Question

A die is thrown. Let 'A' be the event that the number obtained is greater than $3$. Let 'B' be the event that the number obtained is less than $5$. Then, $P(A\cup B)$ is

• A. $1$
• B. $\frac{2}{5}$
• C. $\frac{3}{5}$
• D. $0$

Answer 1

The correct option is A $1$

A die is thrown.

Therefore $S=\{1,2,3,4,5,6\}$

$\Rightarrow n\left(S\right)=6$.

Let $A$ be the event that the number obtained is greater than $3$.

Therefore $A=\{4,5,6\}$

$\Rightarrow n\left(A\right)=3$.

$\Rightarrow P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{3}{6}=\frac{1}{2}$

Let $B$ be the event that the number obtained is less than $5$.

Therefore $B=\{1,2,3,4\}$

$\Rightarrow n\left(B\right)=4$.

$\Rightarrow P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{4}{6}=\frac{2}{3}$

$A\cap B=\left\{4\right\}$

$\Rightarrow n(A\cap B)=1$

$\Rightarrow P(A\cap B)=\frac{n(A\cap B)}{n\left(S\right)}=\frac{1}{6}$

We have to find $P(A\cup B)$.

We know that $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

$=\frac{1}{2}+\frac{2}{3}-\frac{1}{6}$

$=\frac{3+4-1}{6}$

$=\frac{6}{6}$

$=1$

$\therefore P(A\cup B)=1$