JEE Advanced 2015 · Paper 2 · Mathematics Q57

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MathematicsMultiple correct+4 / -2Question 57 of 60

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PARAGRAPH 1

Let $n_1$ and $n_2$ be the number of red and black balls, respectively, in box I. Let $n_3$ and $n_4$ be the number of red and black balls, respectively, in box II.

One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is $\dfrac13$ , then the correct option(s) with the possible values of $n_1$ , $n_2$ , $n_3$ and $n_4$ is(are)

n_1=3,\ n_2=3,\ n_3=5,\ n_4=15

n_1=3,\ n_2=6,\ n_3=10,\ n_4=50

n_1=8,\ n_2=6,\ n_3=5,\ n_4=20

n_1=6,\ n_2=12,\ n_3=5,\ n_4=20

Answer and worked solution

Official answer: A, B

Bayes' theorem.
$P(\text{II}\mid\text{red})=\frac{p_2}{p_1+p_2}=\frac13\implies p_1=2p_2,$
where $p_1=\frac{n_1}{n_1+n_2}$ and $p_2=\frac{n_3}{n_3+n_4}$ .

Check each option.

(A) $\frac12=2\cdot\frac14$ . ✓
(B) $\frac13=2\cdot\frac16$ . ✓
(C) $\frac47\ne2\cdot\frac15$ . ✗
(D) $\frac13\ne2\cdot\frac15$ . ✗

(A), (B)

Topics:Bayes' theorem Conditional probability

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JEE Advanced 2015 · Paper 2, Mathematics question 57