The Addition Theorem

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Pick a number from one to one hundred at random. What is the probability that it is divisible by two, three or five? Fifty numbers are divisible by two, thirty three by three, and twenty by five. Fifty plus thirty three plus twenty gives one hundred and three. So the probability is one hundred and three over one hundred, more than one? That cannot be right, can it? The addition theorem shows exactly what went wrong, and it comes up often in JEE, so let's begin.

0:34Two events: add, then subtract the overlap

So, here are two events, A and B, drawn as two circles. The event, A or B, means at least one of them happens, and that is everything shaded here, okay? Easy, isn't it? Just add the probability of A and the probability of B, right? Hmm, wait. Look at the darker middle part, where both happen. If we simply add, what happens to that part? Yes, exactly, it gets counted twice: once as part of A, and once again as part of B, which is a double count! So we subtract it once. P of A union B equals P of A, plus P of B, minus P of A intersection B, okay? Capital P with round brackets means the probability of. The cup is union, meaning or, and the upside down cup is intersection, meaning and. This is the addition theorem, and it works for any two events. Clear? Now, what if A and B are mutually exclusive, so they never happen together? Then the circles do not overlap, and P of A intersection B is zero, right? So there is nothing to subtract, and P of A union B is just P of A plus P of B, nothing more. But careful, this short form is only for mutually exclusive events. Got it?

1:48Exactly one of the two events

Next, a common twist. Exactly one of A and B means A or B, but not both. That is the two darker outer parts. With me? Now, in P of A plus P of B, the overlap is counted twice. So tell me, for exactly one, how many times should it count? It should count zero times, very good! So we subtract P of A intersection B two times. P of exactly one equals P of A, plus P of B, minus two times P of A intersection B, okay? And here is the trap. Subtract the overlap only once, and you get at least one, which is P of A union B, not exactly one. Don't fall for it, got it?

2:27Three events: add, subtract, add

Okay, now three events, A, B and C, one level up. This big Greek letter is sigma, and it means sum. So sigma P of A is P of A, plus P of B, plus P of C, the three singles, right? And sigma P of A intersection B is the sum over the three pairs: A and B, B and C, and C and A, clear? Now think about the outcomes where all three events happen together, the part common to all three events. In sigma P of A, how many times is that part counted? It is counted three times, absolutely, once in each single! Subtracting the three pairs removes it three times, so now it is counted zero times, and we add it back once. P of A union B union C equals sigma P of A, minus sigma P of A intersection B, plus P of A intersection B intersection C, okay? Add the singles, subtract the pairs, add the triple. Forgetting that last plus term is the most common slip. And for exactly one of the three, take sigma P of A, minus two times sigma P of A intersection B, plus three times the triple. To remember it, the numbers go one, two, three, and the signs go plus, minus, plus. With me?

3:41Bounds: how big, how small?

Now let us look at bounds. Max with curly brackets means the larger of the two, min means the smaller, and this sign is less than or equal to. The union contains both events, so P of A union B is at least the larger probability. And it is at most one, and at most P of A plus P of B, because we only ever subtract the overlap. Right? Now let me ask you about the overlap: how small can P of A intersection B be? Remember, no probability goes above one. Use the addition theorem. P of A, plus P of B, minus the overlap, is P of A union B, and that is at most one. So the overlap is at least P of A, plus P of B, minus one. And of course it can never go below zero either. And the overlap sits inside both events, so it is at most the smaller probability. Clear?

4:35Example 1: P(A or B) = 0.7 and P(exactly one) = 0.5

Alright, an easy one first. P of A is zero point five, P of B is zero point four, and P of A intersection B is zero point two. Find P of A union B, and the probability of exactly one. Pause the video and try it first. Done? Okay, let's check. The zero point two in the middle of the picture is the overlap. Zero point five plus zero point four is zero point nine, and taking away zero point two gives zero point seven. Got it? Now exactly one, the two darker parts. Do we subtract zero point two once, or twice? We subtract it twice, because exactly one leaves out the middle completely. So it is zero point five plus zero point four, minus two times zero point two, right? That is zero point nine minus zero point four, which gives zero point five. On the picture, only A is zero point three, only B is zero point two, and together they give zero point five, perfect!

5:37Example 2: 1 to 100 divisible by 2, 3 or 5, answer 0.74

Now, let's settle our puzzle from the start. We choose a number from one, two, and so on up to one hundred. The three dots mean the numbers carry on. Find the probability that it is divisible by two, three or five. Pause the video and try it first. Done? Okay, let's check. Let A, B and C be the events, divisible by two, three and five, and n of A means how many numbers are in A, okay? Fifty numbers are even. Three times thirty three is ninety nine, so thirty three are multiples of three, and twenty are multiples of five. Now for the pairs: if a number is divisible by both two and three, what must it be divisible by? By six, yes, exactly! The divisor is shown in brackets. Six times sixteen is ninety six, so sixteen numbers. Two and five means ten, which gives ten numbers. Three and five means fifteen, and fifteen times six is ninety, so six numbers, right? All three together means divisible by thirty. Thirty times three is ninety, so three numbers. Clear? Now add the singles, subtract the pairs, add the triple. Fifty plus thirty three plus twenty gives one hundred and three, our puzzle number! The pairs add to thirty two, and one hundred and three minus thirty two leaves seventy one. Adding back the three numbers divisible by thirty gives seventy four. If you stopped at seventy one, you forgot the triple, okay? Every number is equally likely, so we divide by one hundred. The probability is seventy four over one hundred, which is zero point seven four. So our puzzle is solved: the one hundred and three had counted the overlaps more than once.

7:21Quick revision: two events, exactly one, three events

So, quick revision. For two events, P of A union B equals P of A, plus P of B, minus P of A intersection B, right? For exactly one, subtract the overlap twice. And for three events, add the singles, subtract the pairs, and add the triple. Now go and try the practice problem on this page. All the best!