There are some red, green and white marbles in a box. One marble is picked up at random from this box. If the probability of picking up a red marble is $\frac{2}{9}$ and that of picking up a green marble is $\frac{4}{9}$ then find the :
(a) probability of picking up a white marble.
(b) number of green marbles, if total number of marbles is 54.
(c) probability of not picking up a red marble.
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(a) The sum of the probabilities of all possible outcomes is equal to 1.
The possible outcomes are picking a red, green, or white marble.
P(red) + P(green) + P(white) = 1
Given,
P(red) = $\frac{2}{9}$
P(green) = $\frac{4}{9}$
$$\begin{aligned}\Rightarrow \frac{2}{9} + \frac{4}{9} + P(\text{white}) = 1 \\[1em] \Rightarrow \frac{6}{9} + P(\text{white}) = 1 \\[1em] \Rightarrow P(\text{white}) = 1 - \frac{6}{9} \\[1em] \Rightarrow P(\text{white}) = \frac{9 - 6}{9} \\[1em] \Rightarrow P(\text{white}) = \frac{3}{9} = \frac{1}{3}.\end{aligned}$$
Hence, probability of picking up a white marble = $\frac{1}{3}$.
(b) Given,
Total number of marbles = 54
P(green) $=\frac{\text{No of favorable outcomes}}{\text{Total number of outcomes}}$.
$$\begin{aligned}\Rightarrow \frac{4}{9} = \frac{\text{Number of green marbles}}{54} \\[1em] \Rightarrow \frac{4}{9} \times 54 = \text{Number of green marbles} \\[1em] \Rightarrow 4 \times 6 = \text{Number of green marbles} \\[1em] \Rightarrow \text{Number of green marbles} = 24.\end{aligned}$$
Hence, number of green marbles is 24.
(c) Given,
P(red) = $\frac{2}{9}$
P(not picking a red) = 1 - P(red)
= 1 - $\frac{2}{9}$
= $\frac{9 - 2}{9}$
= $\frac{7}{9}$.
Hence, probability of not picking up a red marble is $\frac{7}{9}$.