Conditional Probability

Transcript

0:00Introduction

Hello everyone! Let us start with a famous puzzle about a family with two children. Someone tells you that at least one of them is a boy. Now, what is the probability that both children are boys? Most students say one half at once. The other child is either a boy or a girl, so one half, right? Hmm, hold on to that answer, and we will check it later. Extra information changes the sample space itself, and that is what conditional probability is about. It comes up often in JEE, and this whole chapter is built on it. So, let's begin.

0:35Given B: the sample space shrinks

So, here are two events of one experiment, A and B, drawn as two circles. Before we know anything, every outcome is still possible, okay? Now suppose we are told that the event B has occurred. Then every outcome outside B is impossible, so we throw it away. Only the outcomes inside B remain, and B becomes our new sample space. We call it the reduced sample space. Clear? So tell me, inside this new sample space, where can the event A still happen? Only in the overlap, the darker part, yes, exactly! So the probability of A given B is the share of B that also lies inside A, right? P of A given B equals P of A intersection B, over P of B, okay? The upright bar is read as given, and the upside down cup is intersection, meaning and. And the condition on the right, with the greater than sign, says P of B must be greater than zero, because we cannot divide by zero. With me? And when all the outcomes are equally likely, we simply count. Small n of a set means the number of outcomes in it. So P of A given B equals n of A intersection B, over n of B, okay? We count inside B only, never in the whole sample space. Got it?

1:56Properties of P(A|B)

Now, P of A given B is a probability on the new sample space, so it behaves like any probability. The sign that looks like less-than with a line under it means less than or equal to. So P of A given B always lies between zero and one. And P of B given B equals one, because once B has occurred, B is certain. Right? Next, the complement. The small dash after A is read as A prime, and it means that the event A does not happen. Given B, either A happens or it does not, so P of A prime given B equals one, minus P of A given B, okay? The condition stays the same on both sides. With me? The addition theorem works the same way, too. The cup is union, meaning or. So P of A union C given B equals P of A given B, plus P of C given B, minus P of A intersection C given B, the same formula, with the same condition in every term. Clear? Now a special case. The sideways cup with a line under it means subset, so every outcome of A is also in B, okay? If you are told that B has occurred, does the chance of A go up, or go down? It can only go up, or stay the same, very good! Since A lies inside B, A intersection B is just A, so P of A given B is P of A, over P of B, as shown. And P of B is at most one, so dividing by it cannot make the answer smaller. That is why P of A given B is greater than or equal to P of A, okay?

3:34Example 1: die, even given greater than 3, P(A|B) and P(B|A)

Alright, an easy one first, straight from the course. Throw a die. Event A is getting an even number, and event B is getting a number greater than three. We want two answers, P of A given B and P of B given A, okay? Pause the video and try it first. Done? Okay, let's check. The curly brackets list the outcomes. The even numbers are two, four and six, and the numbers greater than three are four, five and six. In the picture, two is only in A, five is only in B, and four and six sit in both. Right? So A intersection B is four and six, the darker middle part. These are even and also greater than three, okay? Let us find P of A given B first. Once B has happened, what is our new sample space, and how many outcomes does it have? It is B itself, four, five and six, so three outcomes, yes! The six faces are equally likely, so we may count. Of these three, four and six are even. So P of A given B is two over three. Not two over six, because we count only inside B, okay? Now for P of B given A, the roles swap, and A is the new sample space: two, four and six. Of these, four and six are greater than three, so P of B given A is also two over three. But careful, look at the two denominators: n of B in one, n of A in the other. They match here only because A and B both have three outcomes. In general, P of A given B and P of B given A are different, so never swap them!

5:21Example 2: two children, at least one boy, are both boys?

Now, back to our puzzle from the start. Two children in a family, and at least one of them is a boy. Find the probability that both are boys. Pause the video and try it first, and write down the whole sample space. Done? Okay, let's check. Capital B stands for a boy, capital G for a girl, and the first letter is the elder child. So two children give four cases: boy boy, boy girl, girl boy and girl girl. Boy girl and girl boy are different, because the elder child is different. All four are equally likely, okay? Now use the information. At least one is a boy, so girl girl is impossible, and we throw it out. The reduced sample space has three cases: boy boy, boy girl and girl boy. Right? Now, many students argue like this: we know one child is a boy, and the other is a boy or a girl, so the answer is one half. Is that correct? No, no, wait, it is not, and that is the trap! We were never told which child is the boy. Of the three equally likely cases, only boy boy has two boys. The bar again means given, so the probability of both boys, given at least one boy, is one over three, not one half. That settles our puzzle. Clear? Now let us see when one half is right. Compare a similar question from the course: two children, and the elder is a girl. Now we know exactly which child is the girl. The double arrow means that gives. The reduced sample space is girl girl and girl boy, so the probability of both girls, given that the elder is a girl, is one over two. So, what we are told decides the reduced sample space, and that decides the answer. Got it?

7:11Quick revision: reduced sample space, P(A|B) vs P(B|A)

So, quick revision. First, given B, count only inside B, the reduced sample space. Second, P of A given B equals P of A intersection B, over P of B, and only when P of B is greater than zero. And third, P of A given B and P of B given A are different in general, so never swap them. Now go and try the practice problem on this page. All the best!