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: \( E_1 \): 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:

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

Answer 2

To find the probability of drawing a white ball from Bag A given that a white ball has been drawn, we can use Bayes' Theorem. Let's analyze the problem step-by-step:

1. Let \( A_1 \) be the event of selecting Bag A and \( A_2 \) be the event of selecting Bag B.
2. The probability of selecting any bag randomly is: \(P(A_1) = P(A_2) = \frac{1}{2}\)
3. Let \( W \) be the event of drawing a white ball.
4. The probability of drawing a white ball from Bag A is: \(P(W | A_1) = \frac{3}{10}\) because Bag A has 3 white balls out of a total of 10 balls (3 white + 7 red).
5. The probability of drawing a white ball from Bag B is: \(P(W | A_2) = \frac{3}{5}\) because Bag B has 3 white balls out of a total of 5 balls (3 white + 2 red).
6. Using the law of total probability, the total probability of drawing a white ball is: \(P(W) = P(W | A_1) \cdot P(A_1) + P(W | A_2) \cdot P(A_2)\)

Substitute the known values:

1. \(P(W) = \left(\frac{3}{10} \times \frac{1}{2}\right) + \left(\frac{3}{5} \times \frac{1}{2}\right)\)

Simplifying, we get:

1. \(P(W) = \frac{3}{20} + \frac{3}{10} = \frac{3}{20} + \frac{6}{20} = \frac{9}{20}\)
2. Using Bayes' Theorem, the probability that the ball was drawn from Bag A given that it is white is: \(P(A_1 | W) = \frac{P(W | A_1) \cdot P(A_1)}{P(W)}\)

Substitute the known values:

1. \(P(A_1 | W) = \frac{\left(\frac{3}{10} \times \frac{1}{2}\right)}{\frac{9}{20}}\)

Simplifying, we get:

1. \(P(A_1 | W) = \frac{\frac{3}{20}}{\frac{9}{20}} = \frac{3}{20} \times \frac{20}{9} = \frac{3}{9} = \frac{1}{3}\)

Thus, the probability of drawing the ball from Bag A, given that the drawn ball is white, is \(\frac{1}{3}\).