Question

A pair of dice is thrown. Find the probability of getting the sum 8 or more, if 4 appears on the first die.

Answer 1

Consider the given events.
A = 4 appears on first die
B = The sum of the numbers on two dice is 8 or more.

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

B = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6),(5, 3), (5, 4), (5, 5) (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(B) = 15

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