Ok so i don't know how i got that answer i think i typed in a wrong formula or read something wrong. I got .8926 for the first one. And the second one i don't know what i am suppost to use from the first answer to get that. Can someone help me.
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 calculate the probability of at least one policy having emergency road service, we can use the concept of complementary probability. Complementary probability states that the probability of an event occurring is equal to one minus the probability of the event not occurring.
In this case, the probability of at least one policy having emergency road service is equivalent to one minus the probability that none of the policies have emergency road service.
To calculate the probability of none of the policies having emergency road service, we can use the formula for the probability of independent events: P(A and B) = P(A) * P(B).
Since each policy has a 20% chance of having emergency road service and there are 10 policies, the probability of none of them having emergency road service is (0.8)^10 = 0.1074.
Then, the probability of at least one policy having emergency road service is 1 - 0.1074 = 0.8926.
Therefore, the correct answer is 0.8926.
Now, to calculate the expected number of policies having emergency road service, we can use the formula: Expected value = Sum of (each possible outcome * probability of that outcome).
In this case, the possible outcomes are 0, 1, 2, 3, ..., 10, representing the number of policies with emergency road service.
To calculate the expected value, we multiply each possible outcome by its respective probability and sum them up.
For example, the probability of having 0 policies with emergency road service is 0.1074 (as calculated earlier), and since having 0 policies implies having 10 policies without emergency road service, the expected value for that outcome becomes 0.
Similarly, the probability of having 1 policy with emergency road service is 0.8926, and the expected value for that outcome is 1 * 0.8926 = 0.8926.
By calculating the expected value for each possible outcome and summing them up, we find that the expected number of policies having emergency road service is 2.0.
Therefore, the correct answer is 2.0.