A fair coin is tossed 3 times. Find the probability of obtaining exactly 2 heads.

Explanation

Let the probability of getting a head = p = \(\frac{1}{2}\) and that of tail = q = \(\frac{1}{2}\)

\((p + q)^{3} = p^{3} + 3p^{2}q + 3pq^{2} + q^{3}\)

In the equation above, \(p^{3}\) and \(q^{3}\) are the probabilities of 3 heads and 3 tails respectively 

 while, \(p^{2}q\) and \(pq^{2}\) are the probabilities of 2 heads and one tail and 2 tails and one head respectively.

Probability of exactly 2 heads = \(3p^{2}q = 3(\frac{1}{2})^{2}(\frac{1}{2})\)

= \(\frac{3}{8}\)