Mathematics - Probability Question with Solution | TestHub

In order for all balls to be red after three replacements, two of the three balls selected must have been green. There are three possible cases:

Case I: Red, Green, Red. The probability is $P(I)=\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{4}=\frac{1}{16}$.

Case II: Green, Red, Green. The probability is $P(II)=\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{1}{4}=\frac{3}{32}$.

Case III: Green, Green, Red. The probability is $P(III)=\frac{1}{2}\cdot \frac{1}{4}\cdot 1=\frac{1}{8}$.

The required probability is the sum of the probabilities of these three cases:

$$P(total)=P(I)+P(II)+P(III)=\frac{1}{16}+\frac{3}{32}+\frac{1}{8}=\frac{2}{32}+\frac{3}{32}+\frac{4}{32}=\frac{9}{32}$$