A bag contains 20 coins. If the probability that the bag contains exactly 4 biased coin is 3/4 and that of exactly 5 biased coin is 2/3, then the probability that all the biased coin are sorted out from bag is exactly 10 draws is `5/(10)(^(16)C_6)/(^(20)C_9)+1/(11)(^(15)C_5)/(^(20)C_9)` b. `2/(33)[(^(16)C_6+5^(15)C_5)/(^(20)C_9)]` c. `5/(33)(^(16)C_7)/(^(20)C_9)+1/(11)(^(15)C_6)/(^(20)C_9)` d. none of these

Options

A. `5/33(""^(16)C_(6))/(""^(20)C_(9))+1/11(""^(15)C_(5))/(""^(20)C_(9))`
B. `2/33[(""^(16)C_(6)+5^(""^(15))C_(5))/(""^(20)C_(6))]`
C. `5/33(""^(16)C_(7))/(""^(20)C_(9))+1/11(""^(15)C_(6))/(""^(20)C_(9))`
D. None of these

Explanation

To solve the problem, we need to find the probability that all biased coins are sorted out from a bag containing 20 coins, given certain probabilities for the number of biased coins. Let's break down the solution step by step.

### Step 1: Understand the Given Information
We have:
- A bag containing 20 coins.
- The probability that the bag contains exactly 4 biased coins is \( P(4) = \frac{3}{4} \).
- The probability that the bag contains exactly 5 biased coins is \( P(5) = \frac{2}{3} \).

### Step 2: Determine the Probabilities
Since the events of having exactly 4 biased coins and exactly 5 biased coins are mutually exclusive, we can add their probabilities:
\[ P(4) + P(5) = \frac{3}{4} + \frac{2}{3} \]
To add these fractions, we need a common denominator:
- The least common multiple of 4 and 3 is 12.
- Convert the fractions:
\[ P(4) = \frac{3}{4} = \frac{9}{12}, \quad P(5) = \frac{2}{3} = \frac{8}{12} \]
Thus,
\[ P(4) + P(5) = \frac{9}{12} + \frac{8}{12} = \frac{17}{12} \]
This indicates that the probabilities provided are not consistent as they exceed 1. However, we will proceed with the calculations based on the given probabilities.

### Step 3: Calculate the Probability for 10 Draws
We need to find the probability that all biased coins are sorted out in exactly 10 draws. We will consider two cases: when there are 4 biased coins and when there are 5 biased coins.

#### Case 1: Exactly 4 Biased Coins
- We need to choose 3 biased coins from the 4 in the first 9 draws, and the last biased coin must be drawn in the 10th draw.
- The number of ways to choose 3 biased coins from 4 is \( \binom{4}{3} \).
- The remaining coins are 16 (20 total - 4 biased), and we need to choose 6 from these 16.
- Thus, the probability for this case is:
\[ P(4) \cdot \frac{\binom{4}{3} \cdot \binom{16}{6}}{\binom{20}{9}} \cdot \frac{1}{11} \]

#### Case 2: Exactly 5 Biased Coins
- We need to choose 4 biased coins from the 5 in the first 9 draws, and the last biased coin must be drawn in the 10th draw.
- The number of ways to choose 4 biased coins from 5 is \( \binom{5}{4} \).
- The remaining coins are 15 (20 total - 5 biased), and we need to choose 5 from these 15.
- Thus, the probability for this case is:
\[ P(5) \cdot \frac{\binom{5}{4} \cdot \binom{15}{5}}{\binom{20}{9}} \cdot \frac{1}{11} \]

### Step 4: Combine the Probabilities
Now, we combine the probabilities from both cases:
\[ P(\text{all biased coins sorted out}) = P(4) \cdot \frac{\binom{4}{3} \cdot \binom{16}{6}}{\binom{20}{9}} \cdot \frac{1}{11} + P(5) \cdot \frac{\binom{5}{4} \cdot \binom{15}{5}}{\binom{20}{9}} \cdot \frac{1}{11} \]

### Step 5: Substitute the Values
Substituting the values of \( P(4) \) and \( P(5) \):
\[ = \frac{3}{4} \cdot \frac{4 \cdot \binom{16}{6}}{\binom{20}{9}} \cdot \frac{1}{11} + \frac{2}{3} \cdot \frac{5 \cdot \binom{15}{5}}{\binom{20}{9}} \cdot \frac{1}{11} \]

### Step 6: Simplify the Expression
Now, we can simplify the expression:
\[ = \frac{3 \cdot \binom{16}{6}}{44 \cdot \binom{20}{9}} + \frac{10 \cdot \binom{15}{5}}{33 \cdot \binom{20}{9}} \]

### Final Answer
The final expression can be written as:
\[ = \frac{2}{33} \cdot \frac{(5 \cdot \binom{15}{5} + 3 \cdot \binom{16}{6})}{\binom{20}{9}} \]

### Conclusion
Thus, the correct answer is:
**Option B: \( \frac{2}{33} \left( \frac{(16C6 + 5 \cdot 15C5)}{20C9} \right) \)**