# Probability of 9 tails

An experiment is performed to flip a fair coin 10 times and observe the outcome of each flip: heads (labeled 'H') or tails (labeled 'T'). For instance, one outcome, written as a 10-tuple, might be (H,T,T,T,T,T,T,H,H,H).

How many total outcomes are there for this experiment? Explain your reasoning.

There are 1024 different outcomes. You are flipping a (single, 1) fair coin which has 2 sides (heads and tails) so the Sample Space (N(S)) = 1024, 10 consecutive flips, with 2 sides = 2^10=1024 - is this part correct?

How many ways can the result of the experiment show exactly nine tails? Explain your reasoning.

I was thinking that there are 113 different possibilities for have 9 tails? I did 1024/9? Can you help me?

• First part is correct. Why are you dividing 1024 by 9? Maybe you could approach problem by listing down some 10-tuples with 9 tails; see a particular pattern in it? – Mihir Nov 22 '15 at 19:53
• I dare you to show even 13 different ways to get 9 tails .... – Henning Makholm Nov 22 '15 at 19:54
• Would a matrix work? or just a standard table? – Joshua Mathews Nov 22 '15 at 20:05
• You don't need anything as complicated as a matrix. Just start listing the possible outcomes with nine tails. For instance: (H,T,T,T,T,T,T,T,T,T). – Théophile Nov 22 '15 at 20:08
• So there is only 10 possible ways. but how do I show this mathematically? is there an equation? – Joshua Mathews Nov 22 '15 at 20:10

## 1 Answer

Try thinking of it this way: Each head is a "success" and each tail is a "failure." You have ten slots to fill, one for each flip, each of which is either a success or failure. How many ways can you fill ten slots such that exactly one of them is a success?

Hint: have you studied binomial coefficients?