Question

A fair coin is flipped twelve times. What is the probability of the coin landing tails up exactly nine times?

Answer 1

Answer:

$P\left(E\right)=\frac{55}{1024}$

Step-by-step explanation:

Given that a fair coin is flipped twelve times.

It means the number of possible sequences of heads and tails would be:

2¹² = 4096

We can determine the number of ways that such a sequence could contain exactly 9 tails is the number of ways of choosing 9 out of 12, using the formula

$nCr=\frac{n!}{r!\left(n-r\right)!}$

Plug in n = 12 and r = 9

       $=\frac{12!}{9!\left(12-9\right)!}$

       $=\frac{12!}{9!\cdot \:3!}$

       $=\frac{12\cdot \:11\cdot \:10}{3!}$            ∵ $\frac{12!}{9!}=12\cdot \:11\cdot \:10$

       $=\frac{1320}{6}$                   ∵ $3!\:=\:3\times 2\times 1=6$

       $=220$

Thus, the probability will be:

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

         $=\frac{220}{4096}$

         $=\frac{55}{1024}$

Thus, the probability of the coin landing tails up exactly nine times will be:

$P\left(E\right)=\frac{55}{1024}$