JEE Challenger
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MathMedAdv 2015 • Paper 2

Probability of Drawing Red Ball After Transfer Between Boxes

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.

A ball is selected at random from box I and transferred into box II. If the probability of drawing a red ball from box I after this transfer is 13\frac{1}{3}, then the correct choice(s) for the possible values of n1n_1 and n2n_2 is(are)

Options

A

n1=4n_1 = 4 and n2=6n_2 = 6

B

n1=2n_1 = 2 and n2=3n_2 = 3

C

n1=10n_1 = 10 and n2=20n_2 = 20

Correct
D

n1=3n_1 = 3 and n2=6n_2 = 6

Correct

Topics & Concepts

Probability (Advanced)ProbabilityConditional Probability

Step-by-Step Solution

To find the correct values of n1n_1 and n2n_2, we need to calculate the probability of drawing a red ball from Box I after one ball has been transferred from Box I to Box II.

Step 1: Define the Events

Let:

  • E1E_1 be the event that a red ball is drawn from Box I and transferred to Box II.
  • E2E_2 be the event that a black ball is drawn from Box I and transferred to Box II.
  • RR be the event of drawing a red ball from Box I after the transfer.

The initial number of balls in Box I is n1+n2n_1 + n_2. The probabilities of events E1E_1 and E2E_2 occurring during the transfer step are: P(E1)=n1n1+n2P(E_1) = \frac{n_1}{n_1 + n_2} P(E2)=n2n1+n2P(E_2) = \frac{n_2}{n_1 + n_2}

Step 2: Determine Conditional Probabilities

  • If E1E_1 occurs (a red ball is transferred):

    • Remaining red balls in Box I = n1−1n_1 - 1
    • Remaining total balls in Box I = n1+n2−1n_1 + n_2 - 1

    The conditional probability of drawing a red ball now is: P(R∣E1)=n1−1n1+n2−1P(R \mid E_1) = \frac{n_1 - 1}{n_1 + n_2 - 1}

  • If E2E_2 occurs (a black ball is transferred):

    • Remaining red balls in Box I = n1n_1
    • Remaining total balls in Box I = n1+n2−1n_1 + n_2 - 1

    The conditional probability of drawing a red ball now is: P(R∣E2)=n1n1+n2−1P(R \mid E_2) = \frac{n_1}{n_1 + n_2 - 1}

Step 3: Apply the Law of Total Probability

The total probability of drawing a red ball from Box I after the transfer is: P(R)=P(E1)⋅P(R∣E1)+P(E2)⋅P(R∣E2)P(R) = P(E_1) \cdot P(R \mid E_1) + P(E_2) \cdot P(R \mid E_2)

Substitute the expressions into the formula: P(R)=(n1n1+n2)⋅(n1−1n1+n2−1)+(n2n1+n2)⋅(n1n1+n2−1)P(R) = \left(\frac{n_1}{n_1 + n_2}\right) \cdot \left(\frac{n_1 - 1}{n_1 + n_2 - 1}\right) + \left(\frac{n_2}{n_1 + n_2}\right) \cdot \left(\frac{n_1}{n_1 + n_2 - 1}\right)

Factor out common terms in the numerator: P(R)=n1(n1−1)+n2n1(n1+n2)(n1+n2−1)P(R) = \frac{n_1(n_1 - 1) + n_2 n_1}{(n_1 + n_2)(n_1 + n_2 - 1)} P(R)=n1(n1−1+n2)(n1+n2)(n1+n2−1)P(R) = \frac{n_1 (n_1 - 1 + n_2)}{(n_1 + n_2)(n_1 + n_2 - 1)} P(R)=n1(n1+n2−1)(n1+n2)(n1+n2−1)P(R) = \frac{n_1 (n_1 + n_2 - 1)}{(n_1 + n_2)(n_1 + n_2 - 1)}

Canceling out (n1+n2−1)(n_1 + n_2 - 1) from the numerator and denominator gives: P(R)=n1n1+n2P(R) = \frac{n_1}{n_1 + n_2}

Step 4: Solve for n1n_1 and n2n_2

Given that P(R)=13P(R) = \frac{1}{3}: n1n1+n2=13\frac{n_1}{n_1 + n_2} = \frac{1}{3} 3n1=n1+n23n_1 = n_1 + n_2 2n1=n2  ⟹  n1n2=122n_1 = n_2 \implies \frac{n_1}{n_2} = \frac{1}{2}

Step 5: Verify the Options

We test each option to see which satisfies n2=2n1n_2 = 2n_1:

  • Option (A): n1=4n_1 = 4 and n2=6  ⟹  2(4)=8≠6n_2 = 6 \implies 2(4) = 8 \neq 6 (Incorrect)
  • Option (B): n1=2n_1 = 2 and n2=3  ⟹  2(2)=4≠3n_2 = 3 \implies 2(2) = 4 \neq 3 (Incorrect)
  • Option (C): n1=10n_1 = 10 and n2=20  ⟹  2(10)=20n_2 = 20 \implies 2(10) = 20 (Correct)
  • Option (D): n1=3n_1 = 3 and n2=6  ⟹  2(3)=6n_2 = 6 \implies 2(3) = 6 (Correct)

Conclusion

The correct options are (C) and (D).

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