Question

A coin is tossed $3$ times. If $E$ denotes the event in which heads appear at least twice and $F$ denotes the event in which head comes first, then $P(E|F)=$

• A. $\frac{3}{4}$
• B. $\frac{3}{8}$
• C. $\frac{1}{2}$
• D. $\frac{1}{8}$

Answer 1

The correct option is A $\frac{3}{4}$

There are $4$ cases which start with a Head:
(i) $HTH$
(ii)$HHT$
(iii)$HHH$
(iv) $HTT$
Probability of each case = $\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}$
Three of the above four cases have at least two heads.
Hence, required probability = $\frac{3\times \frac{1}{8}}{4\times \frac{1}{8}}=\frac{3}{4}$