Question

There are $18$ beads in a box. Some of them are white and the remaining are black. The probability of drawing a black bead from it is $\frac{1}{3}$. Then
(a) How many black beads are there in the box?
(b) How many white beads are there in the box?
(c) How many white beads should be added to it so that the probability of drawing a black head becomes $\frac{1}{4}$?

Answer 1

Total number of beads $=18$
(a) Let the number of black beads be $a$.
Now. $P$ (black bead) $=\frac{\text{Total number of black beads}}{\text{Total number of beads}}=\frac{1}{3}$
$\Rightarrow \frac{a}{18}=\frac{1}{3}$
$\Rightarrow a=6$
Thus, there are $6$ black beads in the box.
(b) Now, number of white beads $=18-a=18-6=12$
(c) Let $x$ number of white beads be added. Then, total number of beads $=18+x$
For $P$ (black bead) $=\frac{1}{4}$, we have
$\frac{6}{18+x}=\frac{1}{4}$
$\Rightarrow 18+x=24$
$\Rightarrow x=6$
Thus, $6$ white beads should be added.