Question

Two bags marked M and N contain identical pen drives. Bag M contains 10 green pen drives and 8 red pens drives. Beg N contains 7 green pen drives and 6 red pen drives. A pen drive is randomly selected from bag M and put into bag N. Then, another pen drive is randomly selected from bag N.
(a) If a red pen drive is selected from bag N, find the probability a green pen drive is selected from bag M.

Answer 1

Answer on Question #43463 – Math - Statistics and Probability

Two bags marked M and N contain identical pen drives. Bag M contains 10 green pen drives and 8 red pens drives. Beg N contains 7 green pen drives and 6 red pen drives. A pen drive is randomly selected from bag M and put into bag N. Then, another pen drive is randomly selected from bag N.

(a) If a red pen drive is selected from bag N, find the probability a green pen drive is selected from bag M.

Let $H_{1} =$ "a green pen drive is selected from bag M", $H_{2} =$ "a red pen drive is selected from bag M", $D =$ "a red pen drive is selected from bag N", $(H_{1}|D) =$ "a green pen drive is selected from bag M given a red pen drive is selected from bag N"

The corresponding probabilities are $P(H_{1}) = \frac{10}{18}, P(H_{2}) = \frac{8}{18}$.

By total probability formula, $P(D) = P(D|H_1)P(H_1) + P(D|H_2)P(H_2)$.

By Bayes' formula,

$$P(H_{1}|D) = \frac{P(D|H_{1})P(H_{1})}{P(D)} = \frac{P(D|H_{1})P(H_{1})}{P(D|H_{1})P(H_{1}) + P(D|H_{2})P(H_{2})} = \frac{\frac{6}{14} \cdot \frac{10}{18}}{\frac{6}{14} \cdot \frac{10}{18} + \frac{7}{14} \cdot \frac{8}{18}} = \frac{60}{60 + 56} = \frac{60}{116} = \frac{15}{29} \approx 0.52.$$

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