Question

The accompanying Venn diagram illustrates a sample space containing six sample points and three events, A, B, and C. The probabilities of the sample points are P(1) = 0.3, P(2) = 0.2, P(3) = 0.1, P(4) = 0.1, P(5) = 0.1 and P(6) = 0.2.
Find:

Answer 1

Answer:

Step-by-step explanation:

To find the probability of a certain set, we must add the probability of the points that are contained in the specific set.

a) P(A): Add the probability of 1,3,5. That is P(1)+P(3)+P(5) = 0.5

b) $P(A\cap B)$. This set has no points in it, so its probability is 0.

c) P(A\cup B \cup C): Add the probability of 1,2,3,4,5,6. So, it is 1.

d) $P(C^c) = 1- P(C)$. P(C)  is the probability by summing 5,6,2 so P(C) = 0.5. So $P(C^c)=0.5$

e) $P(A\cap C^c)$ That is, all the points that are in A but not in C. So add 1,3. Then the probability is 0.4

e')$P(B|A) = \frac{P(A\cap B )}{P(A)}=\frac{0}{0.5}=0$, using the definition of conditional probabily and results a,b.

f) Two events are mutually exlusive if the probability of their intersection is 0. In this case A and C are not mutually exclusive, since $P(A\cap C)$ is the probability of 5, that is 0.1. Since 0.1>0, they are not mutually exclusive.