Question

Two integers are selected at random from integers $1$ to $9$. If the sum is even, then the probability that both the numbers are odd is

• A. $\frac{5}{8}$
• B. $\frac{6}{7}$
• C. $\frac{5}{9}$
• D. $\frac{1}{3}$

Answer 1

The correct option is A $\frac{5}{8}$

Total odd number=$5$, Total even number=$4$
Sum is even if both numbers are odd or even.
$A=$ Both numbers are odd
$B=$ Sum is even
$P\left(A\right)=\frac{{}^5{C}_2}{{}^9{C}_2},P\left(B\right)=\frac{{}^5{C}_2{+}^4{C}_2}{{}^9{C}_2},P(A\cap B)=\frac{{}^5{C}_2}{{}^9{C}_2}$
So, required probability $=P\left(A/B\right)=\frac{P(A\cap B)}{P\left(B\right)}=\frac{10}{16}=\frac{5}{8}$