Assume that on an average one telephone number out of 15 called between 2 P.M. and 3 P.M. on week days is busy. What is the probability that if six randomly selected telephone numbers are called, at least three of them will be busy ?

Answer

To solve the problem, we need to find the probability that at least 3 out of 6 randomly selected telephone numbers are busy. We can model this situation using the binomial distribution.

### Step-by-Step Solution:

1. **Define the Probability of Busy and Not Busy Calls**:
- The probability \( P \) that a call is busy is given as \( P = \frac{1}{15} \).
- The probability \( Q \) that a call is not busy is \( Q = 1 - P = 1 - \frac{1}{15} = \frac{14}{15} \).

2. **Identify the Number of Trials**:
- The number of trials \( n \) (the number of calls made) is 6.

3. **Determine the Required Probability**:
- We need to find the probability of getting at least 3 busy calls, which can be expressed as:
\[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) \]

4. **Use the Binomial Probability Formula**:
- The binomial probability formula is:
\[ P(X = r) = \binom{n}{r} P^r Q^{n-r} \]
- Here, \( \binom{n}{r} \) is the binomial coefficient, \( P^r \) is the probability of success raised to the number of successes, and \( Q^{n-r} \) is the probability of failure raised to the number of failures.

5. **Calculate Each Probability**:
- For \( r = 3 \):
\[ P(X = 3) = \binom{6}{3} \left(\frac{1}{15}\right)^3 \left(\frac{14}{15}\right)^{3} \]
- For \( r = 4 \):
\[ P(X = 4) = \binom{6}{4} \left(\frac{1}{15}\right)^4 \left(\frac{14}{15}\right)^{2} \]
- For \( r = 5 \):
\[ P(X = 5) = \binom{6}{5} \left(\frac{1}{15}\right)^5 \left(\frac{14}{15}\right)^{1} \]
- For \( r = 6 \):
\[ P(X = 6) = \binom{6}{6} \left(\frac{1}{15}\right)^6 \left(\frac{14}{15}\right)^{0} \]

6. **Calculate the Binomial Coefficients**:
- \( \binom{6}{3} = 20 \)
- \( \binom{6}{4} = 15 \)
- \( \binom{6}{5} = 6 \)
- \( \binom{6}{6} = 1 \)

7. **Substitute and Calculate Each Probability**:
- \( P(X = 3) = 20 \left(\frac{1}{15}\right)^3 \left(\frac{14}{15}\right)^{3} \)
- \( P(X = 4) = 15 \left(\frac{1}{15}\right)^4 \left(\frac{14}{15}\right)^{2} \)
- \( P(X = 5) = 6 \left(\frac{1}{15}\right)^5 \left(\frac{14}{15}\right)^{1} \)
- \( P(X = 6) = 1 \left(\frac{1}{15}\right)^6 \)

8. **Combine the Probabilities**:
- Combine all the probabilities to find \( P(X \geq 3) \):
\[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) \]

9. **Final Calculation**:
- Calculate the values and sum them up:
\[ P(X \geq 3) = \frac{20 \cdot 14^3 + 15 \cdot 14^2 + 6 \cdot 14 + 1}{15^6} \]

### Final Result:
The final probability \( P(X \geq 3) \) can be computed as:
\[ P(X \geq 3) = \frac{57905}{15^6} \]