Question

A box contains 12 balls out of which $x$ are blue. If one ball is drawn at random from the box, what is the probability that it will be a blue ball ? If 6 more blue balls are put in the box , the probability of drawing a black ball is now double of what it was before . Find $x$

• A. (i) $\frac{x}{12}$ (ii) 3
• B. (i) $\frac{x}{6}$ (ii) 4
• C. (i) $\frac{x}{6}$ (ii) 5
• D. (i) $\frac{x}{4}$ (ii) 2

Answer 1

The correct option is A

(i) $\frac{x}{12}$ (ii) 3

Number of blue = $x$

Total number of balls = 12

Hence P ( getting a blue ball ) = $\frac{x}{12}$

(ii) Value of $x$

6 more balls were added . Hence P( getting blue ball ) = $\frac{x+6}{12+6}$

= $\frac{x+6}{18}$

Given the probablity of drawing blue ball is now double

Hence $\frac{x+6}{18}$ = 2 $\times$ $\frac{x}{12}$

$\frac{x+6}{18}$ = 2 $\times$ $\frac{x}{6}$

6$x$ + 36 = 18 $x$

12 $x$ = 36

$x$ = 3