Respuesta :
Answer:
(1) [tex]P(A^C)=0.6[/tex]
The notation [tex]A^C[/tex] means "the complement of event A".
Therefore, here we want to find [tex]P(A^C)[/tex], which is the probability for event A NOT to occur.
Here we are given that the probability of event A is
[tex]P(A)=0.4[/tex]
The probability of the complement of an event X is given by
[tex]P(X^C)=1-P(X)[/tex]
So in this case,
[tex]P(A^C)=1-P(A)=1-0.4=0.6[/tex]
(2) [tex]P(A\cap B)=0.2[/tex]
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,
[tex]P(A\cap B)=P(E4)[/tex]
And since the 5 events are equally likely to occur, the probability of each is 0.2, therefore
[tex]P(A\cap B)=P(E4)=0.2[/tex]
(3) [tex]P(B\cap C)=0.2[/tex]
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,
[tex]P(B\cap C)=P(E3)[/tex]
And since each event E has probability 0.2 to occur,
[tex]P(B\cap C)=P(E3)=0.2[/tex]
(4) [tex]P(A\cup B)=1[/tex]
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
[tex]P(A\cup B)=1[/tex]
(5) [tex]P(B|C)=0.5[/tex]
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:
[tex]P(B|C)=\frac{P(B\cap C)}{P(C)}[/tex]
Here in this problem, we have:
[tex]P(B\cap C)=0.2[/tex] (already calculated in part 3)
[tex]P(C)=0.4[/tex] (given by the problem)
Therefore, we have:
[tex]P(B|C)=\frac{P(B\cap C)}{P(C)}=\frac{0.2}{0.4}=0.5[/tex]
(6) [tex]P(A|B)=0.25[/tex]
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:
[tex]P(A|B)=\frac{P(A\cap B)}{P(B)}[/tex]
In this part of the problem, we have:
[tex]P(A\cap B)=0.2[/tex] (already calculated in part 2)
[tex]P(B)=0.8[/tex] (given by the problem)
And so, we have:
[tex]P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{0.2}{0.8}=0.25[/tex]
(7) [tex]P(A\cup B\cup C)=1[/tex]
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:
[tex]P(A\cup B\cup C)=1[/tex]
(8) [tex]P((A\cap B)^C)=0.8[/tex]
Here we want to find the probability of the complement of the set
[tex]A\cap B[/tex]
The probability of this set was already calculated in part 2, and it was
[tex]P(A\cap B)=0.2[/tex]
We also said that the probability of the complement of a set X is given by
[tex]P(X^C)=1-P(X)[/tex]
Therefore, the probability of the complement of [tex]A\cap B[/tex] is:
[tex]P((A\cap B)^C)=1-P(A\cap B)=1-0.2=0.8[/tex]
