There are three bags $\mathrm{B}_{1}, \mathrm{~B}_{2}$ and $\mathrm{B}_{3}$. The bag $\mathrm{B}_{1}$ contains 5 red and 5 green balls, $\mathrm{B}_{2}$ contains 3 red and 5 green balls, and $\mathrm{B}_{3}$ contains 5 red and 3 green balls, Bags $\mathrm{B}_{1}, \mathrm{~B}_{2}$ and $\mathrm{B}_{3}$ have probabilities $\frac{3}{10}, \frac{3}{10}$ and $\frac{4}{10}$ 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 ?

(B) Probability that the chosen ball is green equals $\frac{39}{80}$; (C) Probability that the chosen ball is green, given that the selected bag is $B_{3}$, equals $\frac{3}{8}$

(A) Probability that the selected bag is $B_{3}$ and the chosen ball is green equals $\frac{3}{10}$

(D) Probability that the selected bag is $B_{3}$, given that the chosen balls is green, equals $\frac{5}{13}$

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Probability XII

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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

\frac{3}{10}, \frac{3}{10}

and

\frac{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 ?

(A)Probability that the selected bag is $B_{3}$ and the chosen ball is green equals $\frac{3}{10}$
(B)Probability that the chosen ball is green equals $\frac{39}{80}$ ✓ Correct
(C)Probability that the chosen ball is green, given that the selected bag is $B_{3}$ , equals $\frac{3}{8}$ ✓ Correct
(D)Probability that the selected bag is $B_{3}$ , given that the chosen balls is green, equals $\frac{5}{13}$

Answer: (B), (C)

Solution

(1)

P (B_{3} \cap G) = P (\frac{G}{B _{3}}) P (B_{3}) = \frac{3}{8} \times \frac{4}{10} = \frac{3}{20}

✗

(2)

P (G) = P (\frac{G}{B _{1}}) P (B_{1}) + P (\frac{G}{B _{2}}) P (B_{2}) + P (\frac{G}{B _{3}}) P (B_{3})

= \frac{1}{2} \cdot \frac{3}{10} + \frac{5}{8} \cdot \frac{3}{10} + \frac{3}{8} \cdot \frac{4}{10} = \frac{3}{20} + \frac{3}{16} + \frac{3}{20} = \frac{39}{80}

✓

(3)

P (\frac{G}{B _{3}}) = \frac{3}{8}

✓

(4)

P (\frac{B _{3}}{G}) = \frac{P ( G \cap B _{3} )}{P ( G )} = \frac{3/20}{39/80} = \frac{4}{13}

✗

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · One or more correct options

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JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Probability XII

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