Two digits are selected at random from the digits 1 through 9. If their sum is even, then the probability…
Two digits are selected at random from the digits 1 through 9. If their sum is even, then the probability that both are odd is
$\frac{3}{8}$
$\frac{1}{2}$
$\frac{5}{8}$
$\frac{3}{4}$
Solution
Let $A=$ Getting two odd numbers
$B=$ Getting the sum as an even number.
$\therefore$ Required probability $P(A / B)=\frac{P(A \cap B)}{P(B)}$
$\Rightarrow P(A / B)=\frac{\frac{{ }^5 C_2}{{ }^9 C_2}}{\frac{{ }^4 C_2+{ }^5 C_2}{{ }^9 C_2}}=\frac{5}{8}$