JEE Challenger
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MathHardAdv 2013 • Paper 2

Probability Box B2 Selected Given White and Red Balls Drawn

Comprehension Passage

A box B1B_1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2B_2 contains 2 white balls, 3 red balls and 4 black balls. A third box B3B_3 contains 3 white balls, 4 red balls and 5 black balls.

If 2 balls are chosen (without replacement) from a randomly chosen box and it is found that one ball is white and the other is red, what is the probability that these 2 balls came from box B2B_2?

Options

A

116181\frac{116}{181}

B

126181\frac{126}{181}

C

65181\frac{65}{181}

D

55181\frac{55}{181}

Correct

Topics & Concepts

Probability (Advanced)ProbabilityBayes' Theorem

Step-by-Step Solution

To find the probability that the two drawn balls came from box B2B_2, given that one ball is white and the other is red, we use Bayes' Theorem.

Step 1: Define the events and probabilities of selecting each box.

Let E1,E2,E_1, E_2, and E3E_3 be the events of choosing boxes B1,B2,B_1, B_2, and B3B_3, respectively. Since a box is chosen at random: P(E1)=P(E2)=P(E3)=13P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}

Let AA be the event that 2 balls chosen without replacement consist of 1 white ball and 1 red ball.

Step 2: Calculate the conditional probabilities P(A∣Ei)P(A|E_i) for each box.

  1. For Box B1B_1:

    • Composition: 1 White,3 Red,2 Black1\text{ White}, 3\text{ Red}, 2\text{ Black} (Total = 66 balls)
    • Total ways to draw 2 balls: (62)=15\binom{6}{2} = 15
    • Favourable ways (1 White and 1 Red): (11)×(31)=3\binom{1}{1} \times \binom{3}{1} = 3 P(A∣E1)=315=15P(A|E_1) = \frac{3}{15} = \frac{1}{5}

  2. For Box B2B_2:

    • Composition: 2 White,3 Red,4 Black2\text{ White}, 3\text{ Red}, 4\text{ Black} (Total = 99 balls)
    • Total ways to draw 2 balls: (92)=36\binom{9}{2} = 36
    • Favourable ways (1 White and 1 Red): (21)×(31)=6\binom{2}{1} \times \binom{3}{1} = 6 P(A∣E2)=636=16P(A|E_2) = \frac{6}{36} = \frac{1}{6}

  3. For Box B3B_3:

    • Composition: 3 White,4 Red,5 Black3\text{ White}, 4\text{ Red}, 5\text{ Black} (Total = 1212 balls)
    • Total ways to draw 2 balls: (122)=66\binom{12}{2} = 66
    • Favourable ways (1 White and 1 Red): (31)×(41)=12\binom{3}{1} \times \binom{4}{1} = 12 P(A∣E3)=1266=211P(A|E_3) = \frac{12}{66} = \frac{2}{11}

Step 3: Apply Bayes' Theorem.

We need to compute P(E2∣A)P(E_2|A): P(E2∣A)=P(E2)⋅P(A∣E2)P(E1)⋅P(A∣E1)+P(E2)⋅P(A∣E2)+P(E3)⋅P(A∣E3)P(E_2|A) = \frac{P(E_2) \cdot P(A|E_2)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2) + P(E_3) \cdot P(A|E_3)}

Since P(E1)=P(E2)=P(E3)P(E_1) = P(E_2) = P(E_3), the equation simplifies to: P(E2∣A)=P(A∣E2)P(A∣E1)+P(A∣E2)+P(A∣E3)P(E_2|A) = \frac{P(A|E_2)}{P(A|E_1) + P(A|E_2) + P(A|E_3)}

Substitute the calculated values into the formula: P(E2∣A)=1615+16+211P(E_2|A) = \frac{\frac{1}{6}}{\frac{1}{5} + \frac{1}{6} + \frac{2}{11}}

Taking the LCM of the denominators 5,6,5, 6, and 1111, which is 330330: P(A∣E1)+P(A∣E2)+P(A∣E3)=66330+55330+60330=181330P(A|E_1) + P(A|E_2) + P(A|E_3) = \frac{66}{330} + \frac{55}{330} + \frac{60}{330} = \frac{181}{330}

Thus: P(E2∣A)=55330181330=55181P(E_2|A) = \frac{\frac{55}{330}}{\frac{181}{330}} = \frac{55}{181}

Correct Answer: (D) 55181\frac{55}{181}

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