A fair coin is tossed four times. If it is known that at least one tail has appeared, then the probability that the first toss shows a tail is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{8}{15}\)
Four tosses give \(2^4 = 16\) equally likely outcomes. Only HHHH has no tail, so the outcomes with at least one tail are
\(16-1 = 15\)
The outcomes with a tail on the first toss are T followed by any of the \(2^3 = 8\) results of the other three tosses. All 8 of them have at least one tail, so they lie inside the reduced sample space:
\( \displaystyle P = \frac{8}{15} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{2}\) — Incorrect. This is the chance of a tail on the first toss when nothing is known; the condition removes HHHH and changes it.
Option (2) \(\displaystyle\frac{7}{15}\) — Incorrect. This is the probability that the first toss is a head, given at least one tail (H followed by the 7 results with a tail).
Option (3) \(\displaystyle\frac{15}{16}\) — Incorrect. This is the probability of at least one tail, the condition itself.