A coin is tossed three times. If the first toss shows a head, then the probability of getting exactly two heads in the three tosses is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{2}\)
Since the first toss is known to be a head, only the outcomes starting with H are possible. This reduced sample space is
HHH, HHT, HTH, HTT
Exactly two heads appear in HHT and HTH, which is 2 of these 4 outcomes.
\( \displaystyle P(\text{exactly two heads} \mid \text{first is H}) = \frac{2}{4} \)
\( \displaystyle = \frac{1}{2} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{3}{4}\) — Incorrect. This counts HHH too, which has three heads, not exactly two.
Option (2) \(\displaystyle\frac{3}{8}\) — Incorrect. This is the probability of exactly two heads without the given condition (HHT, HTH, THH out of 8).
Option (3) \(\displaystyle\frac{1}{4}\) — Incorrect. This is \(\frac{2}{8}\), the probability that both happen together, not divided by \(P(\text{first is H}) = \frac{1}{2}\).