The Binomial Distribution

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Each trial succeeds with probability one quarter. How many trials do you need, to be ninety per cent sure of at least one success? Four quarters make one whole, so four trials, right? Hmm, but then eight trials would give a probability of two, which is impossible! So something is wrong. The binomial distribution gives the real answer, and it comes up often in JEE, so let's begin.

0:31Bernoulli trials and the binomial formula

First, look at the kind of experiment. We repeat a trial a finite number of times, and the trials are independent: one result does not change the chances of another. Each trial has only two outcomes, success and failure. Okay? Next, the success probability, p, stays the same in every trial. The failure probability is q, which is one minus p. Trials like these are called Bernoulli trials. Got it? Now let capital X be the number of successes in n trials. We say that X follows the binomial distribution, capital B of n comma p. The wavy sign is called tilde, read as, follows. Clear? Now, the chance of exactly r successes. Take one fixed order, with r successes and n minus r failures. The trials are independent, so we multiply: p to the power r, times q to the power n minus r. With me? But the successes need not come first. So tell me, in how many different orders can the r successes sit among the n trials? Yes, exactly, in n C r ways, choosing which r trials are the successes. The n over r in tall round brackets is n C r. So P of X equals r is n C r, times p to the power r, times q to the power n minus r, for r from zero up to n; the three dots mean, and so on. Write it down!

1:59Mean and variance of B(n,p)

For B of n comma p, the mean is n times p, and the variance is n times p times q. In the picture, each bar is P of X equals k, for k successes in ten trials. For p equal to one half, the mean is ten times one half, which is five; for one fifth, it is ten times one fifth, which is two. Okay? Now let me ask you this. Can the variance of a binomial distribution ever be bigger than its mean? Very good if you said no, it can never be bigger. The variance is the mean, n times p, multiplied by q, and q is less than one. So the variance is always less than the mean; the pointed sign is read as, less than. Right? So a distribution with mean four and variance five cannot be binomial: its variance is bigger than its mean. Clear? And one more trick. Divide the variance by the mean: n times p cancels, and q is left. So q equals the variance over the mean. Got it?

3:03The most probable value

Next, compare each bar with the one before it. 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. Okay? From this ratio comes the most probable value, the mode. It is the integer part of, n plus one, times p; we just drop the decimal part. In the picture, the tallest bar is at five on the left, and at two on the right. Right? And when n plus one, times p, is a whole number, there are two modes. Clear? Now you try this one: for B of ten comma zero point three, what is the most probable number of successes? Here n plus one is eleven, and eleven times zero point three gives three point three. Its integer part is three, so the mode is three. Got it? When p is one half, the distribution is symmetric, like the left picture. P of X equals r equals P of X equals n minus r, and both are n C r over two to the power n. With me?

4:10At least one success

Now the last fact, and it will solve our puzzle. What is the opposite of getting at least one success? Getting no success at all, that is correct, which means all n trials fail. Each failure has probability q, so P of X equals zero is q to the power n. Okay? So P of at least one success is one minus q to the power n. Remember this one; it comes up often.

4:37Example 1: exactly two sixes in five throws of a die

Let's do the first example. We throw a die five times. Find the probability of exactly two sixes. Pause the video and try it first. Done? Okay, let's check. Each throw is independent, gives a six or not a six, and the chance of a six is always one sixth. So, n is five, p is one sixth, and q is five sixths. Right? Now, we want two sixes and three non-sixes. So the answer is one sixth squared, times five sixths cubed, right? No, no, wait, that is only one order, say six, six, and then three non-sixes. The two sixes can be any two of the five throws, so we multiply by five C two. Leaving it out is a very common mistake. Clear? Five C two is ten, one sixth squared is one over thirty six, and five sixths cubed is one hundred and twenty five over two hundred and sixteen. Ten over thirty six is five over eighteen, so we get five times one hundred and twenty five, over eighteen times two hundred and sixteen. That is six hundred and twenty five over three thousand eight hundred and eighty eight. Got it?

5:57Example 2: least n for a 90% chance of at least one success

Now, let's solve our puzzle from the start. With p equal to one quarter per trial, find the least n for which P of at least one success is at least zero point nine. Pause the video and try it first. Done? Okay, let's check. Here q is three quarters. So one minus three quarters to the power n must be greater than or equal to zero point nine; that is the sign with a line under it. Rearrange, and three quarters to the power n must be less than or equal to zero point one, the same kind of sign, pointing the other way. Okay? So, are the four trials from our puzzle enough? No, it is not even close. Three quarters to the power four is eighty one over two hundred and fifty six, about zero point three two, far bigger than zero point one. The wavy equals sign means approximately. Right? Now jump to n equal to eight: three quarters to the power eight is about zero point one zero zero one. That is basically zero point one, so the answer is eight, right? Oh no, careful, zero point one zero zero one is just above zero point one; the sign pointing this way is read as, greater than. We need less than or equal to zero point one, so eight trials are not enough. Clear? At n equal to nine, three quarters to the power nine is about zero point zero seven five, below zero point one. So nine trials work. So the least n is nine, not four! Remember our puzzle? The chances of separate trials do not simply add up. For ninety per cent, you need nine trials.

7:45Quick revision: formula, np, npq, mode, 1 minus q to the n

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. Second, the mean is n times p, and the variance, n times p times q, is less than the mean. Third, the mode is the integer part of, n plus one, times p, and P of at least one success is one minus q to the power n. Now try the practice problem on this page. All the best!