JEE Advanced, AIEEE and JEE Main (2013–2018) Questions: Random Variables

Transcript

0:00Introduction

Hello everyone! Here is a real JEE Advanced question. Each missile hits its target with probability three quarters, and the target is destroyed only after at least three hits. How many missiles must you fire to be at least ninety five percent sure of destroying it? Is it three missiles, or four, or five? Take a guess and hold on to it; we settle it in Example two. So, let's begin.

0:28Idea 1: spot the binomial

So, idea number one: spot the binomial. We need a fixed number n of independent trials, each a success or a failure, with the same success probability p every time. Then q, the chance of failure, is one minus p. Okay? Then capital X, the number of successes, follows capital B of n comma p, the binomial distribution; the wavy line means follows. P of X equals r is n C r, times p to the power r, times q to the power n minus r, where the tall brackets with n above r are n C r, the ways to pick which r trials succeed. And r runs zero, one, up to n; the dots mean and so on. Right? Now a past-paper question: a pair of dice is thrown three times, and we want a total of nine exactly twice. So what is p here? Each throw of the pair is one trial, and a total of nine is a success. Nine comes as three and six, four and five, five and four, or six and three, so four outcomes out of thirty six. So p, P of a total of nine, is one ninth. Clear? Now exactly two successes in three throws: three C two, times one ninth squared, times eight ninths for the one failure. The dot means times. Three C two is three, so this is eight over two hundred and forty three. Got it?

1:56Idea 2: mean np, variance npq

Idea number two: for the binomial, the mean is n times p, and the variance is n times p times q. They come up often; learn them. With me? Now divide one by the other: variance over mean is just q. Since q is less than one, the variance is less than the mean, the less-than sign on screen. Okay? Here is a JEE Main question from two thousand seventeen. Fifteen green and ten yellow balls are in a box, and ten are drawn with replacement. We want the variance of the number of green balls. Right? Let me ask you this: why do the words with replacement matter here? Yes, exactly! Each ball goes back, so the draws are independent, and the chance of green stays fifteen out of twenty five, three fifths, every time, with q equal to two fifths. Without replacement, p would keep changing, and the binomial would not apply. Clear? So the variance is n p q: ten times three fifths, times two fifths, which gives twelve over five. Got it?

3:03Idea 3: at least? Use the complement

Idea number three: for at least, use the complement. The chance of at least one success is one minus q to the power n, because the only way to miss is to fail all n times. Okay? Now, what about at least two successes? Is it still one minus q to the power n? No, no, careful, that is the trap! One minus q to the power n removes only zero successes; for at least two, remove both zero and one. The greater-than sign with a line under it means at least. So P of X at least k is one minus, in big square brackets, P of X equals zero, plus P of X equals one, up to P of X equals k minus one. Right? And for the least number of trials, test values of n and stop at the first one that works. Clear?

3:53Example 1: P(X = 1) when mean is 4 and variance is 2

Let's start with an easy one, from the two thousand three paper of the exam now called JEE Main. This binomial random variable has mean four and variance two. Find P of X equals one. Pause the video and try it first. Done? Okay, let's check. By idea number two, q equals the variance over the mean, two over four, which is one half. So p is one minus q, also one half. Okay? Now the mean, n times p, is four. The double arrow means so, and with p one half, n must be eight. Right? So one success in eight trials, each with chance one half. Then the answer is just one half to the power eight, right? Oh no, careful, that is the trap! One half to the power eight is one particular pattern, say a success first and then seven failures. But the single success can be any of the eight trials, and that is the eight C one in front. So eight C one, times one half, times one half to the power seven, gives eight over two to the power eight, which is one over thirty two. Clear?

5:07Example 2: least missiles for 95% chance of at least 3 hits

Now, back to our missile puzzle, from JEE Advanced, two thousand twenty. Each missile hits with probability zero point seven five, and at least three hits are needed. Find the least number of missiles for a chance of destruction of at least zero point nine five. Pause the video and try it first. Done? Okay, let's check. With p equal to three quarters and q equal to one quarter, X, the number of hits, follows B of n comma three quarters. We need P of X at least three to be at least zero point nine five. With me? Now, the complement. What is the opposite of at least three hits? Yes, exactly: zero, one or two hits, not just zero! So P of X at least three is one minus, in big brackets, P of zero plus P of one plus P of two, where P of zero is short for P of X equals zero. Okay? Let's test five first; if five is not enough, fewer missiles cannot be either, since fewer missiles only lower the chance. Each term is n C r, times three to the power r, all over four to the power five. So the numerators are one for P of zero, five times three, which is fifteen, for P of one, and ten times nine, which is ninety, for P of two. The less-than sign with a line under it means at most: P of X at most two is one hundred and six over one thousand and twenty four. And the wavy equals sign means approximately, so P of X at least three is about zero point eight nine six. Right? And zero point eight nine six is less than zero point nine five, so even five missiles are not enough. Clear? So try six, now over four to the power six. The numerators are one, then six times three, which is eighteen, then fifteen times nine, which is one hundred and thirty five, since six C two is fifteen. The total is one hundred and fifty four over four thousand and ninety six, so P of X at least three is about zero point nine six two. Got it? Zero point nine six two is at least zero point nine five, so the least number of missiles is six. And our puzzle is settled: not three, four or five, but six missiles! Remember the trap from idea three: subtract only P of zero, and you would wrongly stop at three missiles. Okay?

7:37Quick revision: binomial formula, np, npq, at least k

So, quick revision. First, the binomial formula, for independent trials with the same p. Second, the mean is n p, the variance n p q, and q is the variance over the mean. Third, for at least k, take one minus the cases zero to k minus one, then test values of n. Now go and try the practice problem on this page. All the best!