Question

​​​​​​1. An important part of the customer service responsibilities of a cable company relates to the speed with which trouble in service can be repaired. Historically, the data show that the likelihood is 0.85 that troubles in a residential service can be repaired on the same day. Use the binomial formula to calculate.

c. For the first five troubles reported on a given day, what is the probability that: At least 3 troubles will be repaired on the same day.

Answer 1

Binomial Distribution

If 'X' is the random variable representing the number of successes, the probability of getting ‘r’ successes and ‘n-r’ failures, in 'n' trails, ‘p’ probability of success ‘q’=(1-p) is given by the probability function

$\small P(X=r) = \binom{n}{r}p^{r}q^{n-r}$

p : Probability that a trouble will be repaired on the same day = 0.85

q = 1-p = 1-0.85=0.15

n : Number of troubles reported on a given day : n= 5

X : Number of troubles will be repaired on the same day.

X follows a Binomial with n= 5 and p =0.85

And the probability function of X is

$\small P(X=r) = \binom{5}{r}(0.85)^{r}(0.15)^{5-r}$

Probability that: At least 3 troubles will be repaired on the same day = P(X$\small \geq$3) = P(X=3)+P(X=4)+P(X=5)

$\small P(X=3)=\binom{5}{3} \times0.85^{ 3} \times0.15^{2}=10\times0.614125\times0.0225=0.1382$

$\small P(X=4)=\binom{5}{4} \times0.85^{ 4} \times0.15^{1}=5\times0.52200625\times0.15=0.3915$

$\small P(X=5)=\binom{5}{5} \times0.85^{ 5} \times0.15^{0}=1\times0.4437053125\times1=0.4437$

P(X$\small \geq$3) = P(X=3)+P(X=4)+P(X=5) = 0.1382 + 0.3915 + 0.4437 = 0.9734

Probability that: At least 3 troubles will be repaired on the same day = 0.9734