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 the bag A,if the ball drawn is white,is

Options

A. `1/3`
B. `1/9`
C. `3/10`
D. `1/4`

Answer

To find the probability of drawing a ball from Bag A given that the ball drawn is white, we can use Bayes' theorem. Let's denote the events as follows:

- Let \( E_1 \) be the event that Bag A is chosen.
- Let \( E_2 \) be the event that Bag B is chosen.
- Let \( W \) be the event that a white ball is drawn.

We need to find \( P(E_1 | W) \), the probability of having chosen Bag A given that a white ball is drawn.

### Step-by-step Solution:

1. **Identify the total number of balls in each bag:**
- Bag A contains 3 white and 7 red balls, so the total number of balls in Bag A is \( 3 + 7 = 10 \).
- Bag B contains 3 white and 2 red balls, so the total number of balls in Bag B is \( 3 + 2 = 5 \).

2. **Calculate the prior probabilities of choosing each bag:**
- Since one bag is selected at random, the probability of choosing Bag A is:
\[ P(E_1) = \frac{1}{2} \]
- The probability of choosing Bag B is:
\[ P(E_2) = \frac{1}{2} \]

3. **Calculate the probability of drawing a white ball from each bag:**
- The probability of drawing a white ball from Bag A is:
\[ P(W | E_1) = \frac{3}{10} \]
- The probability of drawing a white ball from Bag B is:
\[ P(W | E_2) = \frac{3}{5} \]

4. **Use the law of total probability to find \( P(W) \):**
- The total probability of drawing a white ball is given by:
\[ P(W) = P(W | E_1) \cdot P(E_1) + P(W | E_2) \cdot P(E_2) \]
- Substituting the values we calculated:
\[ P(W) = \left(\frac{3}{10} \cdot \frac{1}{2}\right) + \left(\frac{3}{5} \cdot \frac{1}{2}\right) \]
- Simplifying this:
\[ P(W) = \frac{3}{20} + \frac{3}{10} = \frac{3}{20} + \frac{6}{20} = \frac{9}{20} \]

5. **Apply Bayes' theorem to find \( P(E_1 | W) \):**
- Bayes' theorem states:
\[ P(E_1 | W) = \frac{P(W | E_1) \cdot P(E_1)}{P(W)} \]
- Substituting the known values:
\[ P(E_1 | W) = \frac{\left(\frac{3}{10}\right) \cdot \left(\frac{1}{2}\right)}{\frac{9}{20}} \]
- Simplifying this:
\[ P(E_1 | W) = \frac{\frac{3}{20}}{\frac{9}{20}} = \frac{3}{9} = \frac{1}{3} \]

### Final Answer:
The probability of drawing the ball from Bag A, given that the ball drawn is white, is \( \frac{1}{3} \).