JEE Advanced 2019 · Paper 1 · Mathematics Q7

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MathematicsMultiple correct+4 / -1Question 43 of 54

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 $\dfrac3{10}$ , $\dfrac3{10}$ and $\dfrac4{10}$ 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?

Probability that the chosen ball is green, given that the selected bag is

B_3

, equals

\dfrac38

Probability that the chosen ball is green equals

\dfrac{39}{80}

Probability that the selected bag is

B_3

, given that the chosen ball is green, equals

\dfrac5{13}

Probability that the selected bag is

B_3

and the chosen ball is green equals

\dfrac3{10}

Answer and worked solution

Official answer: A, B

Probabilities of green.
$P(G\mid B_1)=\tfrac12,\qquad P(G\mid B_2)=\tfrac58,\qquad P(G\mid B_3)=\tfrac38.$
So (A) is true.

(B) By total probability,
$P(G)=0.3\cdot\tfrac12+0.3\cdot\tfrac58+0.4\cdot\tfrac38=0.15+0.1875+0.15=\tfrac{39}{80}.$
True.

(D) $P(B_3\cap G)=0.15=\frac3{20}$ , not $\frac3{10}$ . False.

(C) $P(B_3\mid G)=\dfrac{0.15}{0.4875}=\dfrac4{13}$ , not $\frac5{13}$ . False.

(A), (B)

Topics:Bayes' theorem Conditional probability

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JEE Advanced 2019 · Paper 1, Mathematics question 7