A die is thrown three times, if the first throw is a four, then the probability of getting 15 as the sum is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{1}{18} \)
The first throw is known to be 4, so this is a conditional probability. Only the second and third throws are still uncertain, and they have 6 × 6 = 36 equally likely outcomes.
For a sum of 15, the last two throws must add up to 15 − 4 = 11:
\( \displaystyle (5,6),\ (6,5) \)
\( \displaystyle P(\text{sum}=15\,|\,\text{first throw}=4)=\frac{2}{36}=\frac{1}{18} \)
The same result from the formula \( P(E|F)=\frac{P(E\cap F)}{P(F)} \):
\( \displaystyle \frac{P(E\cap F)}{P(F)}=\frac{\frac{2}{216}}{\frac{1}{6}} \)
\( \displaystyle =\frac{2}{36}=\frac{1}{18} \)
Why the other options are incorrect
Option (2) \( \displaystyle\frac{1}{9} \) — Incorrect. \( \frac{1}{9} \) would need 4 favourable outcomes; only (5, 6) and (6, 5) give 11.
Option (3) \( \displaystyle\frac{1}{36} \) — Incorrect. \( \frac{1}{36} \) counts only one of (5, 6) and (6, 5); they are different outcomes.
Option (4) \( \displaystyle\frac{1}{12} \) — Incorrect. \( \frac{1}{12} \) would need 3 favourable outcomes; there are only 2.