Probability XII — JEE Advanced · 2015 · 24 May 2015 · Paper 2 Question with Solution

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

\frac{1}{3}

, then the correct option(s) with the possible values of

n_{1}

and

n_{2}

is(are)

(A) $n_{1} = 4$ and $n_{2} = 6$
(B) $n_{1} = 2$ and $n_{2} = 3$
(C) $n_{1} = 10$ and $n_{2} = 20$ ✓ Correct
(D) $n_{1} = 3$ and $n_{2} = 6$ ✓ Correct

Answer: (C), (D)

Solution

After transferring one ball from box I to box II, box I has

n_{1} + n_{2} - 1

balls. The probability of drawing a red ball from box I after the transfer is

\frac{1}{3}

. If a red ball was transferred, box I has

n_{1} - 1

red balls out of

n_{1} + n_{2} - 1

total, so

P (Red after) = \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} - 1 ) + n _{1} n _{2}}{( n _{1} + n _{2} ) ( n _{1} + n _{2} - 1 )} = \frac{n _{1} ( n _{1} + n _{2} - 1 )}{( n _{1} + n _{2} ) ( n _{1} + n _{2} - 1 )} = \frac{n _{1}}{n _{1} + n _{2}}

. Wait, this doesn't change. Re-reading: the question asks for the probability of drawing a red ball from box I AFTER the transfer, meaning from the remaining balls in box I. If a red ball was transferred,

P = \frac{n _{1} - 1}{n _{1} + n _{2} - 1}

; if a black ball was transferred,

P = \frac{n _{1}}{n _{1} + n _{2} - 1}

. The overall probability is

\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} - 1 ) + n _{1} n _{2}}{( n _{1} + n _{2} ) ( n _{1} + n _{2} - 1 )} = \frac{n _{1} ( n _{1} + n _{2} - 1 )}{( n _{1} + n _{2} ) ( n _{1} + n _{2} - 1 )} = \frac{n _{1}}{n _{1} + n _{2}} = \frac{1}{3}

. So

3 n_{1} = n_{1} + n_{2}

, giving

n_{2} = 2 n_{1}

. Checking: (A)

6 \neq = 8

✗; (B)

3 \neq = 4

✗; (C)

20 = 20

✓; (D)

6 = 6

✓.

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · One or more correct options

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Probability XII

Solution

More JEE Advanced questions on Probability XII