Two numbers are selected random from integers 1 through $9 .$ If the sum if even, what is the probability that both numbers are odd?A. $\frac{1}{6}$B. $\frac{2}{3}$C. $\frac{4}{9}$D. $\frac{5}{8}$

Home » Two numbers are selected random from integers 1 through If the sum if even, what is the probability that both numbers are odd?A. B. C. D.

Two numbers are selected random from integers 1 through $9 .$ If the sum if even, what is the probability that both numbers are odd?
A. $\frac{1}{6}$
B. $\frac{2}{3}$
C. $\frac{4}{9}$
D. $\frac{5}{8}$

Two numbers are selected random from integers 1 through $9 .$ If the sum if even, what is the probability that both numbers are odd?
A. $\frac{1}{6}$
B. $\frac{2}{3}$
C. $\frac{4}{9}$
D. $\frac{5}{8}$

The sum will be even when; both numbers are either even or odd,
i.e. for both numbers to be even, the total cases ${ }^{5} \mathrm{C}_{1} \mathrm{X}^{4} \mathrm{C}_{1}$ (Both the numbers are odd) $+{ }^{4} \mathrm{C}_{1} \mathrm{x}^{3} \mathrm{C}_{1}$ (Both the numbers are even) $=32$
The favourable number of cases will be,
Both odd, i.e. selecting numbers from $1,3,5,7$ , or 9, i.e.
${ }^{5} \mathrm{C} 1 \mathrm{X}^{4} \mathrm{C}_{1}=20$
Thus, the probability that both numbers are odd will be =
$\begin{array}{l} =\frac{\text { Favorable outcomes }}{\text { Total outcomes }} \\ \Rightarrow \frac{20}{32}=\frac{5}{8} \end{array}$
Hence, the correct option is d.