Two brothers $X$ and $Y$ appeared for an exam. Let $A$ be the event that $X$ has passed the exam and $B$ is…
Two brothers $X$ and $Y$ appeared for an exam. Let $A$ be the event that $X$ has passed the exam and $B$ is the event that $Y$ has passed. The probability of $A$ is $\frac{1}{7}$ and of $B$ is $\frac{2}{9}$. Then, the probability that both of them pass the exam is
$\frac{1}{63}$
$\frac{2}{35}$
$\frac{2}{63}$
$\frac{9}{14}$
Solution
Given, $A$ be the event that $x$ is selected and $B$ is the event that $y$ is selected.
$P(A)=\frac{1}{7}, P(B)=\frac{2}{9}$
Let $C$ be the event that both are selected.
$P(C)=P(A) \times P(B)$ as $A$ and $B$ are independent events
$=\frac{1}{7} \times\left(\frac{2}{9}\right)=\frac{2}{63}$