Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
Chapter Test

Exercise 19 Page 791

What is the probability of a compound event when the events are mutually exclusive?

Yes, see solution.

Practice makes perfect

We are given the formula for the probability of a compound event. P(AorB) = P(A) + P(B) - P(AandB) This formula is used for calculating the probability of overlapping events. We want to determine whether we can also use it to calculate the probability of mutually exclusive events. Let's use a Venn diagram to illustrate the probabilities of overlapping and mutually exclusive events.

Mutually exclusive events have no outcomes in common — which implies that the probability of both events occurring at the same time is 0. lEventsAandBare mutually exclusive. ⇒ P(AandB)=0 We can now substitute P(AandB)= 0 into the given formula.

P(AorB) = P(A) + P(B) - P(AandB)
P(AorB) = P(A) + P(B) - 0
P(AorB) = P(A) + P(B)

Notice that we obtained the formula for the probability of mutually exclusive events. This means that we can use the given formula for both overlapping events and mutually exclusive events.

Mutually Exclusive Events Example

Let's consider a scenario that we randomly draw a 2 or a 10 from a standard deck of cards. Since a card cannot be both a 2 and a 10 at the same time, these are mutually exclusive events. Let's draw these events with a Venn diagram!

Since there are 52 cards in a deck, and 4 of each 2 and of each 10, we will divide 4 by 52 to find the probability of each individual event. ccc Drawing a2: && Drawing a10: [0.5em] P(2) = 4/52 && P(10) = 4/52 We know that there are no favorable outcomes for both events. Therefore, the probability of drawing a card that is both a 2 and a 10 is 0. P(2 and10) = 0 Now we will find the probability that we draw either a 2 or a 10. To do so, let's use the given formula.

P(2 or10) = P(2) + P(10) - P(2and10)
P(2 or10) = 4/52+ 4/52 - 0
P(2 or10) = 4/52+ 4/52
P(2 or10) = 8/52

We can confirm this probability seeing that 8 out of the 52 cards are our favorable outcomes. This probability can be further simplified to 213.