Question

11) Forty percent of the homes constructed in the Quail Creek area include a security system. Three homes are selected at random: 1. What is the probability all three of the selected homes have a security system? 2. What is the probability none of the three selected homes have a security system? 3. Did you assume the events to be dependent or independent?

Answer 1

The probability all three homes have a security system is 0.064, the probability none have it is 0.216, and the events were assumed to be independent.

Step-by-step explanation:

The question pertains to probability theory, specifically calculating the probability of independent events occurring consecutively. We're focusing on the scenario where 40% of homes in Quail Creek have a security system and we are selecting three homes at random.

1. The probability that all three of the selected homes have a security system can be calculated by multiplying the individual probabilities for each home having a system since the events are independent. This is 0.40 * 0.40 * 0.40, which equals to 0.064.
2. The probability that none of the three selected homes have a security system is calculated by considering the probability that a home does not have a security system, which is 1 - 0.40. Therefore, the probability is (1 - 0.40)^3, which equals to 0.216.
3. In answering questions 1 and 2, I assumed the events to be independent because the selection of one home does not influence the likelihood of a security system being in another home.

Answer 2

1) Pr(all three) = 0.064

2) Pr(none) = 0.216

3) Independent event

Step-by-step explanation:

Let Probability of homes constructed in the Quail Creek area with a security system = Pr (security)

Where Pr = probability

This is a binomial distribution problem. There ate only two outcomes in this distribution: a success and a failure

Where p = success, q = failure

For n trials,

Pr(X = x) = n!/[(n-x)!x!] × p^x × q^(n-x)

Pr (security) = 40% = 0.4

p = 0.4

q = 1-p = 1-0.4 = 0.6

Numbers selected at random = n = 3

See attachment for more details on the workings.

1) Probability all three of the selected homes have a security system = Pr(all three)

Pr(all three) = 1× (0.4)³ (0.6)^0 = 0.4³

= 0.4×0.4×0.4

Pr(all three) = 0.064

2) Probability none of the three selected homes have a security system

= Pr(none)

Pr(none) = 1 × (0.4)^0 × (0.6)³

= 0.6³ = 0.6×0.6×0.6

Pr(none) = 0.216

3) The events were assumed to be independent as the selection of one house doesn't affect the outcome of the selection of another.