Question

An experiment can result in one of five equally likely simple events, E1, E2, , E5. Events A, B, and C are defined as follows. A: E2, E4 P(A) = 0.4 B: E1, E3, E4, E5 P(B) = 0.8 C: E2, E3 P(C) = 0.4 Find the probabilities associated with the following events by listing the simple events in each.

Answer 1

Answer:

(1) $P(A^C)=0.6$

The notation $A^C$ means "the complement of event A".

Therefore, here we want to find $P(A^C)$, which is the probability for event A NOT to occur.

Here we are given that the probability of event A is

$P(A)=0.4$

The probability of the complement of an event X is given by

$P(X^C)=1-P(X)$

So in this case,

$P(A^C)=1-P(A)=1-0.4=0.6$

(2) $P(A\cap B)=0.2$

Event A is defined as either event E2 or event E4 to occur.

Event B is defined as either one of the following events to occur: E1, E3, E4, E5.

Here we want to find the intersection between the sets of events A and B.

We see that this intersection consists of the events that belong simultaneously to both sets: therefore, only E4.

Therefore,

$P(A\cap B)=P(E4)$

And since the 5 events are equally likely to occur, the probability of each is 0.2, therefore

$P(A\cap B)=P(E4)=0.2$

(3) $P(B\cap C)=0.2$

Event B is defined as either one of the following events to occur: E1, E3, E4, E5.

Event C is defined as either E2 or E3 to occur.

Here we want to find the intersection between the sets of events B and C.

This intersection consists of the events that belong simultaneously to both sets, so, only E3.

Therefore,

$P(B\cap C)=P(E3)$

And since each event E has probability 0.2 to occur,

$P(B\cap C)=P(E3)=0.2$

(4) $P(A\cup B)=1$

Here we want to find the probability of the set consisting of the union of A and B.

We have:

Event A is defined as either event E2 or event E4 to occur.

Event B is defined as either one of the following events to occur: E1, E3, E4, E5.

So the union of the two sets is: E1, E2, E3, E4, E5

But these events represents all the possible events of the experiment. Therefore, their total probability is 1, so

$P(A\cup B)=1$

(5) $P(B|C)=0.5$

Here we want to find the conditional probability of B given C: that is, the probability of B to occur, given that C has occurred.

This probability can be calculated as:

$P(B|C)=\frac{P(B\cap C)}{P(C)}$

Here in this problem, we have:

$P(B\cap C)=0.2$ (already calculated in part 3)

$P(C)=0.4$ (given by the problem)

Therefore, we have:

$P(B|C)=\frac{P(B\cap C)}{P(C)}=\frac{0.2}{0.4}=0.5$

(6) $P(A|B)=0.25$

Here we want to find the conditional probability of A given B: that is, the probability of A to occur, given that B has occurred.

This probability can be calculated as:

$P(A|B)=\frac{P(A\cap B)}{P(B)}$

In this part of the problem, we have:

$P(A\cap B)=0.2$ (already calculated in part 2)

$P(B)=0.8$ (given by the problem)

And so, we have:

$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{0.2}{0.8}=0.25$

(7) $P(A\cup B\cup C)=1$

Here we want to find the probability that either A, either B or either C occurs.

Event A is defined as either event E2 or event E4 to occur.

Event B is defined as either one of the following events to occur: E1, E3, E4, E5.

Event C is defined as either E2 or E3 to occur.

We see that the 3 events A, B and C contain all the possible events of the experiment: E1, E2, E3, E4 or E5. Therefore, if any of these 5 events occurs, then either A, B or C has occurred as well.

Therefore, the total probability is 1:

$P(A\cup B\cup C)=1$

(8) $P((A\cap B)^C)=0.8$

Here we want to find the probability of the complement of the set

$A\cap B$

The probability of this set was already calculated in part 2, and it was

$P(A\cap B)=0.2$

We also said that the probability of the complement of a set X is given by

$P(X^C)=1-P(X)$

Therefore, the probability of the complement of $A\cap B$ is:

$P((A\cap B)^C)=1-P(A\cap B)=1-0.2=0.8$