# Probability of getting 2 head and 2 tail

If a fair coin is tossed 4 times what is the probability that two heads and two tails will result ?

My calculation is. no. of ways of getting exactly 2 head and 2 tails .will be $6$ out of $8$. Eg $$HHTT,THHT,TTHH,HTTH,HTHT,THTH,HHHT,TTTH$$

• You mean, 6 out of 16, right?
– Did
Jul 3, 2016 at 14:53
• There are $16$ equally probable possible throws. Of these $\binom 42=6$ will have exactly two $H's$.
– lulu
Jul 3, 2016 at 14:54
• ... don't forget $HHTH, HTHH, THHH, TTHT, THTT, HTTT, HHHH, TTTT$ Jul 3, 2016 at 15:01
• $$HHHH,HHHT,HHTH,HTHH,THHH , TTTT , TTTH , TTHT ,THTT,HTTT,\color{red}{HHTT,HTHT,HTTH,TTHH,THTH,THHT}$$ Jul 3, 2016 at 15:02

Let the random variable $X$ be the number of heads that come up when a fair coin is tossed $4$ times. Then $X \sim B\left(4, \frac{1}{2}\right)$.
We want there to be exactly two heads (forcing the other two tosses to be tails), so $$\mathbb{P}(X = 2) = \binom{4}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^2 = \frac{3}{8}.$$
There are $2^4=16$ possible outcomes:
$HHHH$ $\ \$ $HHHT$ $\ \$ $HHTH$ $\ \$ $HTHH$ $\ \$ $THHH$ $\ \$ $\color{red}{HHTT}$ $\ \$ $\color{red}{HTHT}$ $\ \$ $\color{red}{THHT}$ $\ \$ $\color{red}{HTTH}$ $\ \$ $\color{red}{THHT}$ $\ \$ $\color{red}{TTHH}$ $\ \$ $HTTT$ $\ \$ $THTT$ $\ \$ $TTHT$ $\ \$ $TTTH$ $\ \$ $TTTT$ $\ \$
$\frac {\text{No. of favorable outcomes}}{\text{No. of possible outcomes}}=\frac{6}{16}=\frac38$