Question

Assume there are 14 homes in the Quail Creek area and 5 of them have a security system. Three homes are selected at random.

A.) What is the probability all three of the selected homes have a security system? (Round to 4 decimal places)

B.) What is the probability none of the three selected homes has a security system?

C.) What is the probability at least one of the selected homes has a security system?

Answer 1

Total number of homes = 14

Homes with security system = 5

Homes without security system = 9

Number of homes selected = 3

a) Probability of all three homes have security system = 5C₃/ 14C₃

5C₃ = 5*4*3*2*1/(5-3)! 3! = (5*4*3*2*1)/((2*1)(2*3*1))

5C₃ = 10

14C₃ = 14! /(3! (14-3)!) = 364

Probability = 10/364= 0.027

b) Probability of three homes not having security system = 9C₃/14C₃

9C₃ = 9!/((9-3)!*3!) = 84

Probability = 84/364 = 0.23

c) Chances of selecting one home with security system = 5C₁ ways = 5

Chances of getting remaining homes without security system = 9C2 = 36

Probability = (5*36)/14C₃

Probability of at least one home with security system = (5*36)/(364) = 0.494