Assume there are 21 homes in the Quail Creek area and 9 of t

Assume there are 21 homes in the Quail Creek area and 9 of them have a security system. Four homes are selected at random:

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

What is the probability none of the four selected homes has a security system? (Round your answer to 4 decimal places.)

What is the probability at least one of the selected homes has a security system? (Round your answer to 4 decimal places.)

Solution

a) total 21 homes and 9 of them have security

probability of home having security = 9/21 = 3/7

probability of home not having security = 12/21 = 4/7

4 are selected and probability of having 4 security homes = 4C4*(3/7)^4*(4/7)^0 = 0.0337

B)NONE HAVE SECURITY = 4C0*(3/7)^0*(4/7)^4 = 0.1066

C) ATLEAST ONE HAS SECURITY = 1 - NONE HAVE SECURITY

NONE HAVE SECURITY IS WE FOUND IN PART B

HENCE PROBABILITY AT LEAST ONE HAVING SECURITY IS = 1-0.1066 = 0.8934