Exercise 27 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 the card drawn is a jack and

B

be the card drawn is a spade. There is a jack of spades in a standard deck of cards. Therefore,

A

and

B

are not mutually exclusive events. Let's start by calculating

P (A),

P (B),

and

P (A and B) .

P (A) P (B) P (A and B) = \frac{4}{5 2} \leftarrow \leftarrow jacks in a deck total number of cards = \frac{1 3}{5 2} \leftarrow \leftarrow spades in a deck total number of cards = \frac{1}{5 2} \leftarrow \leftarrow jack of spades in a deck total number of cards

Finally, considering that

A

and

B

are not mutually exclusive events, let's calculate

P (A or B) .

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

SubstituteValues

Substitute values

P (A or B) = \frac{4}{5 2} + \frac{1 3}{5 2} - \frac{1}{5 2}

AddSubFrac

Add and subtract fractions

P (A or B) = \frac{1 6}{5 2}

ReduceFrac

$\frac{a}{b} = \frac{a / 4}{b / 4}$

P (A or B) = \frac{4}{1 3}

We found that

P (jack or spade) = \frac{4}{1 3} .

We can also write this probability as a percentage.

P (A or B) = \frac{4}{1 3}

UseCalc

Use a calculator

P (A or B) = 0.307692 \dots

RoundDec

Round to $2$ decimal place(s)

P (A or B) \approx 0.31

WritePercent

Convert to percent

P (A or B) \approx 31 %

The probability is about

31 % .

Hints

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