Question

Bag \( B_1 \) contains 4 white and 2 black balls. Bag \( B_2 \) contains 3 white and 4 black balls. A bag is chosen at random and a ball is drawn from it at random, then the probability that the ball drawn is white, is:

• A. \( \frac{1}{42} \)
• B. \( \frac{42}{32} \)
• C. \( \frac{33}{42} \)
• D. \( \frac{23}{42} \)

Hint:

For probability problems with multiple outcomes, use the law of total probability by considering all possible events and their corresponding probabilities.

Answer 1

The Correct Option is D

We are given two bags, \( B_1 \) and \( B_2 \). Bag \( B_1 \) contains 4 white and 2 black balls, and Bag \( B_2 \) contains 3 white and 4 black balls. A bag is chosen randomly, and a ball is drawn from it. We need to find the probability that the ball drawn is white. The probability of choosing Bag \( B_1 \) is \( \frac{1}{2} \) and similarly, the probability of choosing Bag \( B_2 \) is also \( \frac{1}{2} \). Now, the probability of drawing a white ball from Bag \( B_1 \) is: \[ P(\text{white from } B_1) = \frac{4}{6} = \frac{2}{3} \] And the probability of drawing a white ball from Bag \( B_2 \) is: \[ P(\text{white from } B_2) = \frac{3}{7} \] The total probability of drawing a white ball is the sum of the probabilities of drawing a white ball from each bag: \[ P(\text{white}) = \left( \frac{1}{2} \times \frac{2}{3} \right) + \left( \frac{1}{2} \times \frac{3}{7} \right) \] \[ P(\text{white}) = \frac{1}{2} \times \left( \frac{2}{3} + \frac{3}{7} \right) \] \[ P(\text{white}) = \frac{1}{2} \times \frac{23}{21} = \frac{23}{42} \] Thus, the probability that the ball drawn is white is \( \frac{23}{42} \).