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 $\dfrac{2}{9}$ and that of picking up a green marble is $\dfrac{4}{9}$ 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.

(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) = $\dfrac{2}{9}$

P(green) = $\dfrac{4}{9}$

$\Rightarrow \dfrac{2}{9} + \dfrac{4}{9} + P(\text{white}) = 1 \\[1em] \Rightarrow \dfrac{6}{9} + P(\text{white}) = 1 \\[1em] \Rightarrow P(\text{white}) = 1 - \dfrac{6}{9} \\[1em] \Rightarrow P(\text{white}) = \dfrac{9 - 6}{9} \\[1em] \Rightarrow P(\text{white}) = \dfrac{3}{9} = \dfrac{1}{3}.$

Hence, probability of picking up a white marble = $\dfrac{1}{3}$.

(b) Given,

Total number of marbles = 54

P(green) $=\dfrac{\text{No of favorable outcomes}}{\text{Total number of outcomes}}$.

$\Rightarrow \dfrac{4}{9} = \dfrac{\text{Number of green marbles}}{54} \\[1em] \Rightarrow \dfrac{4}{9} \times 54 = \text{Number of green marbles} \\[1em] \Rightarrow 4 \times 6 = \text{Number of green marbles} \\[1em] \Rightarrow \text{Number of green marbles} = 24.$

Hence, number of green marbles is 24.

(c) Given,

P(red) = $\dfrac{2}{9}$

P(not picking a red) = 1 - P(red)

= 1 - $\dfrac{2}{9}$

= $\dfrac{9 - 2}{9}$

= $\dfrac{7}{9}$.

Hence, probability of not picking up a red marble is $\dfrac{7}{9}$.