Mixed Problems: Set 3 (Random Variables and the Binomial Distribution)

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Someone gives you just two numbers about a binomial distribution: the mean is eight, and the variance is six. Nothing else, no number of trials, and no probability of success. Can you still find both of them? It sounds impossible, right? Keep your answer ready, because Example one settles it in a few short lines. This is Set three of our mixed problems in the JEE Main pattern, on random variables and the binomial distribution, and questions like these come up often. So, let's begin.

0:37Set 3: what it holds

So, here is what Set three holds. Fifteen solved problems, on random variables, the binomial distribution, and waiting for the first success. They are all mixed up, so you choose the tool yourself, just like in the exam. Okay? And the course target is one to two minutes per problem. So first, let us sharpen three tools from Chapter five, and then we solve two problems together. With me?

1:02Tool 1: the binomial formula

Tool number one, from Chapter five: the binomial formula. First, the kind of experiment. We need a fixed number, n, of independent trials, where each trial is either a success or a failure. Okay? And the success probability, p, must stay the same every time. The failure probability is q, which is one minus p. Trials like these are called Bernoulli trials. Right? Now let capital X be the number of successes. Then X follows the binomial distribution, capital B of n comma p. The wavy sign is called tilde, and we read it as, follows. Clear? Now, the chance of exactly r successes in the n trials. The successes need not come first, right? So tell me, in how many different orders can the r successes sit among the n trials? In n C r ways, yes, exactly, by choosing which r trials succeed. So P of X equals r is n C r, times p to the power r, times q to the power n minus r. The n over r in tall round brackets is n C r, the small raised letters are powers, and r runs from zero up to n; the three dots mean, and so on. Got it?

2:20Tool 2: mean and variance

Tool number two. For B of n comma p, the mean is n times p, and the variance is n times p times q. These two formulas solve many problems in this set. Okay? Now let me ask you this. If you divide the variance by the mean, what is left? Just q, yes, exactly! The n times p cancels from the top and the bottom. So q equals the variance over the mean. Keep this one ready, because our puzzle needs it. Right? And since q is less than one, the variance is always less than the mean; the pointed sign is read as, less than. So a variance bigger than the mean means the distribution cannot be binomial. Clear?

3:06Tool 3: compare neighbouring terms

Tool number three. In the picture, each bar is P of X equals k, the chance of k successes in ten trials. The left one, B of ten comma one half, is symmetric, with its tallest bar, the mode, at five. The right one, B of ten comma one fifth, is skewed, with its mode at two. Okay? Now, many questions compare two neighbouring bars, like P of X equals two and P of X equals one. Do you really have to expand both, with all those powers? No, you do not. Just divide, and almost everything cancels. P of X equals r plus one, over P of X equals r, is n minus r over r plus one, times p over q; the dot means times. The n C r parts give n minus r over r plus one, and the one extra p and one fewer q give p over q. With me? And one habit that saves marks. The smaller value always sits at the bottom, and that value is r, while the top is r plus one. Got it?

4:06Example 1: find n and p from mean 8 and variance 6

Now, remember our puzzle? Here it is as Example one, marked easy. We are told that a binomial distribution has mean eight and variance six. Find both n and p from these two numbers. Pause the video and try it first. Done? Okay, let's check. Write down what we know: the mean, n times p, is eight, and the variance, n times p times q, is six. Okay? Now we divide to get q. Many students write the mean over the variance, eight over six. Is that right? Oh no, careful, that is the trap! Eight over six is bigger than one, and no probability can be bigger than one. Tool two says q is the variance over the mean: n times p times q, over n times p, which is six over eight, and that gives three quarters. Right? Now careful once more, because three quarters is q, the failure probability, and not p. So p is one minus three quarters, which is one quarter. Clear? And since n times p is eight, n is eight over p, which is eight times four, and that gives thirty two. So X follows B of thirty two comma one quarter. With me? Quick check before we move on: thirty two times one quarter is eight, the mean, and eight times three quarters is six, the variance. Both match, so n is thirty two and p is one quarter, the course answer. And our puzzle is solved: just two numbers gave back the whole distribution. Got it?

5:42Example 2: find p in B(10, p) when P(X=2) = 2P(X=1)

Now, Example two, and this one is marked hard. X follows B of ten comma p, and P of X equals two is twice P of X equals one. Find p, the success probability. Pause the video and try it first. Done? Okay, let's check. Two neighbouring probabilities are being compared, so this is a job for Tool three. But tell me, what is r here, one or two? It is one, the smaller value, which sits at the bottom. So the ratio is ten minus one, over one plus one, times p over q. That is nine over two, times p over q. Okay? The question says this ratio equals two. So nine over two, times p over q, equals two. Multiply both sides by two over nine, and p over q is four over nine; the double arrow means, which gives. Right? So the answer is four over nine, isn't it? No, no, wait, that is the trap in this question! Four over nine is p over q, not p itself. Cross multiply, and nine times p equals four times q, which is four times, bracket one minus p. Open the bracket and bring the p terms together, and that gives thirteen times p equals four. Clear? So p is four over thirteen, the course answer. It lies between zero and one, as a probability must. And see, we never expanded a single power of p or q, just one ratio. That is the power of Tool three. Got it?

7:19Quick revision: binomial formula, np, npq, term ratio

So, quick revision. First, for Bernoulli trials, P of X equals r is n C r, times p to the power r, times q to the power n minus r, with mean n times p and variance n times p times q. Second, q is the variance over the mean, and q is the failure probability, not p. Third, to compare neighbouring terms, use the ratio n minus r over r plus one, times p over q, with the smaller value as r. Now try the practice problem on this page, and then attempt the rest of Set three, at one to two minutes each. All the best!