Independent Events and Repeated Trials

Transcript

0:00Introduction

Hello everyone! Let us start with a small game. Two friends, A and B, toss a fair coin in turns, A first, and the first head wins. Is this game fair, fifty fifty for each friend? But player A always gets the first chance. Does that matter? Let's find out, with independent events and repeated trials, which come up often in JEE papers.

0:26Independent events: just multiply

So, what does independent mean? Two events are independent when one happening does not change the chance of the other. The dark middle is A intersection B, and the cap-shaped symbol means intersection. So, for independent events, what is its probability? Yes, exactly, we just multiply! P of A intersection B equals P of A times P of B, and that is the definition of independence. Capital P with round brackets means the probability of the event inside. Clear? Equivalently, P of A given B equals P of A, where the vertical bar is read as given. This needs P of B not equal to zero, the equals sign with a slash. Okay? And here is a bonus. The little dash is called prime, and A prime, the complement, means A does not happen. If A and B are independent, so are A prime and B, A and B prime, the shaded part, and A prime and B prime. Got it?

1:25The trap: mutually exclusive is not independent

Now here is a classic trap. These two circles do not overlap, so the events never happen together, and we call them mutually exclusive. They look completely separate, so they must be independent, right? No, no, careful, that is the trap! Mutually exclusive means A intersection B is the empty set, this circle with a slash, so P of A intersection B is zero. With me? But if P of A and P of B are both non-zero, their product is positive, and this sign means greater than. So the product rule fails, and such events are never independent.

2:03Many events: mutual independence and at least one

Clear? Now, more than two events: A one, A two, up to A n. The small numbers below are subscripts, and the dots mean and so on. They are mutually independent when the product rule holds for every subcollection: every pair, every three, and all of them together. Okay? If the rule holds only for the pairs, they are pairwise independent, and that is weaker. The pairs can pass while all three together fail. Right? Now the most useful formula. For independent events, we want the chance that at least one happens. Simply adding would count the overlaps more than once, so let me ask you this: what is the opposite of at least one? Yes, exactly, none of them! The misses are independent too, so P of none is the product of one minus P of A sub i, and capital pi, the Greek letter, means product. So P of at least one is one minus that product. Got it?

3:01Repeated trials: waiting for the first success

Now let us look at repeated trials. There are two conditions: the trials are independent, and each succeeds with the same probability, small p. Small q, one minus p, is the chance of failure. With me? So tell me, if the first success comes exactly on trial n, what must happen before it? Correct, every earlier trial must fail! That is n minus one failures, then one success. So P is q to the power n minus one, times p, and the raised part is the power. Okay? And no success in n trials means every trial fails, so P is q to the power n. Clear?

3:43Taking turns: who wins?

Now, back to our coin game. Two players take turns, the first to succeed wins, and each turn succeeds with probability small p. So, on which turns can the starter win? Yes, on turns one, three, five, and so on! Turn one gives p. Turn three needs two failures first, so q squared times p. Then q to the power four, times p, and so on. Right? This is a geometric series with ratio q squared, less than one, so its sum is p over one minus q squared. And one minus q squared is one minus q, times one plus q, where one minus q is p. Cancel the p, and the starter wins with probability one over one plus q. Beautiful, isn't it? Now let us answer our puzzle. For a fair coin, p and q are both one half, so P is one over one plus one half, which is two thirds. So the game is not fair! The first player wins two thirds of the time. Got it?

4:48Example 1: three shooters, target hit and exactly one hit

Let's start with the easier example. Three people, A, B and C, hit a target independently with probabilities point six, point five and point four. Find the probability that the target is hit, and that exactly one of them hits it. Pause the video, and try it first. Done? Okay, let's check. The target is hit when at least one hits it. So just add: point six plus point five plus point four gives one point five, right? Oh no, careful, a probability can never be more than one! Use the opposite, everyone misses. Misses are independent, so multiply point four, point five and point six, and the cross means times. That gives point one two. So the target is hit with probability one minus point one two, which is point eight eight. With me? Now let us find exactly one. Only A means A hits while B and C miss, so point six times point five times point six, which gives point one eight. Okay? Similarly, only B is point four times point five times point six, which is point one two. And only C is point four times point five times point four, which is point zero eight. Right? These cases never happen together, so we add them. Point one eight, plus point one two, plus point zero eight, gives point three eight. Clear?

6:19Example 2: A, B, C throw a die in turn, chance C wins

Now let us try a harder one. Three players, A, B and C, throw a die in that order, and the first to throw a six wins. Find the probability that C wins. Here is a hint: think in rounds. Pause the video, and try it first. Done? Okay, let's check. Three players, so each one has a one third chance, right? No, no, wait, remember the coin game! Going earlier is an advantage, so the order matters. Success is a six, so p is one sixth, and q is five sixths. Clear? In the first round, C wins if player A fails, player B fails, and then C throws a six. So five sixths squared, times one sixth, gives twenty five over two hundred and sixteen, and the dot means times. Right? Now, the chance that nobody throws a six in a round is five sixths cubed, which is one hundred and twenty five over two hundred and sixteen. Then the game starts again, since the throws are independent. With me? So, P of C wins is twenty five over two hundred and sixteen, all over one minus one hundred and twenty five over two hundred and sixteen, and the slashes mean over. The bottom is ninety one over two hundred and sixteen, so the answer is twenty five over ninety one, less than one third, since C throws last. Got it?

7:51Quick revision: product rule, at least one, taking turns

So, quick revision. Independent means P of A intersection B equals P of A times P of B, and mutually exclusive events with non-zero probabilities are never independent. At least one is one minus the product of the misses, and the first success on trial n has probability q to the power n minus one, times p. With two players taking turns, the starter wins with probability one over one plus q. Now try the practice problem on this page. All the best!