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

$\frac{2}{5}$

Consider the following events:

A = Number 4 appears at least once,

B = The sum of the numbers appearing is 6.

∴ Required probability

$= P(A/ B)$

= Probability of occurrence of A when B is taken as the sample space

Number of elementary events favourable to

$=\frac{\text{A which are favourable to B}}{\text{Number of elementary events favourable to B}}=\frac{2}{5}$