Question

A box contains 4 red and 6 white marbles. Two successive draws of 3 balls are made without replacement. The probability that in first draw all the 3 balls are white and in second draw all the 3 balls are red, is

• A. \( \frac{2}{105} \)
• B. \( \frac{1}{70} \)
• C. \( \frac{4}{105} \)
• D. \( \frac{3}{105} \)
• E. \( \frac{1}{35} \)

Hint:

For successive draws without replacement, multiply conditional probabilities.

Answer 1

The Correct Option is A

Concept: Use conditional probability.

Step 1: First draw (3 white).
\[ P_1 = \frac{{^6C_3}}{{^{10}C_3}} = \frac{20}{120} = \frac{1}{6} \]

Step 2: Second draw (3 red).
Remaining: 4 red, 3 white \[ P_2 = \frac{{^4C_3}}{{^7C_3}} = \frac{4}{35} \]

Step 3: Total probability.
\[ P = \frac{1}{6} \cdot \frac{4}{35} = \frac{2}{105} \]