JEE Challenger
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MathHardAdv 2015 • Paper 2

Probability of Drawing Red Ball from Box II

Comprehension Passage

Let n1n_1 and n2n_2 denote the number of red and black balls, respectively, in box I. Let n3n_3 and n4n_4 denote the number of red and black balls, respectively, in box II.

Consider two boxes, box I and box II. One of the boxes is chosen at random and a ball is drawn from it. The drawn ball is found to be red. If the probability that this red ball came from box II is 13\frac{1}{3}, then the correct choice(s) with the possible values of n1,n2,n3n_1, n_2, n_3 and n4n_4 is(are)

Options

A

n1=3,n2=3,n3=5,n4=15n_1 = 3, n_2 = 3, n_3 = 5, n_4 = 15

Correct
B

n1=3,n2=6,n3=10,n4=50n_1 = 3, n_2 = 6, n_3 = 10, n_4 = 50

Correct
C

n1=8,n2=6,n3=5,n4=20n_1 = 8, n_2 = 6, n_3 = 5, n_4 = 20

D

n1=6,n2=12,n3=5,n4=20n_1 = 6, n_2 = 12, n_3 = 5, n_4 = 20

Topics & Concepts

Probability (Advanced)ProbabilityConditional ProbabilityBayes' Theorem

Step-by-Step Solution

To determine the correct option(s), we apply Bayes' Theorem to find the conditional probability that a drawn red ball came from Box II.

Let the events be defined as:

  • B1B_1: Box I is chosen.
  • B2B_2: Box II is chosen.
  • RR: The drawn ball is red.

Since one of the two boxes is chosen at random: P(B1)=P(B2)=12P(B_1) = P(B_2) = \frac{1}{2}

The probability of drawing a red ball from Box I is: P(R∣B1)=n1n1+n2P(R \mid B_1) = \frac{n_1}{n_1 + n_2}

The probability of drawing a red ball from Box II is: P(R∣B2)=n3n3+n4P(R \mid B_2) = \frac{n_3}{n_3 + n_4}

Using Bayes' Theorem, the probability that the drawn red ball came from Box II is given by: P(B2∣R)=P(B2)P(R∣B2)P(B1)P(R∣B1)+P(B2)P(R∣B2)P(B_2 \mid R) = \frac{P(B_2) P(R \mid B_2)}{P(B_1) P(R \mid B_1) + P(B_2) P(R \mid B_2)}

Substitute P(B1)=P(B2)=12P(B_1) = P(B_2) = \frac{1}{2} into the formula: P(B2∣R)=n3n3+n4n1n1+n2+n3n3+n4P(B_2 \mid R) = \frac{\frac{n_3}{n_3 + n_4}}{\frac{n_1}{n_1 + n_2} + \frac{n_3}{n_3 + n_4}}

We are given that P(B2∣R)=13P(B_2 \mid R) = \frac{1}{3}. Therefore: n3n3+n4n1n1+n2+n3n3+n4=13\frac{\frac{n_3}{n_3 + n_4}}{\frac{n_1}{n_1 + n_2} + \frac{n_3}{n_3 + n_4}} = \frac{1}{3}

Cross-multiplying yields: 3(n3n3+n4)=n1n1+n2+n3n3+n43 \left( \frac{n_3}{n_3 + n_4} \right) = \frac{n_1}{n_1 + n_2} + \frac{n_3}{n_3 + n_4}

2(n3n3+n4)=n1n1+n22 \left( \frac{n_3}{n_3 + n_4} \right) = \frac{n_1}{n_1 + n_2}

  ⟹  n1n1+n2=2⋅(n3n3+n4)\implies \frac{n_1}{n_1 + n_2} = 2 \cdot \left( \frac{n_3}{n_3 + n_4} \right)

Now, let's test each option against this required condition:

  • Option (A): n1=3,n2=3,n3=5,n4=15n_1 = 3, n_2 = 3, n_3 = 5, n_4 = 15 n1n1+n2=33+3=12\frac{n_1}{n_1 + n_2} = \frac{3}{3 + 3} = \frac{1}{2} 2⋅(n3n3+n4)=2⋅(55+15)=2⋅520=122 \cdot \left( \frac{n_3}{n_3 + n_4} \right) = 2 \cdot \left( \frac{5}{5 + 15} \right) = 2 \cdot \frac{5}{20} = \frac{1}{2} This satisfies the condition. Thus, Option (A) is correct.

  • Option (B): n1=3,n2=6,n3=10,n4=50n_1 = 3, n_2 = 6, n_3 = 10, n_4 = 50 n1n1+n2=33+6=13\frac{n_1}{n_1 + n_2} = \frac{3}{3 + 6} = \frac{1}{3} 2⋅(n3n3+n4)=2⋅(1010+50)=2⋅1060=132 \cdot \left( \frac{n_3}{n_3 + n_4} \right) = 2 \cdot \left( \frac{10}{10 + 50} \right) = 2 \cdot \frac{10}{60} = \frac{1}{3} This satisfies the condition. Thus, Option (B) is correct.

  • Option (C): n1=8,n2=6,n3=5,n4=20n_1 = 8, n_2 = 6, n_3 = 5, n_4 = 20 n1n1+n2=88+6=47\frac{n_1}{n_1 + n_2} = \frac{8}{8 + 6} = \frac{4}{7} 2⋅(n3n3+n4)=2⋅(55+20)=252 \cdot \left( \frac{n_3}{n_3 + n_4} \right) = 2 \cdot \left( \frac{5}{5 + 20} \right) = \frac{2}{5} Since 47≠25\frac{4}{7} \neq \frac{2}{5}, Option (C) is incorrect.

  • Option (D): n1=6,n2=12,n3=5,n4=20n_1 = 6, n_2 = 12, n_3 = 5, n_4 = 20 n1n1+n2=66+12=13\frac{n_1}{n_1 + n_2} = \frac{6}{6 + 12} = \frac{1}{3} 2⋅(n3n3+n4)=2⋅(55+20)=252 \cdot \left( \frac{n_3}{n_3 + n_4} \right) = 2 \cdot \left( \frac{5}{5 + 20} \right) = \frac{2}{5} Since 13≠25\frac{1}{3} \neq \frac{2}{5}, Option (D) is incorrect.

Hence, the correct options are A and B.

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