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

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

\frac{1}{3}

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

n_{1}, n_{2}, n_{3}

and

n_{4}

is(are)

(A) $n_{1} = 3, n_{2} = 3, n_{3} = 5, n_{4} = 15$ ✓ Correct
(B) $n_{1} = 3, n_{2} = 6, n_{3} = 10, n_{4} = 50$ ✓ Correct
(C) $n_{1} = 8, n_{2} = 6, n_{3} = 5, n_{4} = 20$
(D) $n_{1} = 6, n_{2} = 12, n_{3} = 5, n_{4} = 20$

Answer: (A), (B)

Solution

By Bayes' theorem,

P (Box II ∣ Red) = \frac{P ( Red ∣ Box II ) \cdot P ( Box II )}{P ( Red )} = \frac{\frac{n _{3}}{n _{3} + n _{4}} \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{n _{1}}{n _{1} + n _{2}} + \frac{1}{2} \cdot \frac{n _{3}}{n _{3} + n _{4}}} = \frac{n _{3} / ( n _{3} + n _{4} )}{n _{1} / ( n _{1} + n _{2} ) + n _{3} / ( n _{3} + n _{4} )} = \frac{1}{3}

. This gives

3 n_{3} (n_{1} + n_{2}) = (n_{3} + n_{4}) [n_{1} (n_{3} + n_{4}) + n_{3} (n_{1} + n_{2})]

. Simplifying:

2 n_{3} (n_{1} + n_{2}) = n_{1} (n_{3} + n_{4})

. Checking options: (A)

2 \cdot 5 \cdot 6 = 60 = 3 \cdot 20

✓; (B)

2 \cdot 10 \cdot 9 = 180 = 3 \cdot 60

✓; (C)

2 \cdot 5 \cdot 14 = 140 \neq = 8 \cdot 25 = 200

✗; (D)

2 \cdot 5 \cdot 18 = 180 \neq = 6 \cdot 25 = 150

✗.

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

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

Solution

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