48% of all pets sold at a pet store are dogs. One day, 9 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 only has 5 dogs, what is the probability that they have sufficient dogs for sale that day?

Answer 1

(a) By the PMF for the binomial distribution, the desired probability is C(9,3) (0.48)³ (1-0.48)⁶, or about 18.366%.

(b) We require the probability that at most 5 of the 9 customers will elect to purchase a dog. By the PMF for the binomial distribution. this is

C(9,0)(0.48)⁰(1-0.48)⁹

+ C(9,1)(0.48)¹(1-0.48)⁸

+ C(9,2)(0.48)²(1-0.48)⁷

+C(9,3)(0.48)³ (1-0.48)⁶

+C(9,4)(0.48)⁴(1-0.48)⁵

+C(9,5)(0.48)⁵(1-0.48)⁴

= 78.386%

Hope this helps!