A pair of unbiased dice is thrown together and sum of the numbers appearing is observed. The probability that the sum of numbers on the dice was 10, given that the observed sum was at least 9, is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{3}{10} \)
There are 36 equally likely outcomes. Let E = sum is 10 and F = sum is at least 9.
\( \displaystyle \text{Sum } 9:\ (3,6),(4,5),(5,4),(6,3)\ \to 4 \)
\( \displaystyle \text{Sum } 10:\ (4,6),(5,5),(6,4)\ \to 3 \)
\( \displaystyle \text{Sum } 11:\ (5,6),(6,5)\ \to 2 \)
\( \displaystyle \text{Sum } 12:\ (6,6)\ \to 1 \)
So n(F) = 4 + 3 + 2 + 1 = 10, and E ∩ F = E has 3 outcomes. By conditional probability:
\( \displaystyle P(E\,|\,F)=\frac{n(E\cap F)}{n(F)} \)
\( \displaystyle =\frac{3}{10} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{5}{18} \) — Incorrect. \( \frac{5}{18}=\frac{10}{36} \) is P(F), the probability that the sum is at least 9, not the conditional probability.
Option (3) \( \displaystyle\frac{1}{12} \) — Incorrect. \( \frac{1}{12}=\frac{3}{36} \) is P(E) without the condition; given F, the sample space shrinks to 10 outcomes.
Option (4) 1 — Incorrect. A sum of at least 9 does not force the sum to be 10; it could be 9, 11 or 12.