An insurance sales representative selects 3 policies to review. The group of policies she can select from contains 8 life policies, 5 automobile policies, and 2 homeowner policies. Find the probability of selecting”
(a) All life policies
(b) Both homeowner policies
(c) All automobile policies
(d) 1of each policy
(e) 2 life policies and 1 automobile policy
I will formulate one problem to indicate the process and leave the rest to you.
(a) 3 life policies.
First 8/15, second (no replacement) 7/14, third 6/13.
Final probability for all found by multiplying the individual probabilities.
This should give you a start.
a. 12/65
To find the probability of selecting certain policies, we need to calculate the ratio of the number of desired outcomes to the total number of possible outcomes.
(a) To find the probability of selecting all life policies, we need to select all 3 policies from the 8 life policies available. The total number of possible outcomes is the number of ways to select 3 policies from the total pool of 15 policies (8 life policies + 5 automobile policies + 2 homeowner policies). We can use the combination formula to calculate this: C(n, r) = n! / (r!(n-r)!). Therefore, the probability of selecting all life policies is C(8, 3) / C(15, 3).
(b) To find the probability of selecting both homeowner policies, we need to select 2 policies from the 2 homeowner policies available. The total number of possible outcomes is the number of ways to select 2 policies from the total pool of 15 policies. Therefore, the probability of selecting both homeowner policies is C(2, 2) / C(15, 2).
(c) To find the probability of selecting all automobile policies, we need to select all 3 policies from the 5 automobile policies available. The total number of possible outcomes is the number of ways to select 3 policies from the total pool of 15 policies. Therefore, the probability of selecting all automobile policies is C(5, 3) / C(15, 3).
(d) To find the probability of selecting 1 of each policy, we need to select 1 life policy, 1 automobile policy, and 1 homeowner policy. The total number of possible outcomes is the number of ways to select 1 policy from each category: C(8, 1) * C(5, 1) * C(2, 1) / C(15, 3).
(e) To find the probability of selecting 2 life policies and 1 automobile policy, we need to select 2 policies from the 8 life policies available and 1 policy from the 5 automobile policies available. The total number of possible outcomes is the number of ways to select 3 policies from the total pool of 15 policies. Therefore, the probability of selecting 2 life policies and 1 automobile policy is C(8, 2) * C(5, 1) / C(15, 3).
Now you can substitute the values in the formula and calculate the probabilities for each scenario.