Question

A dice 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?

Answer 1

Consider the given events.
A = 4 appears on the die
B = The sum of the numbers on two dice is 6.

Clearly,
A = {(1, 4) (2, 4), (3, 4),(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 4), (6, 4)}
B = {(1, 5), (5, 1), (2, 4), (4, 2),(3, 3)}

$$\begin{gathered} \mathrm{Now}, \\ A\cap B=\left\{\left(2,4\right)\mathrm{and}\left(4,2\right)\right\} \\ \therefore \mathrm{Required}\mathrm{probability}=P\left(A/B\right)=\frac{n\left(A\cap B\right)}{n\left(B\right)}=\frac{2}{5} \end{gathered}$$