Sandy's pet center grooms large and small dogs. It takes Sandy 40 minutes to groom a small dog and 70 minutes to groom a large dog. Large dogs account for 20% of Sandy's business. Sandy has 5 appointments tomorrow. a. What is the probability that all 5 appointments tomorrow are for small dogs? b. What is the probability that two of the appointments tomorrow are for large dogs? c. What is the expected amount of time to finish all five dogs tomorrow?

(a)  The probability that all 5 appointments tomorrow are for small dogs is 0.3277

(b) The probability that two of the appointments tomorrow are for large dogs is 0.2048

(c) The expected amount of time to finish all five dogs tomorrow is 230 minutes

Step-by-step explanation:

Let L denote large dogs and S denote small dogs. The probabilities given are:

P(L) = 20% = 0.2

P(S) = 1 - 0.2 = 0.8

(a) We need to compute the probability that all 5 dogs that have an appointment tomorrow are small. So,

P(S=5) = 0.8 * 0.8 * 0.8 * 0.8 * 0.8

           = 0.32768

P(S=5) = 0.3277

(b) We need to compute the probability that two of the appointments are for large dogs. Using the binomial distribution formula:

P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ

Where n = total no. of trials

           x = no. of successful trials

           p = probability of success

           q = probability of failure = 1-p

P(L=2) = ⁵C₂ (0.2)² (0.8)³

           = (10)*(0.04)*(0.512)

P(L=2) = 0.2048

(c) It takes 40 minutes to groom a small dog with a probability of 0.8 and 70 minutes to groom a large dog with a probability of 0.2.

The expected value of the amount of time to finish one dog can be computed as:

E(x) = ∑x*P(x)

       = 40(0.8) + 70(0.2)

E(x) = 46 minutes

The expected time to finish all 5 dogs = 5 * E(x)

                                                                = 5 * 46

The expected time to finish all 5 dogs = 230 minutes