Sample space ={1,2,3,4,5}

p(1)=0.1, p(2)=0.2 and p(3)=p(4)=0.05. Assume that A={1,3,5} and B={2,3,4}.
Find p(5).
Find p(A^c).

I need help with this question. It is a probability question.
please help.

To find the probability of an event when given the probability of other events, we can use basic probability rules.

1. To find p(5), we can use the fact that the sum of the probabilities of all possible outcomes in a sample space is equal to 1.

Given that p(1) = 0.1, p(2) = 0.2, p(3) = p(4) = 0.05, and A = {1, 3, 5}, we can find p(5) as follows:

Since p(1) = 0.1, p(2) = 0.2, p(3) = p(4) = 0.05, and A = {1, 3, 5}, the probabilities sum up to p(1) + p(2) + p(3) + p(4) + p(5) = 0.1 + 0.2 + 0.05 + 0.05 + p(5) = 0.4 + p(5) = 1.

Rearranging the equation: 0.4 + p(5) = 1, we can solve for p(5):

p(5) = 1 - 0.4 = 0.6.

Therefore, p(5) = 0.6.

2. To find p(A^c) (the probability of the complement of A), we need to find the probability of all the outcomes in the sample space that are not in A.

Given that A = {1, 3, 5}, the complement of A (A^c) would consist of all the outcomes in the sample space that are not in A, which are {2, 4}.

To find p(A^c), we sum up the probabilities of the outcomes in A^c:

p(A^c) = p(2) + p(4) = 0.2 + 0.05 = 0.25.

Therefore, p(A^c) = 0.25.