An insurance sales representative selects three policies to review.the group of policies he can select from contains eight life policies,five automobile policies and two homeowner's policies.find the probability of selecting:

I.all life policies.
II.both homeowner's policies.
III.one of each policy.

To find the probability of selecting different combinations of policies, we need to know the total number of possible outcomes and the number of favorable outcomes for each case.

Total number of policies available = 8 (life policies) + 5 (automobile policies) + 2 (homeowner's policies) = 15

I. Probability of selecting all life policies:
Number of outcomes: 3
Number of favorable outcomes: 8 (life policies)
P(I) = Number of favorable outcomes / Total number of outcomes
P(I) = 8C3 / 15C3

II. Probability of selecting both homeowner's policies:
Number of outcomes: 3
Number of favorable outcomes: 2 (homeowner's policies)
P(II) = Number of favorable outcomes / Total number of outcomes
P(II) = 2C2 / 15C3

III. Probability of selecting one of each policy:
Number of outcomes: 3
Number of favorable outcomes: 8 (for life policies) * 5 (for automobile policies) * 2 (for homeowner's policies)
P(III) = Number of favorable outcomes / Total number of outcomes
P(III) = (8C1 * 5C1 * 2C1) / 15C3

Simplifying the probabilities:
P(I) = 8C3 / 15C3 ≈ 0.0667
P(II) = 2C2 / 15C3 ≈ 0.0044
P(III) = (8C1 * 5C1 * 2C1) / 15C3 ≈ 0.1429

Therefore, the probabilities are:
I. P(I) ≈ 0.0667
II. P(II) ≈ 0.0044
III. P(III) ≈ 0.1429

To find the probability of selecting certain policies, we need to know the total number of possible outcomes and the number of favorable outcomes.

I. Probabilty of selecting all life policies:
There are 8 life policies to choose from, and the representative needs to select 3 of them. To calculate the probability, we use the concept of combinations.

Total number of possible outcomes = Total life policies available = 8
Number of favorable outcomes = Number of ways to select 3 life policies from 8 = C(8, 3) = 8! / (3!(8-3)!) = 8! / (3!5!) = (8*7*6) / (3*2*1) = 56

Probability of selecting all life policies = Number of favorable outcomes / Total number of possible outcomes
= 56 / 8
= 7/1
= 7

So, the probability of selecting all life policies is 7/1 or simply 7.

II. Probability of selecting both homeowner's policies:
There are 2 homeowner's policies in the selection, and the representative needs to select both of them. Since the representative needs to select the specific two policies, the probability is simply the number of favorable outcomes divided by the total number of possible outcomes.

Total number of possible outcomes = Total homeowner's policies available = 2
Number of favorable outcomes = Number of ways to select 2 homeowner's policies from 2 = C(2, 2) = 2! / (2!(2-2)!) = 2! / (2!0!) = (2*1) / (2*1) = 1

Probability of selecting both homeowner's policies = Number of favorable outcomes / Total number of possible outcomes
= 1 / 2
= 1/2

So, the probability of selecting both homeowner's policies is 1/2.

III. Probability of selecting one of each policy:
To find the probability of selecting one of each policy, we need to calculate the total number of possible outcomes and the number of favorable outcomes.

Total number of possible outcomes = Total number of combinations of selecting policies from each category = C(8,1) * C(5,1) * C(2,1) = (8 * 5 * 2) = 80

Number of favorable outcomes = Number of ways to select 1 life policy from 8 * Number of ways to select 1 automobile policy from 5 * Number of ways to select 1 homeowner's policy from 2 = (C(8,1) * C(5,1) * C(2,1)) = (8 * 5 * 2) = 80

Probability of selecting one of each policy = Number of favorable outcomes / Total number of possible outcomes
= 80 / 80
= 1

So, the probability of selecting one of each policy is 1, which means it is certain to happen.