A die is thrown three times. If the first throw is a five, the probability of getting 14 as the sum is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{9}\)
Since the first throw is a five, the sum is 14 exactly when the last two throws add up to 9:
\(14 - 5 = 9\)
The last two throws have 6 × 6 = 36 equally likely ordered outcomes. Those with sum 9 are (3, 6), (4, 5), (5, 4) and (6, 3), that is, 4 outcomes.
\(\displaystyle P = \frac{4}{36} = \frac{1}{9}\)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{1}{18}\) — Incorrect. This counts only the 2 unordered pairs {3, 6} and {4, 5}, giving 2/36. The order of the throws matters: (3, 6) and (6, 3) are different outcomes.
Option (3) \(\displaystyle\frac{1}{36}\) — Incorrect. This would need only 1 favourable outcome out of 36; there are 4.
Option (4) \(\displaystyle\frac{1}{54}\) — Incorrect. This is \(\frac{1}{6} \times \frac{4}{36} = \frac{4}{216}\), the probability that the first throw is a five and the sum is 14. The first throw is given as a five, so the question asks for a conditional probability.