Probably Not, 2nd Edition

6The Binomial Theorem

Introduction

Suppose you have ten coins that are numbered 1, 2, …, 10 and you want to know the probability of flipping a head on the first coin, a tail on the second coin, a …, etc. This is not a difficult problem. Actually, this isn't a problem, it's ten independent problems lumped together in one sentence. The probability of a given result with each coin is .5, so the probability of all ten results is simply .

But, what if you're interested in a slightly different problem, the problem of getting exactly 6 heads and 4 tails out of 10 flips? You don't care which of the flips gives you heads and which gives you tails so long as when you're done, there are 6 heads and 4 tails. Remember that this is the same as saying you toss ten coins into the air at once to get six heads (and four tails).

One way of solving this problem is to write down every possible result of flipping ten coins and then to count the number of ways that you could get 6 heads and 4 tails. The probability would just be this number of ways multiplied by the probability of getting any one of them, , or equivalently this number of ways divided by 1024. This was the procedure in Chapter 1 when we wanted to see how many ways we could get, say, a total of 6 when rolling a pair of dice (Table 1.2 ...