Question

Two digit number is selected randomly from the digit 1 to 9. If sum of digit is even, the probability that both the numbers are odd is

• A. $\frac{4}{9}$
• B. $\frac{5}{9}$
• C. $\frac{3}{4}$
• D. None of these

Answer 1

Answer: (D)

None of these

Sum is even provided both are even or both are odd. Let A denotes even, B denotes odd.

$n\left(S\right)=n\left(AorB\right)=n\left(A\right)+n\left(B\right).=n\left(S\right)=n\left(bothodd\right)+n\left(botheven\right)$ ${=}^5{C}_2{+}^4{C}_2=16$

$\therefore$ Required probability $=\frac{n\left(bothodd\right)}{n\left(S\right)}$ $=\frac{10}{16}=\frac{5}{8}$