I need help with this problem for school. For the first one i got .3758 and second 1.6, but i don't think they are right. My calculations are off
It is known that about 20% of car insurance policies include emergency road service. Suppose we randomly select 10 car insurance policies.
What is the probability that at least one policy will have emergency road service?
Answer
0.6242
0.3758
0.8926
0.2684
0.1074
What is the expected number of policies having emergency road service?
Answer
3.5
8.0
1.6
2.0
5.0
To solve this problem, we will use the concept of binomial probability. The probability of success (p) is given as 20% or 0.2, and the number of trials (n) is 10. We need to find the probability of at least one success (i.e., one or more policies with emergency road service) and the expected number of policies with emergency road service.
To find the probability of at least one success, we can calculate the complement of the probability of no successes. So, the probability of at least one success is equal to 1 minus the probability of no successes.
To find the probability of no successes (i.e., no policies with emergency road service), we can use the formula for binomial probability:
P(X = k) = (n choose k) * (p^k) * ((1-p)^(n-k))
Where:
n = the number of trials
k = the number of successes
p = the probability of success
Using this formula, we can calculate the probability of no successes for k = 0 (i.e., no policies with emergency road service):
P(X = 0) = (10 choose 0) * (0.2^0) * ((1-0.2)^(10-0))
= 1 * 1 * 0.8^10
= 0.8^10
≈ 0.1074
Now, we can calculate the probability of at least one success:
P(at least one success) = 1 - P(X = 0)
= 1 - 0.1074
≈ 0.8926
Therefore, the probability that at least one policy will have emergency road service is approximately 0.8926.
To find the expected number of policies having emergency road service, we can use the formula:
E(X) = n * p
Where:
E(X) = expected value of X (number of policies with emergency road service)
n = number of trials
p = probability of success
Using this formula, we can calculate the expected number of policies with emergency road service:
E(X) = 10 * 0.2
= 2
Therefore, the expected number of policies having emergency road service is 2.
So, the correct answers are:
1. The probability that at least one policy will have emergency road service is approximately 0.8926.
2. The expected number of policies having emergency road service is 2.0.