Probability of coin heads

Question

Four fair coins are flipped. If the outcomes are assumed independent, what is the probability that two heads and two tails are obtained?

I calculated the answer as $\frac{11}{16}$ via using brute-force computations; is there any logical approach for this?

Answer 1

The correct answer is $\frac6{16}$. In general, for $n$ fair coins and obtaining $k$ heads and $n-k$ tails the probability is $\binom nk/2^n$.

Answer 2

The number $X$ of Heads in four independent tosses of fair coins has $X \sim \mathsf{Binom}(n=4, p=1/2).$ You seek $P(X = 2) = {4 \choose 2}(1/2)^4 = 6/16 = 0.375 \ne 11/16.$

In R statistical software, where dbinom is a binomial PDF:

dbinom(2, 4, .5)
[1] 0.375

By 'brute force' listing:

TTTT
TTTH, TTHT, THTT, HTTT
TTHH, THTH, HTTH, HHTT, HTHT, THHT
HHHT, HHTH, HTHH, THHH
HHHH

You want the third row.

Answer 3

You would have $$P=\frac{\mbox{good cases}}{\mbox{total cases}}=\frac{\frac{4!}{2!2!}}{\sum_{i=0}^4\frac{4!}{i!(4-i)!}}=\frac{6}{16}$$