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There are three bags $B_{1}, B_{2}$ and $B_{3}$ . The bag $B_{1}$ contains 5 red and 5 green balls, $B_{2}$ contains 3 red and 5 green balls, and $B_{3}$ contains 5 red and 3 green balls. Bags $B_{1}, B_{2}$ and $B_{3}$ have probabilities 3/10, 3/10 and 4/10 respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?

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Probability that the selected bag is $B_{3}$ , given that the chosen ball is green, equals 5/13.

Probability that the chosen ball is green, given that the selected bag is $B_{3}$ , equals 3/8.

Probability that the selected bag is $B_{3}$ and the chosen ball is green equals 3/10

Probability that the chosen ball is green equals 39/80

answer is A, C.

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Detailed Solution

The bag $B_{1}$ has 5 red and 5 green balls.

Bag $B_{2}$ has 3 red and 5 green balls

Bag $B_{3}$ has 5 red and 3 green balls

Clearly, $P (\frac{G}{B_{3}}) = \frac{3}{8}$

$P (G) = P (B_{1}) \cdot P (\frac{G}{B_{1}}) + P (B_{2}) \cdot P (\frac{G}{B_{2}}) + P (B_{3}) \cdot P (\frac{G}{B_{3}})$

$= \frac{3}{10} \cdot \frac{1}{2} + \frac{3}{10} \cdot \frac{5}{8} + \frac{4}{10} \cdot \frac{3}{8}$

$= \frac{3}{20} + \frac{3}{16} + \frac{3}{20} = \frac{12 + 15 + 12}{80} = \frac{39}{80}$

Now, $P (G \cap B_{3}) = P (\frac{G}{B_{3}}) P (B_{3}) = \frac{3}{8} \times \frac{4}{10} = \frac{3}{20}$

and $P (\frac{B_{3}}{G}) = \frac{P (G \cap B_{3})}{P (G)} = \frac{3 / 20}{39 / 80} = \frac{4}{13}$

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