A die is thrown twice and the sum of the numbers appearing is observed to be 6. Then the probability that the number 4 has appeared at least once, is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{2}{5} \)
The sum is known to be 6, so this is a conditional probability. Only the outcomes with sum 6 are possible.
Let E = the sum is 6 and F = the number 4 appears at least once.
\( \displaystyle E=\{(1,5),\ (2,4),\ (3,3),\ (4,2),\ (5,1)\} \)
\( \displaystyle n(E)=5 \)
\( \displaystyle E\cap F=\{(2,4),\ (4,2)\} \)
\( \displaystyle n(E\cap F)=2 \)
\( \displaystyle P(F|E)=\frac{n(E\cap F)}{n(E)} \)
\( \displaystyle =\frac{2}{5} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{4}{5} \) — Incorrect. Only 2 of the 5 outcomes with sum 6 contain a 4, not 4 of them.
Option (2) \( \displaystyle\frac{1}{5} \) — Incorrect. (2, 4) and (4, 2) are different outcomes (first throw 2, second 4, and the reverse), so both must be counted.
Option (3) \( \displaystyle\frac{3}{5} \) — Incorrect. This is the probability that 4 does not appear: (1, 5), (3, 3) and (5, 1).
NCERT (2026-27 edition):
Conditional Probability
