1. **Problem statement:** We are given that on average, 1 out of 10 telephones is busy, so the probability a telephone is busy is $p=\frac{1}{10}=0.1$. We select 6 telephones at random and want to find the probability that exactly 4 of them are busy. 2. **Formula used:** This is a binomial probability problem. The probability of exactly $k$ successes in $n$ independent trials with success probability $p$ is given by: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. 3. **Apply values:** Here, $n=6$, $k=4$, $p=0.1$, and $1-p=0.9$. 4. **Calculate binomial coefficient:** $$\binom{6}{4} = \frac{6!}{4!2!} = \frac{720}{24 \times 2} = \frac{720}{48} = 15$$ 5. **Calculate probability:** $$P(X=4) = 15 \times (0.1)^4 \times (0.9)^2$$ 6. **Calculate powers:** $$(0.1)^4 = 0.0001$$ $$(0.9)^2 = 0.81$$ 7. **Multiply all terms:** $$P(X=4) = 15 \times 0.0001 \times 0.81 = 15 \times 0.000081 = 0.001215$$ 8. **Final answer:** The probability that exactly 4 out of 6 telephones are busy is approximately $0.001215$ or about 0.12%.