A jar contains 54 marbles, each of which some are blue, some are green and some are white. The probability of selecting a blue marble at random is $\frac{1}{3}$ and the probability of selecting a green marble at random is $\frac{4}{9}$. How many white marbles does the jar contain?
Total number of marbles = 54.
It is given that, $P$ (getting a blue marble) $=\frac{1}{3}$ and $P$ (getting a green marble) $=\frac{4}{9}$
Let P(getting a white marble) be x.
Since, there are only 3 types of marbles in the jar, the sum of probabilities of all three marbles must be 1.
$\therefore \frac{1}{3}+\frac{4}{9}+x=1$
$\Rightarrow \frac{3+4}{9}+x=1$
$\Rightarrow x=1-\frac{7}{9}$
$\Rightarrow x=\frac{2}{9}$
$\therefore \mathrm{P}($ getting a white marble $)=\frac{2}{9} \quad \ldots(1)$
Let the number of white marbles be n.
Then, $\mathrm{P}$ (getting a white marble) $=\frac{n}{54}$
From (1) and (2),
$\frac{n}{54}=\frac{2}{9}$
$\Rightarrow n=\frac{2 \times 54}{9}$
$\Rightarrow n=12$
Thus, there are 12 white marbles in the jar.