A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

$\frac{2}{5}$

The correct answer is Option (2) → $\frac{2}{5}$ ##

Let $E$ be the event that ‘number 4 appears at least once’ and $F$ be the event that ‘the sum of the numbers appearing is 6’.

Then, $E = \{(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)\}$

and $F = \{(1,5), (2,4), (3,3), (4,2), (5,1)\}$

We have $P(E) = \frac{11}{36}$ and $P(F) = \frac{5}{36}$

Also $E \cap F = \{(2,4), (4,2)\}$

Therefore $P(E \cap F) = \frac{2}{36}$

Hence, the required probability

$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{2}{36}}{\frac{5}{36}} = \frac{2}{5}$