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

Answer: (C)

$\frac{1}{3}$

Answer 2

Define events: $E_1$: Bag $A$ is selected. $E_2$: Bag $B$ is selected. $E$: A white ball is drawn.

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:

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

Using Bayes’ theorem:

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

Substituting 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:

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