A fair coin is tossed three times in succession. If the first toss produces a head, then find the probability of getting exactly two heads in three tosses.

A fair coin is tossed three times in succession. If the first toss produces a head, then the probability of getting exactly two heads in three tosses is 

(A) $\dfrac{1}{8}$ 

(B) $\dfrac{1}{2}$ 

(C) $\dfrac{3}{8}$ 

(D) $\dfrac{3}{4}$ 

(GATE 2005) 

Answer: (B) $\dfrac{1}{2}$ 

Explanation: 

Here, the first produces a head. 

Hence, the sample space, $S = \{HHH, HHT, HTH, HTT\}$. 

So, the number of sample points, $n(S) = 4$. 

Let, event $E = \{HHT, HTH\}$. 

So, the number of event points, $n(E) = 2$. 

Therefore, the required probability is 

$P(E) = \dfrac{n(E)}{n(S)} = \dfrac{2}{4} = \dfrac{1}{2}$