a three is obtained for the second time before the sixth row.

[00:14:10]For the second time before the sixth row.

[00:14:12]What does that mean?

[00:14:17]For the second time before the sixth row. Now, it could be on the fifth row, it could be on the fourth row, on the third row, on the second row.

[00:14:28]We'll have to consider all of those cases.

[00:14:30]Yes, we'll have to make cases for this.

[00:14:33]So, what are the possible cases now?

[00:14:35]If it If the second success comes before the sixth row, the possible cases could be that it comes on the fifth row. So, you would say first, second, third, fourth, fifth. You have the second success on the fifth row.

[00:14:56]That means in the first four, you have one out of four. One S out of four, and then fifth one is a success.

[00:15:10]What is this going to be?

[00:15:15]What is this going to be?

[00:15:21]This part is binomial.

[00:15:24]That's 4 C 1 into P, probability of success, what was that?

[00:15:32]1 over 6 raised to power 1 into 5 over 6 raised to power 3.

[00:15:41]And then you have a success on the fifth attempt.

[00:15:46]That is 1 over 6 and that will give you something. That's your first case.

[00:15:52]And then you'll have to make more cases like this. This another possibility could be the second success comes on the fourth attempt.

[00:16:05]Second success comes before the sixth throw. That means the second success could come on the fifth throw or it could come on the fourth or it could come on the third or the second.

[00:16:17]So I have to consider all of them.

[00:16:23]But the second time it comes, the second time the success comes, they're saying that happens before the sixth throw.

[00:16:34]The second success comes before the sixth throw. Now the second success could either come on the fifth attempt it or it could come on the fourth attempt. That will be another case that we consider. That's why we're making cases in the first place. So we're dividing them into multiple cases.

[00:16:55]The first case is that the second success comes on the fifth attempt.

[00:17:00]That's the first case that we're considering.

[00:17:02]Then we consider another case that the second success comes on the fourth attempt. In that case, we would need to have one success out of three in the first three attempts.

[00:17:16]So that will give you 3 C 1 into 1 over 6 raised to power 1 into 5 over 6 raised to power 2 and then a success on the fourth attempt, that's 1 over 6 and that will give you something. And then the third possibility is that you have the second success on the third attempt and then from the first two, you have one success as well.

[00:17:44]One S out of two.

[00:17:48]That's your third case.

[00:17:51]This will be become 2 C 1 into 1 over 6 raised to power 1 into 5 over 6 raised to power one.

[00:18:03]And then a success on the third attempt, that will be 1 over 6 and that will give you something.

[00:18:09]And then finally, the second success comes on the second attempt. That means the first is a success and then the second is also a success.

[00:18:18]That will just be 1 over 6 into 1 over 6.

[00:18:29]Find all of these probabilities. It could be the first case, the second case, the third case, the fourth case and then we just add all of them up and that's going to give us the result. You can see what that is.

[00:18:53]Is this okay?

[00:18:57]0.196, that's the final answer?

[00:19:00]Uh but you should write the individual answers as well of these cases.

[00:19:09]What are those individual values?

[00:19:19]Has anyone done this? 0.0643 0.05787 0.046 296 and 0.02778.

[00:19:54]Thank you. And then we just add all of them up.

[00:19:58]And that gives us the result.

[00:20:03]Is that okay?

[00:20:06]Again, that's quite tricky.

[00:20:09]This is how we would approach this.

[00:20:15]No problems?

[00:20:24]Subtract the cases of no threes and one three and five throws.

[00:20:39]Should we get no threes in the first five? Okay.

[00:21:04]Okay, yeah. That could be another way of thinking about this.

[00:21:09]What you were saying is uh the total probability is one.

[00:21:15]If the second success comes before the sixth row.

[00:21:21]If the second success comes before the sixth row, that means it's either the second row or the third or the fourth or the fifth.

[00:21:33]Now, the total probability for anything is one. What if we subtract from one the probability that we have no success.

[00:21:45]So, if you think of this this as a binomial distribution with n equal to five.

[00:21:51]Binomial with n equal to five.

[00:21:58]That we're considering the first five throws. In those five throws probability that there are no successes and probability that there is one success out of the first five throws.

[00:22:14]If you subtract these two cases from one in all the other cases, we would have two successes in the first five outcomes.

[00:22:25]Right? We have two successes out in the first five outcomes if we take out these two cases.

[00:22:33]And that will be the same thing that you're asking about here. Because if there are m- uh at least two successes, then that means the second success comes in the first five throws.

[00:22:48]So, that will actually that will actually be a much more efficient way of doing this.

[00:22:52]A little bit harder to think of.

[00:22:55]1 minus 5 C 0 and then 1 over 6 raised to power 0 into 5 over 6 raised to power 5. Also subtract the case when you have one success 5 c 1 into 1 over 6 raised to power 1 into 5 over 6 raised to power 4.

[00:23:15]If you evaluate this, that should also give you 0.196.

[00:23:25]Is that all right?