Two integers are drawn at random from the set $\{5,6, . ., \ldots 35\}$. What is the probability that their…
Two integers are drawn at random from the set $\{5,6, . ., \ldots 35\}$. What is the probability that their difference is odd?
$\frac{15}{62}$
$\frac{8}{31}$
$\frac{15}{31}$
$\frac{16}{31}$
Solution
Two integers are drawn from $\{5,6, \ldots \ldots, 35\}$
Given, integer are 31
Total ways of choosing 2 integer from 31 integers
$
={ }^{31} C_2
$
Two integer difference will be odd only when one is even and other is odd.
Let $A$ be the event of selecting two integers whose difference is odd.
Now, $P(A)+P(\bar{A})=1$
$P(\bar{A})=$ Probability of selecting two integers whose difference is not odd.
Case I When two integers are even.
Total ways of choosing 2 integer (even) from 15 even integers $={ }^{15} C_2$
Case II When two integers are odd Total ways $={ }^{16} C_2$
$
\begin{aligned}
P(\bar{A}) & =\frac{{ }^{15} C_2+{ }^{16} C_2}{{ }^{31} C_2}=\frac{\frac{15 \times 14}{2}+\frac{16 \times 15}{2}}{\frac{31 \times 30}{2}} \\
P \overline{(A}) & =\frac{(15 \times 14)+(16 \times 15)}{31 \times 30} \\
& =\frac{14+16}{31 \times 2}=\frac{30}{2 \times 31}=\frac{15}{31} \\
\therefore \quad P(A) & =1-P(\bar{A})=1-\frac{15}{31}=\frac{31-15}{31}=\frac{16}{31}
\end{aligned}
$