JEE Advanced 2015 · Paper 2 · Mathematics Q58

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MathematicsMultiple correct+4 / -2Question 58 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.

A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is $\dfrac13$ , then the correct option(s) with the possible values of $n_1$ and $n_2$ is(are)

n_1=4

and

n_2=6

n_1=2

and

n_2=3

n_1=10

and

n_2=20

n_1=3

and

n_2=6

Answer and worked solution

Official answer: C, D

Total probability. After a random ball is moved out of box I,
$P(\text{red}) = \frac{n_1}{n_1+n_2}\cdot\frac{n_1-1}{n_1+n_2-1}+\frac{n_2}{n_1+n_2}\cdot\frac{n_1}{n_1+n_2-1}=\frac{n_1}{n_1+n_2}.$

Condition. This equals $\frac13$ , so $n_2=2n_1$ .

Check each option.

(C) (10, 20). ✓
(D) (3, 6). ✓

(C), (D)

Topics:Conditional probability Classical probability

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JEE Advanced 2015 · Paper 2, Mathematics question 58