The probabilities of `A`, `B` and `C` solving a problem independently are respectively `(1)/(4)`, `(1)/(5)`, `(1)/(6)`. If `21` such problems are given to `A`, `B` and `C` then the probability that at least `11` problems can be solved by them is

Options

A. `"^(21)C_(11)((1)/(2))^(11)`
B. `(1)/(2)`
C. `((1)/(2))^(11)`
D. none of these

Explanation

To solve the problem, we need to find the probability that at least 11 problems can be solved by A, B, and C, given their individual probabilities of solving a problem. Let's break it down step by step.

### Step 1: Determine the probabilities of not solving the problem
- Probability of A solving a problem, \( P(A) = \frac{1}{4} \)
- Probability of A not solving a problem, \( P(A') = 1 - P(A) = 1 - \frac{1}{4} = \frac{3}{4} \)

- Probability of B solving a problem, \( P(B) = \frac{1}{5} \)
- Probability of B not solving a problem, \( P(B') = 1 - P(B) = 1 - \frac{1}{5} = \frac{4}{5} \)

- Probability of C solving a problem, \( P(C) = \frac{1}{6} \)
- Probability of C not solving a problem, \( P(C') = 1 - P(C) = 1 - \frac{1}{6} = \frac{5}{6} \)

### Step 2: Calculate the probability of at least one of them solving the problem
To find the probability that at least one of A, B, or C solves the problem, we can use the formula:
\[ P(A \cup B \cup C) = 1 - P(A' \cap B' \cap C') \]
Since A, B, and C are independent, we have:
\[ P(A' \cap B' \cap C') = P(A') \cdot P(B') \cdot P(C') \]
Calculating this gives:
\[ P(A') \cdot P(B') \cdot P(C') = \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \]
Calculating this step-by-step:
- First, \( \frac{3}{4} \cdot \frac{4}{5} = \frac{3}{5} \)
- Then, \( \frac{3}{5} \cdot \frac{5}{6} = \frac{3}{6} = \frac{1}{2} \)

Thus,
\[ P(A' \cap B' \cap C') = \frac{1}{2} \]
Now, substituting back:
\[ P(A \cup B \cup C) = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 3: Set up the binomial distribution
Let \( p = P(A \cup B \cup C) = \frac{1}{2} \) be the probability of success (solving a problem), and \( q = 1 - p = \frac{1}{2} \) be the probability of failure (not solving a problem).

We have \( n = 21 \) problems, and we want to find the probability of solving at least 11 problems:
\[ P(X \geq 11) = P(X = 11) + P(X = 12) + ... + P(X = 21) \]

Using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
Thus,
\[ P(X \geq 11) = \sum_{k=11}^{21} \binom{21}{k} \left(\frac{1}{2}\right)^{21} \]

### Step 4: Simplify the sum
Factoring out \( \left(\frac{1}{2}\right)^{21} \):
\[ P(X \geq 11) = \left(\frac{1}{2}\right)^{21} \sum_{k=11}^{21} \binom{21}{k} \]

### Step 5: Use the binomial theorem
The sum of binomial coefficients can be simplified using the binomial theorem:
\[ \sum_{k=0}^{n} \binom{n}{k} = 2^n \]
Thus,
\[ \sum_{k=0}^{21} \binom{21}{k} = 2^{21} \]
Using symmetry in binomial coefficients:
\[ \sum_{k=11}^{21} \binom{21}{k} = \sum_{k=0}^{10} \binom{21}{k} = \frac{1}{2} \cdot 2^{21} = 2^{20} \]

### Step 6: Final probability calculation
Substituting back:
\[ P(X \geq 11) = \left(\frac{1}{2}\right)^{21} \cdot 2^{20} = \frac{2^{20}}{2^{21}} = \frac{1}{2} \]

Thus, the final probability that at least 11 problems can be solved by A, B, and C is:
\[ \boxed{\frac{1}{2}} \]