55% of all pets sold at a pet store are dogs. One day, 10 people arrive at the store and purchase pets. (Note: One pet is purchased per person.)

Question

(a) What is the probability that exactly 3 of them are dogs? (b) If the store has only 7 dogs, what is the probability that they have sufficient dogs for sale that day?

Answer 1

a) Let us first calculate the probability that the first three purchases are dogs, and the next 4 are not dogs:

p = (.55)³ * (1-.55)⁷ = (.55)³ * (.45)⁷

Now we must count how many ways that exactly 3 dogs can be purchased among 10 pets:

C(10,3) = 10! / (3! * 7! ) = 10 * 9 * 8 / (2 * 3) = 120

Our answer is

C(10,3) * (.55)³ * (.45)⁷ = 120 * 0.0068224 = 0.81689 = 81.689%

b) Let P(n) = Probability that n dogs will be purchased on the day in question.

P(n) = C(10,n) * (0.55)ⁿ * (0.45)^10-n

The answer will be the sum P(0) + P(1) + P(2) + ... + P(7).

Or, we can find Complimentary probability, then subtracting it from 1. The Complimentary probability is the probability of there being 8, 9 , or 10 dog buyers, which is equal to the sum P(8) + P(9) + P(10).

P(8) = 7.63%

P(9) = 2.07%

P(10)=0.25%

Sum = 9.95%

1-Sum = 90.05%, ans.