Throwback maths 11

throwback 11
Solution

The probability of drawing Green, then Red, then Red is $$P(GRR) = \frac{3}{5}\times \frac{2}{4} \times \frac{1}{3} = \frac{6}{60} = \frac{1}{10}$$

The probability of drawing Red, then Green, then Green is $$P(RGG) = \frac{2}{5}\times \frac{3}{4} \times \frac{2}{3} = \frac{12}{60} = \frac{1}{5}$$

It follows that outcome (A) is the most likely.