Question

Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from bag A, if the ball drawn is white, is:

• A. \( \frac{1}{4} \)
• B. \( \frac{1}{9} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{3}{10} \)

Answer 1

The Correct Option is C

Define events: \( E_1 \): Selection of Bag \( A \). \( E_2 \): Selection of Bag \( B \). \( E \): Drawing of a white ball.

Calculate probabilities: Probability of selecting Bag \( A \):

\[ P(E_1) = \frac{1}{2} \]

Probability of selecting Bag \( B \):

\[ P(E_2) = \frac{1}{2} \]

Probability of drawing a white ball given Bag \( A \) is selected:

\[ P(E|E_1) = \frac{3}{10} \]

Probability of drawing a white ball given Bag \( B \) is selected:

\[ P(E|E_2) = \frac{3}{5} \]

Apply Bayes’ theorem:

\[ P(E_1|E) = \frac{P(E|E_1) \times P(E_1)}{P(E|E_1) \times P(E_1) + P(E|E_2) \times P(E_2)} \]

Substitute values:

\[ P(E_1|E) = \frac{\frac{3}{10} \times \frac{1}{2}}{\frac{3}{10} \times \frac{1}{2} + \frac{3}{5} \times \frac{1}{2}} \]

Simplify:

\[ = \frac{\frac{3}{20}}{\frac{3}{20} + \frac{3}{10}} = \frac{\frac{3}{20}}{\frac{3}{20} + \frac{6}{20}} = \frac{3}{9} = \frac{1}{3} \]