What is the probability of seeing $7$ heads in the toss of $12$ coins?

Question

Problem: What is the probability of seeing $7$ heads in the toss of $12$ coins?

My answer: If $C(n,k)$ is the number of combinations of $k$ things taken $n$ at a time, then, since order does not matter, the number of successful outcomes is

$$ C(12,7) $$

while the number of total outcomes is $2^{12}$. Hence the probability is the quotient

$$ \frac{C(12,7)}{2^{12}}. $$

Book answer: $\displaystyle \frac{C(12,5)}{2^{12}}.$

Question: Typo?

Book: "Methods of Mathematics" by Hamming

Answer 1

It's not a typo. Your answer and the book answer evaluate to the same number, $\frac{99}{512} = 0.193359375$. This follows from the identity ${n \choose k} = {n \choose {n-k}}$.

This can be easily seen by rephrasing the question as “What is the probability of seeing 5 tails in the toss of 12 coins?”

Answer 2

You can break it up into more fundamental concepts as well. There are $2^{12}=4096$ possible flip outcomes. The flip outcomes of interest have the form TTTTTHHHHHHH.

There are $12!$ ways of arranging 12 distinct objects. If they aren't distinct, you need to divide by the number of ways to rearrange the identical items. Since there are 7 heads and 5 tails the result is $\frac{12!}{5!7!}= C(12,7)=C(12,5)$

So $C(12,7)/2^{12}$ is the answer, but the formula follows form more basic arguments.