Question

Box I contains 3 red and 2 blue marbles while Box II contains 2 red and 8 blue marbles. A fair coin is tossed. If the coin turns up heads a marble is chosen from Box I, if it turns up tails a marble is chosen from Box II. Find the probability that a red marble is chosen.

Answer 1

Answer on Question #65390 - Math – Statistics and Probability

Box I contains 3 red and 2 blue marbles while Box II contains 2 red and 8 blue marbles. A fair coin is tossed. If the coin turns up heads a marble is chosen from Box I, if it turns up tails a marble is chosen from Box II. Find the probability that a red marble is chosen.

Solution. Denote $A = \{a \text{ red marble is chosen}\}$,

$B_1 = \{Box\ I \text{ is chosen}\}, B_2 = \{Box\ II \text{ is chosen}\}$

By the formula for total probability (e.g. [1], p.23, (3)),

$$P(A) = P(A|B_1)P(B_1) + P(A|B_2)P(B_2).$$ $$P(B_1) = P(B_2) = \frac{1}{2}, \quad P(A|B_1) = \frac{3}{5}, \quad P(A|B_2) = \frac{2}{10} = \frac{1}{5}.$$

Then

$$P(A) = \frac{1}{2} \cdot \frac{3}{5} + \frac{1}{2} \cdot \frac{1}{5} = \frac{4}{10} = \frac{2}{5}.$$

Answer. $\frac{2}{5}$.

References

1. Shiryaev A.N. Probability-1. Third edition. Springer. 2016.

Answer provided by www.AssignmentExpert.com