Question

An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

• A. \(\frac{9}{14}\)
• B. \(\frac{27}{28}\)
• C. \(\frac{3}{7}\)
• D. \(\frac{2}{3}\)

Hint:

Alternatively, use combinations: \(P = \frac{\binom{10}{2}}{\binom{15}{2}} = \frac{\frac{10 \times 9}{2}}{\frac{15 \times 14}{2}} = \frac{45}{105} = \frac{3}{7}\). Both methods yield identical results quickly in competitive exams.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The problem deals with conditional probability and drawing without replacement from a finite sample space.
When items are drawn without replacement, the total number of items and the count of the chosen category decrease after each draw.
Key Formula or Approach:
\[ P(A \cap B) = P(A) \cdot P(B|A) \]
where \(P(A)\) is the probability of the first event, and \(P(B|A)\) is the conditional probability of the second event given that the first has occurred.

Step 2: Detailed Explanation:
Total number of balls initially in the urn:
\[ N_1 = 10 \text{ (black)} + 5 \text{ (white)} = 15 \text{ balls} \]
Probability of drawing a black ball on the first draw:
\[ P(B_1) = \frac{10}{15} = \frac{2}{3} \]
After drawing one black ball, 9 black balls and 5 white balls remain in the urn (total 14 balls).
Conditional probability of drawing a black ball on the second draw:
\[ P(B_2 | B_1) = \frac{9}{14} \]
The combined probability that both drawn balls are black is:
\[ P(B_1 \cap B_2) = \frac{10}{15} \times \frac{9}{14} = \frac{2}{3} \times \frac{9}{14} = \frac{18}{42} = \frac{3}{7} \]

Step 3: Final Answer:
Thus, the probability that both drawn balls are black is \(\frac{3}{7}\), which corresponds to option (C).