Question

A bag P contains 4 red and 5 black balls, another bag Q contains 3 red and 6 black balls. If one ball is drawn at random from bag P and two balls are drawn from bag Q, then the probability that out of the three balls drawn two are black and one is red, is

• A. \( \frac{25}{54} \)
• B. \( \frac{25}{64} \)
• C. \( \frac{27}{64} \)
• D. \( \frac{35}{54} \)

Hint:

For mixed events involving different bags or stages, break the event into distinct mutually exclusive cases and sum their individual probabilities.

Answer 1

The Correct Option is A

Step 1: Bag P has 4 red and 5 black balls. So total = 9 balls. Bag Q has 3 red and 6 black balls. So total = 9 balls. We draw 1 ball from P and 2 balls from Q. We want the probability that among the three balls, two are black and one is red. Step 2: The favorable cases are: - Red from P and 2 blacks from Q - Black from P and (1 red + 1 black) from Q Case 1: Red from P, 2 blacks from Q \[ P(\text{Red from P}) = \frac{4}{9},
P(2 blacks from Q) = \frac{6}{9} . \frac{5}{8} \] \[ \Rightarrow \frac{4}{9} . \frac{6}{9} . \frac{5}{8} = \frac{120}{648} \] Case 2: Black from P, 1 red and 1 black from Q \[ P(\text{Black from P}) = \frac{5}{9},
P(\text{1 red and 1 black from Q}) = 2 . \frac{3}{9} . \frac{6}{8} \] \[ \Rightarrow \frac{5}{9} . \left( \frac{3}{9} . \frac{6}{8} . 2 \right) = \frac{5}{9} . \frac{36}{72} = \frac{180}{648} \] Step 3: Total probability = \( \frac{120 + 180}{648} = \frac{300}{648} = \frac{25}{54} \)