Random Variables and Probability Distributions

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Toss two fair coins and count the heads. You can get zero heads, one head, or two heads. So a quick student says: three possible answers, so each one has a chance of one third. Hmm, is that really right? Today we learn random variables and their probability distributions. This idea is the base for mean, variance and the binomial distribution, and these come up often in JEE, so let's begin.

0:32Random variable: a number for each outcome

So, first the sample space. Capital S, with curly brackets, lists all four outcomes: H H, H T, T H and T T, where H is a head and T is a tail. All four are equally likely, because the coins are fair. Okay? Now, we do not care which coin showed what. We only care about the number of heads. So we give it a name, capital X: X equals the number of heads. Right? Let me ask you this. Take the outcome H T, a head on the first coin and a tail on the second. What number does X give to it? One, yes, exactly! X of H T, written with round brackets, is one, and X of T H is also one. Then X of H H is two, and X of T T is zero. So X takes each outcome and gives it a number. Clear? And that is all a random variable is. It is written with a capital letter like X, and it assigns a number to each outcome of the experiment. Strange name, I know. It is really a rule that gives every outcome a number. Got it?

1:42Probability distribution: values with their chances

Now, the probability distribution. It lists the values of X, written small x with a subscript i, the small letter at the bottom, together with their probabilities. Small p i, written as capital P of X equals x i, is the probability that capital X takes the value x i. Okay? So let's do it for our two coins. Zero heads comes only from T T, which is one outcome out of four. Now tell me, how many of the four outcomes give exactly one head? Two outcomes, correct, H T and T H! So P of X equals zero is one quarter, and P of X equals one is two quarters, which is one half. And P of X equals two is one quarter. With me? So the quick student's one third is wrong. Classical probability needs equally likely outcomes. The four outcomes are equally likely, but the three values zero, one and two are not. The value one has two outcomes behind it, so it is twice as likely. Right?

2:46Two conditions every distribution must satisfy

Now, the definition comes with two conditions. First, each p i is greater than or equal to zero; that is the symbol with a line under the greater-than sign. No probability can be negative. Okay? Now for the second condition, think about this: every single time, X takes some value from the list. So what must all the p i add up to? They add up to one, absolutely! The big Greek letter is capital sigma, and it means, add them all up. So sigma p i equals one. Clear? Check our coins: one quarter, plus one half, plus one quarter, gives one. Both conditions hold, so this is a genuine probability distribution. Got it? And here is a useful tip for the exam. When a table has an unknown constant like k, the sum condition gives you the equation for k straight away. Then check that no probability comes out negative. That is exactly our first example. With me?

3:46Example 1: find k in the table, then P(X at least 2)

Now, our first example from the course. X takes the values zero, one, two and three. Their probabilities are zero point one, then k, then two times k, and then zero point three. Find k, and the probability that X is greater than or equal to two. Pause the video and try it first. Done? Okay, let's check. All the probabilities must add up to one. So zero point one, plus k, plus two times k, plus zero point three, equals one. The double arrow means, this gives. Zero point one plus zero point three is zero point four, so three times k equals zero point six. Okay? So k is zero point two. Now the four probabilities are zero point one, zero point two, zero point four and zero point three. None of them is negative, so both conditions hold. Right? Now the probability that X is greater than or equal to two. Some students take only the value above two, X equals three, and write zero point three. Is that right? No, and that is the trap: greater than or equal to includes two itself. So we add P of X equals two and P of X equals three. Zero point four plus zero point three gives zero point seven. Clear? And a quick check, the topper's way. The opposite of X greater than or equal to two is X less than or equal to one, the symbol with a line under the less-than sign. So one minus the bracket, zero point one plus zero point two, is one minus zero point three, which is zero point seven. Same answer. Got it?

5:30Example 2: number of aces in two cards drawn from 52

Now a harder one. Two cards are drawn, without replacement, from a pack of fifty two cards, and X is the number of aces. Find the probability distribution of X, okay? Pause the video and try it first. Done? Okay, let's check. Step one is always the same. Which values can X take here? Zero, one or two aces, yes! Now the total. The fifty two above the two, in round brackets, is fifty two C two, the number of ways to choose two cards from fifty two. Fifty two times fifty one, over two, gives one thousand three hundred and twenty six. Okay? For zero aces, both cards come from the forty eight cards that are not aces. Forty eight C two is one thousand one hundred and twenty eight. Divide the top and the bottom by six, and P of X equals zero is one hundred and eighty eight over two hundred and twenty one. Right? Now exactly one ace. In how many ways can we choose one ace and one card that is not an ace? Four times forty eight, correct! Four choices for the ace and forty eight for the other card; the dot means times. That gives one hundred and ninety two. Dividing by six again, P of X equals one is thirty two over two hundred and twenty one. Clear? And two aces: four C two, which is six ways. So P of X equals two is six over one thousand three hundred and twenty six, which is one over two hundred and twenty one. With me? Now a final check: do they add up to one? One hundred and eighty eight, plus thirty two, plus one, gives two hundred and twenty one. So the total is two hundred and twenty one over two hundred and twenty one, which is exactly one. So our distribution is correct, and that is the whole method!

7:29Quick revision: random variable, each p at least 0, sum 1

So, quick revision. First, a random variable gives a number to each outcome. Second, its distribution lists the values x i with their probabilities p i. Third, every p i is at least zero, and they all add up to one; use this to find k, and to check your answer. Now go and try the practice problem on this page. All the best!