A jar contains 54 marbles, each of which some are blue, some are green and some are white.
Question:

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?

 

Solution:

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.

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