A bag contains 5 white balls, 3 black balls, 4 yellow balls. A ball is drawn from the bag, its colour is noted and put back into the bag with 5 additional balls of the same colour. The process is repeated. The probability that a yellow ball is drawn in the `1^("st")` draw given that a white ball is drawn in the `2^("nd")` draw is

Options

A. `(9)/(55)`
B. `(4)/(17)`
C. `(1)/(5)`
D. `(2)/(11)`

Answer

To solve the problem, we need to find the probability that a yellow ball is drawn in the first draw given that a white ball is drawn in the second draw. We will use the concept of conditional probability.

Let:
- \( A \) be the event that a yellow ball is drawn in the first draw.
- \( B \) be the event that a white ball is drawn in the second draw.

We are looking for \( P(A | B) \), which can be calculated using Bayes' theorem:

\[ P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} \]

### Step 1: Calculate \( P(A) \)
The total number of balls in the bag initially is:
- White balls = 5
- Black balls = 3
- Yellow balls = 4

Total balls = \( 5 + 3 + 4 = 12 \)

The probability of drawing a yellow ball first is:
\[ P(A) = \frac{4}{12} = \frac{1}{3} \]

**Hint:** To find the probability of an event, divide the number of favorable outcomes by the total number of outcomes.

### Step 2: Calculate \( P(B | A) \)
If a yellow ball is drawn first, we add 5 yellow balls back to the bag. The new composition of the bag will be:
- White balls = 5
- Black balls = 3
- Yellow balls = \( 4 + 5 = 9 \)

Total balls now = \( 5 + 3 + 9 = 17 \)

Now, the probability of drawing a white ball in the second draw is:
\[ P(B | A) = \frac{5}{17} \]

**Hint:** After drawing a ball and replacing it with more of the same color, recalculate the total number of balls and the number of each color.

### Step 3: Calculate \( P(B) \)
To find \( P(B) \), we need to consider all possible scenarios for the first draw:

1. **First draw is white:**
- Probability = \( P(W) = \frac{5}{12} \)
- New composition: White = 10, Black = 3, Yellow = 4
- Total = 17
- Probability of drawing white second = \( P(B | W) = \frac{10}{17} \)

2. **First draw is black:**
- Probability = \( P(B) = \frac{3}{12} \)
- New composition: White = 5, Black = 8, Yellow = 4
- Total = 17
- Probability of drawing white second = \( P(B | B) = \frac{5}{17} \)

3. **First draw is yellow:**
- Probability = \( P(Y) = \frac{4}{12} \)
- New composition: White = 5, Black = 3, Yellow = 9
- Total = 17
- Probability of drawing white second = \( P(B | Y) = \frac{5}{17} \)

Now we can calculate \( P(B) \):
\[ P(B) = P(W) \cdot P(B | W) + P(B) \cdot P(B | B) + P(Y) \cdot P(B | Y) \]
\[ P(B) = \left(\frac{5}{12} \cdot \frac{10}{17}\right) + \left(\frac{3}{12} \cdot \frac{5}{17}\right) + \left(\frac{4}{12} \cdot \frac{5}{17}\right) \]
\[ = \frac{50}{204} + \frac{15}{204} + \frac{20}{204} = \frac{85}{204} \]

**Hint:** To find the total probability of an event, consider all possible ways that event can occur and sum their probabilities.

### Step 4: Calculate \( P(A | B) \)
Now we can substitute back into Bayes' theorem:
\[ P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} = \frac{\frac{5}{17} \cdot \frac{1}{3}}{\frac{85}{204}} \]
\[ = \frac{\frac{5}{51}}{\frac{85}{204}} = \frac{5 \cdot 204}{51 \cdot 85} = \frac{1020}{4335} \]
Simplifying gives:
\[ = \frac{4}{17} \]

Thus, the final answer is:
\[ \boxed{\frac{4}{17}} \]