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.
(base on probability and combinatorial analysis )
To find the probabilities of the different scenarios, we need to use combinatorial analysis.
I. Probability of selecting all life policies:
Since there are 8 life policies and we need to select all 3, we can use combinations to calculate this probability. The formula for combinations is nCr = (n!)/(r!(n-r)!). So the probability of selecting all life policies is 8C3 / (15C3), where 15 is the total number of policies available.
II. Probability of selecting both homeowner's policies:
Since there are 2 homeowner's policies and we need to select both, the probability would be calculated as 2C2 / (15C3), where 15 is the total number of policies available.
III. Probability of selecting one of each policy:
To calculate this probability, we need to consider the number of ways to select one life policy, one automobile policy, and one homeowner's policy. The number of ways to select one life policy from 8 choices is 8C1, the number of ways to select one automobile policy from 5 choices is 5C1, and the number of ways to select one homeowner's policy from 2 choices is 2C1. Therefore, the probability is (8C1 * 5C1 * 2C1) / (15C3).
Note: In all the above calculations, nCr represents the number of combinations of choosing r objects from a set of n objects.
You can use these formulas and calculate the probabilities based on the specific values in your question to get the exact probabilities in decimal or fractional form.
To find the probability of selecting specific policies, we need to determine the total number of possible outcomes and the number of desired outcomes.
I. Probability of selecting all life policies:
There are a total of 8 life policies, and the sales representative needs to select all 3.
The number of ways to select 3 life policies from 8 is given by the combination formula: 8C3 = (8!)/(3!(8-3)!) = 56.
The total number of possible outcomes is the combination of all policies, which is (8+5+2)C3 = (15)C3 = 455.
Therefore, the probability of selecting all life policies is 56/455 ≈ 0.1231.
II. Probability of selecting both homeowner's policies:
There are a total of 2 homeowner's policies, and the sales representative needs to select both.
The number of ways to select 2 homeowner's policies from 2 is given by the combination formula: 2C2 = (2!)/(2!(2-2)!) = 1.
The total number of possible outcomes is still 455.
Therefore, the probability of selecting both homeowner's policies is 1/455 ≈ 0.0022.
III. Probability of selecting one of each policy:
To calculate this probability, we need to consider the different combinations of policies.
The number of ways to select 1 life policy out of 8 is 8C1 = 8.
The number of ways to select 1 automobile policy out of 5 is 5C1 = 5.
The number of ways to select 1 homeowner's policy out of 2 is 2C1 = 2.
To find the total number of possible outcomes, we need to multiply these values together: 8 * 5 * 2 = 80.
Therefore, the probability of selecting one of each policy is 80/455 ≈ 0.1758.
Note: In all cases, the assumption is made that each policy is equally likely to be chosen.
prob = C(8,3)xC(5,0)xC(2,0)/C(15,3)
= 56/455
= 8/65
prob(both homeowners plus one of the others)
= C(2,2) x (13,1)/C(15,3) = 13/455 = 1/35
one of each:
C(8,1) x C(5,1) x C(2,1)/C(15,3)
= 80/455
= 19/91