The Multiplication Theorem

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Take a bag with five red balls and three blue balls. You take out one ball, keep it aside, and then take out another. What is the chance that both are red? The first ball is red with probability five over eight, easy. And the second ball, given that the first was red? Five over eight again, right? Hmm, wait. One red ball is already outside, so the bag has changed! The multiplication theorem handles this situation, and questions on it come up often in JEE, so let's begin.

0:36From conditional probability to multiplication

So, first a quick recall from conditional probability, with event A shaded in the diagram. Capital P with round brackets means the probability of, and the vertical bar is read as given. P of B given A equals P of A intersection B, over P of A, where the upside down cup is intersection, both events happening, the darker part. And this needs P of A to be greater than zero, which is the greater than sign you see on screen. Clear? Now let me ask you this. Suppose we want P of A intersection B, the chance that both events happen. What one step pulls it out of this equation? Multiply both sides by P of A, yes, exactly! P of A intersection B equals P of A, times P of B given A, and this is the multiplication theorem. Read it like a story: first, A must happen, and then B must happen, given that A has already happened. Okay? And the same idea works the other way round, now with B shaded. P of A intersection B also equals P of B, times P of A given B, when P of B is greater than zero. Both orders give the same answer, so pick the one whose pieces are easier to find. With me?

1:53Many events: multiply step by step

Now, what if there are more than two events, one after another? Let us call them A one, A two, and so on. The small numbers at the bottom are subscripts; they just say first, second and third. For two events, it is the same theorem: P of A one intersection A two equals P of A one, times P of A two given A one, okay? Now add a third event. We have P of A one, then P of A two given A one. So tell me, the third factor is P of A three given what? Given both A one and A two, correct! So P of A one intersection A two intersection A three equals P of A one, times P of A two given A one, times P of A three given A one A two. Writing A one and A two side by side after the bar means both of them have happened. Right? And for n events, the chain keeps going. The three dots in the middle and at the end mean, continue in the same pattern, up to the last event, A subscript n. Okay? So here is the whole rule in one sentence: each factor is given everything before it. The first event alone, the second given the first, the third given the first two, and so on.

3:09Draws without replacement

Got it? So where does this theorem shine? In draws without replacement. Without replacement means that a ball, once drawn, stays outside the bag. So after every draw, the bag changes, okay? Now think about the next draw. After one ball comes out, which numbers in our probability fraction change? The total drops by one, yes! And the count of the colour we just drew also drops by one. So each factor is computed from what is left in the bag, not from the original bag. Clear? So every factor is favourable balls left, over total balls left. This works because each ball still in the bag is equally likely to be picked, as classical probability needs. Here is a topper's trick: write one fraction per draw, in order, and the bottoms go down by one each time. Simple, right?

4:02Example 1: both balls red, drawn without replacement

Now, remember our puzzle from the start? Let's finish it. The bag has five red balls and three blue balls. Two are drawn without replacement. Find the probability that both are red. Pause the video and try it first. Done? Okay, let's check. Let R one be the event that the first ball is red, and R two the event that the second ball is red. The small one and two are subscripts again, okay? The first draw is easy. There are five red balls out of eight, so P of R one is five over eight. Now the second draw. Many students write five over eight again, and multiply to get twenty five over sixty four. Is that right? No, that is the trap, because five over eight again would mean the first red ball went back into the bag. Here it stays out. So the bag now has four red balls, and seven balls in all. P of R two given R one is four over seven. With me? Now we multiply the two. Five over eight times four over seven gives twenty over fifty six, which is five over fourteen. The dot between the fractions just means times. So the answer to our puzzle is five over fourteen, not twenty five over sixty four. Mystery solved! Here is a quick check by counting. The tall brackets with five on top and two below mean five C two, the number of ways to choose two red balls from five, and that is ten. Eight C two, the ways to choose any two balls from eight, is twenty eight. Ten over twenty eight is again five over fourteen. Both methods agree, perfect!

5:43Example 2: first white ball on the third draw

Now a harder one, with three draws. An urn has four white balls and six black balls. Balls are drawn one by one, without replacement. Find the probability that the first white ball appears on the third draw. Pause the video and try it first. Done? Okay, let's check. Many students look only at the third draw. But let me ask you this: if the first white ball comes on the third draw, what must the first two draws be? Both black, yes, exactly! Otherwise a white ball would have come earlier. So the event is black on the first draw, B one, black on the second, B two, and white on the third, W three, all in that order. Okay? First draw: there are six black balls out of ten, so P of B one is six over ten. Now one black ball is out. There are five black balls left, and nine balls in all. So P of B two given B one is five over nine, right? Now the third draw. Two black balls are gone, so all four white balls are still there, out of eight. P of W three given B one B two is four over eight. Careful here: the white count did not change, only the total did. Clear? Multiply along the chain. Six times five times four is one hundred and twenty, and ten times nine times eight is seven hundred and twenty. One hundred and twenty over seven hundred and twenty gives one over six. So the answer is one sixth. See how each fraction came from what was left in the urn?

7:24Quick revision: P(A and B) = P(A) P(B given A), chain rule

So, quick revision. First, P of A intersection B equals P of A times P of B given A, when P of A is greater than zero. Second, for many events, multiply along the chain, with each factor given everything before it. And third, without replacement, compute each factor from what is left in the bag. Now go and try the practice problem on this page. All the best!