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JEE Mathematics 2024 Question with Solution

An urn contains 66 white and 99 black balls. Two successive draws of 44 balls are made without replacement. The probability that the first draw gives all white balls and the second draw gives all black balls is:

  • A

    5256\frac{5}{256}

  • B

    5715\frac{5}{715}

  • C

    3715\frac{3}{715}

  • D

    3256\frac{3}{256}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The urn contains 66 white balls and 99 black balls, so total balls are 1515.

Find: The probability that the first draw of 44 balls contains all white balls and the second draw of 44 balls contains all black balls.

For the first draw, the number of ways to choose 44 white balls from 66 white balls is

(64)=15\binom{6}{4} = 15

The total number of ways to choose any 44 balls from 1515 balls is

(154)=1365\binom{15}{4} = 1365

Therefore, the probability that the first draw gives all white balls is

(64)(154)=151365\frac{\binom{6}{4}}{\binom{15}{4}} = \frac{15}{1365}

Step-by-step Computation

After the first draw removes 44 white balls, the remaining balls are 22 white and 99 black, so total remaining balls are 1111.

For the second draw, the number of ways to choose 44 black balls from 99 black balls is

(94)=126\binom{9}{4} = 126

The total number of ways to choose any 44 balls from the remaining 1111 balls is

(114)=330\binom{11}{4} = 330

So, the probability that the second draw gives all black balls is

(94)(114)=126330\frac{\binom{9}{4}}{\binom{11}{4}} = \frac{126}{330}

Direct Multiplication

Multiply the probabilities of the two successive required events:

(64)(154)×(94)(114)=151365×126330\frac{\binom{6}{4}}{\binom{15}{4}} \times \frac{\binom{9}{4}}{\binom{11}{4}} = \frac{15}{1365} \times \frac{126}{330}

Now simplify:

15×1261365×330=3715\frac{15 \times 126}{1365 \times 330} = \frac{3}{715}

Therefore, the correct option is C, that is 3715\frac{3}{715}.

Common mistakes

  • A common mistake is to use (158)\binom{15}{8} or treat both draws as one combined selection. This is wrong because the question asks for an ordered event: first all white, then all black. Compute the two successive probabilities separately and then multiply.

  • Another mistake is to use (154)\binom{15}{4} again for the second draw. This is wrong because the first draw is without replacement, so only 1111 balls remain. For the second draw, the denominator must be (114)\binom{11}{4}.

  • Students may incorrectly say the two events are independent in the usual sense and ignore the changed composition after the first draw. The composition of the urn changes after removing balls, so the second probability must be evaluated using the remaining balls.

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