Two natural numbers are chosen at random from 1 to 100 and are multiplied. If A is the event that the…
Two natural numbers are chosen at random from 1 to 100 and are multiplied. If A is the event that the product is an even number and $B$ is the event that the product is divisible by 4 , then $P(A \cap \bar{B})=$
$\frac{25}{198}$
$\frac{49}{198}$
$\frac{25}{99}$
$\frac{50}{99}$
Solution
Total number of pairs from 1 to $100={ }^{100} \mathrm{C}_2$
$n(\mathrm{~A})=$ Total number of pair whose product is even
$=\underbrace{50}_{\begin{array}{c}
\text { one odds } \\
\text { one even }
\end{array}} \mathrm{C}_1 \times{ }^{50} \mathrm{C}_1+\underbrace{{ }^{50} \mathrm{C}_2}_{\begin{array}{c}
\text { both } \\
\text { even }
\end{array}}$
$n(B)=$ Total number of pair whose product is divisible by 4
$\begin{aligned}
& \underbrace{5^0 \mathrm{C}_2}_{\begin{array}{c}
\text { both are } \\
\text { even }
\end{array}}+\underbrace{n(\mathrm{~A} \cap \mathrm{~B})={ }^{50} \mathrm{C}_2+{ }^{50} \mathrm{C}_1 \times 25}_{\begin{array}{c}
\text { one odd } \& \\
\text { one }\{4,8,12 \ldots 100\} \\
50 \\
\mathrm{C}_1 \times 25
\end{array}}
\end{aligned}$
$\therefore$ Required probability
$=\frac{{ }^{50} \mathrm{C}_1 \times{ }^{50} \mathrm{C}_1+{ }^{50} \mathrm{C}_2-{ }^{50} \mathrm{C}_2-{ }^{50} \mathrm{C}_1+25}{{ }^{100} \mathrm{C}_2}=\frac{25}{99}$