Exercise 16 Page 798

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.

Addition Rules for Probability
If $A$ and $B$ are mutually exclusive events, the probability that $A$ or $B$ will occur is $P (A or B) = P (A) + P (B) .$	If $A$ and $B$ are not mutually exclusive events, the probability that $A$ or $B$ will occur is $P (A or B) = P (A) + P (B) - P (A and B) .$

Let

A

be Dario selects a baseball card and

B

be Dario selects a football card. A card cannot be a baseball card and a football card at the same time. Thus,

A

and

B

are mutually exclusive events.

P (A) P (B) = \frac{1 4 5}{2 9 5} \leftarrow \leftarrow number of baseball cards total number of cards = \frac{1 0 2}{2 9 5} \leftarrow \leftarrow number of football cards total number of cards

Now let's calculate

P (A or B) .

P (A or B) = P (A) + P (B)

SubstituteII

$P (A) = \frac{1 4 5}{2 9 5}$ , $P (B) = \frac{1 0 2}{2 9 5}$

P (A or B) = \frac{1 4 5}{2 9 5} + \frac{1 0 2}{2 9 5}

AddFrac

Add fractions

P (A or B) = \frac{2 4 7}{2 9 5}

We found that the probability that Dario randomly selects a baseball or a football card is

\frac{2 4 7}{2 9 5} .

We can also write this probability as a percentage.

P (A or B) = \frac{2 4 7}{2 9 5}

UseCalc

Use a calculator

P (A or B) = 0.837288 \dots

RoundDec

Round to $2$ decimal place(s)

P (A or B) \approx 0.84

WritePercent

Convert to percent

P (A or B) \approx 84 %

The probability is about

84 % .

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