On dialling certain telephone numbers, assume that on an average one telephone number out of five is busy, ten telephone numbers are randomly selected and dialled. Find the probability that at least threeof them will be busy

Answer

To solve the problem, we will use the binomial probability formula. The problem states that on average, one telephone number out of five is busy. We need to find the probability that at least three out of ten randomly selected telephone numbers are busy.

### Step-by-Step Solution:

1. **Define the Probability of Busy and Not Busy:**
- Let \( p \) be the probability that a telephone number is busy. Since one out of five is busy, we have:
\[ p = \frac{1}{5} = 0.2 \]
- The probability that a telephone number is not busy, denoted as \( q \), is:
\[ q = 1 - p = 1 - \frac{1}{5} = \frac{4}{5} = 0.8 \]

2. **Identify the Number of Trials:**
- We are selecting \( n = 10 \) telephone numbers.

3. **Define the Required Probability:**
- We need to find the probability that at least 3 of them will be busy. This can be expressed as:
\[ P(X \geq 3) = 1 - P(X < 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2)) \]

4. **Calculate \( P(X = k) \) Using the Binomial Formula:**
- The binomial probability formula is given by:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
- We will calculate \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \).

5. **Calculate \( P(X = 0) \):**
\[ P(X = 0) = \binom{10}{0} (0.2)^0 (0.8)^{10} = 1 \cdot 1 \cdot (0.8)^{10} = 0.1073741824 \]

6. **Calculate \( P(X = 1) \):**
\[ P(X = 1) = \binom{10}{1} (0.2)^1 (0.8)^{9} = 10 \cdot 0.2 \cdot (0.8)^{9} = 10 \cdot 0.2 \cdot 0.134217728 = 0.268435456 \]

7. **Calculate \( P(X = 2) \):**
\[ P(X = 2) = \binom{10}{2} (0.2)^2 (0.8)^{8} = 45 \cdot (0.2)^2 \cdot (0.8)^{8} = 45 \cdot 0.04 \cdot 0.16777216 = 0.301989888 \]

8. **Sum the Probabilities for \( P(X < 3) \):**
\[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \]
\[ P(X < 3) = 0.1073741824 + 0.268435456 + 0.301989888 = 0.6777995264 \]

9. **Calculate the Final Probability:**
\[ P(X \geq 3) = 1 - P(X < 3) = 1 - 0.6777995264 = 0.3222004736 \]

### Final Answer:
The probability that at least three of the ten telephone numbers will be busy is approximately:
\[ \boxed{0.3222} \]