On average, every one out of 10 telephones is found busy. Six telephone numbers are selected at random. Find the probability that four of them will be busy.

To find the probability that four out of six telephone numbers will be busy, we can use the concept of binomial probability.

The probability of a telephone being busy is 1 out of 10, or 1/10. Therefore, the probability of a telephone not being busy is 9 out of 10, or 9/10.

To calculate the probability of four out of six telephones being busy, we can use the binomial probability formula:

P(X=k) = (nCk) * p^k * (1-p)^(n-k)

Where: P(X=k) is the probability of getting exactly k successes (in this case, k busy telephones) n is the total number of trials (in this case, 6 telephone numbers selected) k is the number of successful outcomes (in this case, 4 busy telephones) p is the probability of success (in this case, 1/10)

Using this formula, we can calculate the probability as follows:

P(X=4) = (6C4) * (1/10)^4 * (9/10)^(6-4)

Calculating the values:

(6C4) = 6! / (4! * (6-4)!) = 6! / (4! * 2!) = (6 * 5) / (2 * 1) = 15

(1/10)^4 = 1/10 * 1/10 * 1/10 * 1/10 = 1/10,000

(9/10)^(6-4) = (9/10)^2 = 81/100

Now, we can substitute these values into the formula:

P(X=4) = 15 * (1/10,000) * (81/100) = 0.01215

Therefore, the probability that four out of six telephone numbers will be busy is approximately 0.01215, or 1.215%.