Three heads out of seven coins

Text

Seven fair coins are tossed. What is the probability of getting exactly $3$ heads (and therefore $4$ tails)?

Solution

The number $X$ of heads out of $7$ independent tosses is binomial $X\sim B(7;\,\tfrac12)$: $P(X=k) = \binom{7}{k}\left(\frac12\right)^{k}\left(\frac12\right)^{7-k} = \frac{\binom{7}{k}}{2^{7}}.$ For $k=3$: $P(X=3) = \frac{\binom{7}{3}}{2^{7}} = \frac{35}{128} \approx \boxed{0.273}.$