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?

0.6242

0.3758

0.8926

0.2684

0.1074

What is the expected number of policies having emergency road service?

3.5

8.0

1.6

2.0

5.0

To calculate the probability and expected number of policies having emergency road service, we need to use the concept of binomial probability.

For the first question, to find the probability that at least one policy will have emergency road service, we can use the complement rule. The complement of "at least one policy having emergency road service" is "no policy having emergency road service." The probability of no policy having emergency road service can be calculated as (1 - probability of having emergency road service). So let's calculate this:

Probability of no policy having emergency road service:
= (1 - probability of having emergency road service)^10
= (1 - 0.20)^10
= 0.8^10
≈ 0.1074

Therefore, the probability that at least one policy will have emergency road service is 1 - 0.1074, which is approximately 0.8926.

So the correct answer option for the first question is 0.8926.

Now, moving on to the second question, the expected number of policies having emergency road service can be calculated using the formula:

Expected number = Total number of trials × Probability of success

In this case, the total number of trials is 10 (since we are randomly selecting 10 car insurance policies). And the probability of success is 0.20 (since 20% of car insurance policies include emergency road service).

Expected number = 10 × 0.20
= 2.0

Therefore, the expected number of policies having emergency road service is 2.0.

So the correct answer option for the second question is 2.0.