Question

An urn contains 6 white and 9 black balls. Two successive draws of 4 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. \( \frac{5}{256} \)
• B. \( \frac{2}{715} \)
• C. \( \frac{3}{715} \)
• D. \( \frac{3}{256} \)

Answer 1

The Correct Option is C

To solve this problem, we need to find the probability that the first draw gives all white balls, and the second draw gives all black balls. This requires understanding the concept of probability with respect to combinations.

The urn contains: 

• 6 white balls
• 9 black balls
• Total balls = 6 + 9 = 15

We are drawing 4 balls in two successive draws without replacement. Let's calculate the probability step by step:

1. First Draw (4 white balls):
   • The number of ways to choose 4 white balls out of 6 is given by the combination formula: \(\binom{6}{4}\).
   • \(\binom{6}{4} = \frac{6 \times 5}{2 \times 1} = 15\).
2. Total ways to choose any 4 balls from 15 is:
   • \(\binom{15}{4} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365\).
3. Probability of first draw giving all white balls is:
   • \(\frac{\binom{6}{4}}{\binom{15}{4}} = \frac{15}{1365}\).
4. Second Draw (4 black balls remaining):
   • After the first draw, 4 white balls are removed, leaving 2 white and 9 black balls.
   • The ways to choose 4 black balls from 9 is: \(\binom{9}{4}\).
   • \(\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126\).
5. Total ways to choose any 4 balls from remaining 11 is:
   • \(\binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330\).
6. Probability of the second draw giving all black balls is:
   • \(\frac{\binom{9}{4}}{\binom{11}{4}} = \frac{126}{330}\).
7. Overall probability:
   • Multiply the probabilities of the two events since they are independent: \(\frac{15}{1365} \times \frac{126}{330}\).
   • This results in: \(\frac{15 \times 126}{1365 \times 330}\).
   • Simplifying, we get \(\frac{3}{715}\).

The correct answer is therefore \(\frac{3}{715}\).

Answer 2

Probability of drawing 4 white balls in the first draw:
\(\frac{\binom{6}{4}}{\binom{15}{4}} = \frac{15}{1365}.\)

After removing 4 white balls, there are 9 black balls left. Probability of drawing 4 black balls in the second draw:  
\(\frac{\binom{9}{4}}{\binom{11}{4}} = \frac{126}{330}.\)

The required probability is:  
\(\frac{15}{1365} \times \frac{126}{330} = \frac{3}{715}.\)

The Correct answer is: \( \frac{3}{715} \)