
Concept explainers
(a)
Whether A and B are mutually exclusive, and to draw a diagram to support answer.
(a)

Answer to Problem 75E
Thus for mutually exclusive event is
1.34
Explanation of Solution
Given information:
Let A and B be two events from the same
Calculation:
Consider two events A and B from the same sample space such that
P(A)=0.76 and P(B)=0.58
Event A and B are exclusive if A and B have no sample point in common or if
A∩B is empty.
If A∩B is empty, then
P(A∩B)=P(ϕ)=0
Thus for mutually exclusive event is
P(A∪B)=P(A)+P(B)−P(A∩B)=0.76+0.58−0=1.34
But the probability P(A∪B) is not possible since the probability of any event cannot be greater than one
Thus, the events A and B are not mutually exclusive since the sum of probability is greater than one
The diagram showing the events A and B is shown below.
(b)
Whether A′ and B′ are mutually exclusive, and to draw a diagram to support answer.
(b)

Answer to Problem 75E
The sum of probability of the complementary events A′ and B′ is
0.66
Explanation of Solution
Given information:
Let A and B be two events from the same sample space such that P(A)=0.76 and P(B)=0.58 .
Calculation:
The formula of the complement of the event is
P(X′)=1−P(X)
Therefore, the complement of P(A) and P(B) is
P(A′)=1−P(A)=1−0.76=0.24
And
P(B′)=1−P(B)=1−0.58=0.42
Thus, the sum of probability of the complementary events A′ and B′ is
P(A′) and P(B′)
=0.24+0.42=0.66
Thus, the events A′ and B′ are mutually exclusive since the sum of probability is less than one
The diagram showing the events A′ and B′ is shown below
(c)
To find: the possible range of P(A∪B) .
(c)

Answer to Problem 75E
The range of P(A∪B) is 0.76≤P(A∪B)≤1
Explanation of Solution
Given information:
Let A and B be two events from the same sample space such that P(A)=0.76 and P(B)=0.58 .
Calculation:
Note that the probability of an event A or B is always greater than the individual probabilities of the events A and B
That is
P(A∪B)≥P(A) and P(A∪B)≥P(B)
Thus,
P(A∪B)≥0.76
Since the probability of an event is always smaller than 1, therefore
P(A∪B)≤1
Therefore, the range of P(A∪B) is 0.76≤P(A∪B)≤1
Chapter 8 Solutions
Precalculus with Limits: A Graphing Approach
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