Question

A bag contains $4$ white, $3$ red and $4$ green balls. A ball is drawn at random, then the probability that the ball drawn is white or green is :

• A. $\frac{6}{11}$
• B. $\frac{7}{11}$
• C. $\frac{8}{11}$
• D. $\frac{9}{11}$

Answer 1

The correct option is C $\frac{8}{11}$

Total number of balls in bag is $11.$
Let $W$ be the event that ball drawn is white.
$\Rightarrow P\left(W\right)=\frac{4}{11}$
Let $G$ be the event that ball drawn is green.
$P\left(G\right)=\frac{4}{11}$
So, the probability that the ball drawn is white or green is
$P(W\cup G)=P\left(W\right)+P\left(G\right)-P(W\cap G)$
$=\frac{4}{11}+\frac{4}{11}-0=\frac{8}{11}$