Question

A pair of dice is thrown. Find the probability of getting 7 as the sum, if it is known that the second die always exhibits an odd number.

Answer 1

Consider the given events.
A = Number appearing on second die is odd
B = The sum of the numbers on two dice is 7.

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

$$\begin{gathered} \mathrm{Now}, \\ A\cap B=\left\{\left(2,5\right),\left(4,3\right),\left(6,1\right)\right\} \\ \therefore \mathrm{Required}\mathrm{probability}=P\left(B/A\right)=\frac{n\left(A\cap B\right)}{n\left(A\right)}=\frac{3}{18}=\frac{1}{6} \end{gathered}$$