Question

A bag contains $40$ balls out of which some are red, some are blue and remaining are black. If the probability of drawing a red ball is $\frac{11}{20}$ and that of blue ball is $\frac{1}{5}$, then the number of black balls is

• A. $5$
• B. $25$
• C. $10$
• D. $30$

Answer 1

The correct option is C $10$

Total balls = $40$

Let $A$ be the event of red ball, $B$ be the event of blue ball and $C$ be the event of blue ball.

given, $P\left(A\right)=$ $\frac{11}{20}$

$P\left(B\right)=$ $\frac{1}{5}$

since, $P\left(A\right)+P\left(B\right)+P\left(C\right)=$ $1$

$\Rightarrow \frac{11}{20}+\frac{1}{5}$ $+P\left(C\right)=$ $1$

$\Rightarrow \frac{11+4}{20}$ $+P\left(C\right)=$ $1$

$\Rightarrow \frac{15}{20}$ $+P\left(C\right)=$ $1$

$\Rightarrow$ $P\left(C\right)=$ $1-\frac{15}{20}$

$\Rightarrow$ $P\left(C\right)=$ $\frac{5}{20}$

$\Rightarrow$ $P\left(C\right)=$$\frac{1}{4}$ ..........(i)

Now, let no. of black balls are $x$.

$\therefore$ $P\left(C\right)=$$\frac{x}{40}$

but from (i)

$\frac{x}{40}=\frac{1}{4}$

$4x=40$

$x=10$

So, no. of black balls are $10$.

Option C is correct.