Question

If four coins are tossed$,$ find the chance that there should be two heads and two tails.

Answer 1

Coin is tossed four times

So total number of outcomes $=2^4=16$

Number of ways of getting two heads ${=}^4{C}_2=\frac{4!}{(4-2)!2!}=\frac{4!}{2!2!}=6$

Now number of ways of getting two tails ${=}^2{C}_2=1$

$\therefore$ required probability $=\frac{6}{16}=\frac{3}{8}$