To find the probabilities related to the repair of troubles in the residential service, we can use the binomial probability formula. In this case, the probability of a trouble being repaired on the same day is 70% or 0.7, and the probability of it not being repaired on the same day is 30% or 0.3.
a. To find the probability that all five troubles will be repaired on the same day, we can use the binomial probability formula:
P(X = k) = nCk * p^k * (1-p)^(n-k)
Where:
n = total number of trials
k = number of successes
p = probability of success
(1-p) = probability of failure
For this case, n = 5, k = 5, p = 0.7, and (1-p) = 0.3.
P(all five repaired on the same day) = 5C5 * 0.7^5 * 0.3^(5-5)
P(all five repaired on the same day) = 1 * 0.7^5 * 0.3^0
P(all five repaired on the same day) = 0.7^5 * 1
P(all five repaired on the same day) = 0.16807
Therefore, the probability that all five troubles will be repaired on the same day is 0.16807 or 16.807%.
b. To find the probability that at least three troubles will be repaired on the same day, we need to sum the probabilities of three, four, and five successes.
P(at least three repaired on the same day) = P(X = 3) + P(X = 4) + P(X = 5)
P(at least three repaired on the same day) = 5C3 * 0.7^3 * 0.3^(5-3) + 5C4 * 0.7^4 * 0.3^(5-4) + 5C5 * 0.7^5 * 0.3^(5-5)
P(at least three repaired on the same day) = 10 * 0.7^3 * 0.3^2 + 5 * 0.7^4 * 0.3^1 + 1 * 0.7^5 * 0.3^0
P(at least three repaired on the same day) = 0.3087 + 0.36015 + 0.16807
P(at least three repaired on the same day) = 0.83692
Therefore, the probability that at least three troubles will be repaired on the same day is 0.83692 or 83.692%.
c. To find the probability that fewer than two troubles will be repaired on the same day, we need to sum the probabilities of zero and one success.
P(fewer than two repaired on the same day) = P(X = 0) + P(X = 1)
P(fewer than two repaired on the same day) = 5C0 * 0.7^0 * 0.3^(5-0) + 5C1 * 0.7^1 * 0.3^(5-1)
P(fewer than two repaired on the same day) = 1 * 0.7^0 * 0.3^5 + 5 * 0.7^1 * 0.3^4
P(fewer than two repaired on the same day) = 0.00243 + 0.03615
P(fewer than two repaired on the same day) = 0.03858
Therefore, the probability that fewer than two troubles will be repaired on the same day is 0.03858 or 3.858%.
d. To find the mean and standard deviation of this probability distribution, we can use the formulas:
Mean (μ) = n * p
Standard deviation (σ) = sqrt(n * p * (1-p))
For this case, n = 5 and p = 0.7.
Mean (μ) = 5 * 0.7 = 3.5
Standard deviation (σ) = sqrt(5 * 0.7 * 0.3) = sqrt(1.05) ≈ 1.0247
Therefore, the mean of this probability distribution is 3.5 and the standard deviation is approximately 1.0247.