McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
6. Probabilities of Mutually Exclusive Events
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Exercise 10 Page 929

Can both events occur at the same time?

Mutually exclusive or not? Not mutually exclusive.
Probability: 1320 or 65 %

Practice makes perfect

We want to determine whether the events are mutually exclusive and find the probability of their union. We will do these things one at a time.

Mutually Exclusive

Events that cannot occur at the same time are mutually exclusive. Mutually exclusive events have no outcomes in common. For example, because it is not possible to toss a coin and obtain heads and tails at the same time, these two events are mutually exclusive.

Rules for Probability
If A and B are mutually exclusive events, the probability that A and B will occur is P(AandB)=0. If A and B are not mutually exclusive events, the probability that A and B will occur is P(AandB)≠0.
We select a number at random from integers 1 to 20. Let A be selecting an even number and B be selecting a number divisible by 3.

Three out of the ten even numbers are divisible by 3, so it is possible to be both an even number and a number divisible by 3. Therefore, A and B are not mutually exclusive events.

Probability

Now, we can find the probability that we select an even number or a number divisible by 3.

Addition Rules for Probability
If A and B are mutually exclusive events, the probability that A or B will occur is P(AorB)=P(A)+P(B). If A and B are not mutually exclusive events, the probability that A or B will occur is P(AorB)=P(A)+P(B)-P(AandB).
When we select a number from integers 1 to 20, there are 20 numbers to choose from. Out of these, there are 10 even numbers, 6 numbers divisible by three, and 3 even numbers divisible by three. Now, we can calculate P(A), P(B), and P(AandB). P(A)&=10/20 l← ← l Even Total numbers [0.5em] P(B)&=6/20 l← ← l Divisible by3 Total numbers [0.5em] P(AandB)&=3/20 l← ← l Even number divisible by3 Total numbers Finally, considering that A and B are not mutually exclusive events, let's calculate P(AorB).
P(AorB)=P(A)+P(B)-P(AandB)
P(AorB)=10/20+ 6/20- 3/20
P(AorB)=13/20
Convert to percent
P(AorB)=0.65
P(AorB)=65 %
We found that P(Even or divisible by3) is equal to 1320, which can be also written as 65 %.