A bag contains some white and some black balls, all combinations of balls being equally likely. The total number of balls in the bag is 10. If there ball are drawn at random without replacement and all of them are found to be black, the probability that eh bag contains 1 white and 9 black balls is `14//55` b. `12//55` c. `2//11` d. `8//55`

Options

A. `14//55`
B. `15//55`
C. `2//11`
D. `8//55`

Explanation

To solve the problem, we need to find the probability that the bag contains 1 white ball and 9 black balls given that 3 balls drawn at random without replacement are all black. We will use Bayes' theorem to find this probability.

### Step-by-Step Solution:

1. **Define the Events**:
- Let \( E_i \) be the event that there are \( i \) white balls in the bag, where \( i = 0, 1, 2, \ldots, 10 \). Therefore, the number of black balls will be \( 10 - i \).
- Let \( A \) be the event that 3 balls drawn are all black.

2. **Calculate the Total Probability of Drawing 3 Black Balls**:
- We need to calculate \( P(A | E_i) \) for each possible value of \( i \):
- If \( i = 1 \) (1 white, 9 black):
\[ P(A | E_1) = \frac{\binom{9}{3}}{\binom{10}{3}} = \frac{84}{120} = \frac{7}{10} \]
- If \( i = 0 \) (0 white, 10 black):
\[ P(A | E_0) = \frac{\binom{10}{3}}{\binom{10}{3}} = 1 \]
- If \( i = 2 \) (2 white, 8 black):
\[ P(A | E_2) = \frac{\binom{8}{3}}{\binom{10}{3}} = \frac{56}{120} = \frac{14}{30} = \frac{7}{15} \]
- If \( i = 3 \) (3 white, 7 black):
\[ P(A | E_3) = \frac{\binom{7}{3}}{\binom{10}{3}} = \frac{35}{120} = \frac{7}{24} \]
- Continuing this way, we can calculate \( P(A | E_i) \) for \( i = 4, 5, \ldots, 10 \).

3. **Calculate the Prior Probabilities**:
- Since all combinations of balls are equally likely, the prior probabilities \( P(E_i) \) for \( i = 0, 1, 2, \ldots, 10 \) are:
\[ P(E_i) = \frac{1}{11} \quad \text{for } i = 0, 1, 2, \ldots, 10 \]

4. **Use Bayes' Theorem**:
- We want to find \( P(E_1 | A) \):
\[ P(E_1 | A) = \frac{P(A | E_1) P(E_1)}{P(A)} \]
- To find \( P(A) \), we sum over all possible \( i \):
\[ P(A) = \sum_{i=0}^{10} P(A | E_i) P(E_i) \]
- Calculate \( P(A) \):
\[ P(A) = \frac{1}{11} \left( 1 + \frac{7}{10} + \frac{7}{15} + \frac{7}{24} + \ldots \right) \]
- After calculating all terms, we find \( P(A) \).

5. **Final Calculation**:
- Substitute back into Bayes' theorem:
\[ P(E_1 | A) = \frac{P(A | E_1) P(E_1)}{P(A)} \]
- Calculate the final value to find the probability that the bag contains 1 white and 9 black balls.

After performing all calculations, we find that the probability \( P(E_1 | A) \) is \( \frac{14}{55} \).

### Final Answer:
The probability that the bag contains 1 white and 9 black balls is \( \frac{14}{55} \).