Geometry

Probability of Compound Events

A single event is called a simple event. These events have fairly simple

probabilities.

Chance of rain next Saturday in Thousand Oaks

20%

Chance of rain next Saturday in

Chicago60%

P(Thousand Oaks) =

0.2 P(Chicago) = 0.6

When flipping a coin twice (or flipping two coins at the same time) having a result of Tails on the first coin toss and Heads on the second is a simple event. However getting one Tail and one Head is a compound event as there are two ways in which that can happen. Look the Sample Space: {HH, TH, HT, TT}

The weather in Thousand Oaks, CA, doesn’t affect the weather in Chicago, IL.

These two events are called independent events because the outcome of one doesn’t affect the outcome of the other one.

Similarly the result of one coin toss does not effect the result of another toss.

QuickTime™ and aTIFF (Uncompressed) decompressor

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What if we looked at the probability of rain occurring in both cities on Saturday?

When two or more simple events are combined, it is considered a compound

event. (Like the HT and TH outcomes when tossing two coins form the compound event of

tossing one Head and one Tail)Probability of Compound Events

If 2 events, A and B, are independent, then the probability of both events

occurring is the product of the probability of A and the probability of B.

( ) ( ) ( )P A andB P A P B= ⋅

( ) ( ) ( )P T.O. and Chicago P T.O. P Chicago= ⋅

0.2 0.6= ⋅0.12=

The probability of rain occurring Saturday in both T.O. and Chicago is 12%.

The event of it not raining in Chicago next Saturday is a special kind of event called a Complement. In this case it is the complement of it raining in Chicago next Saturday. You can think of Complements as two events that are opposites in terms of all the possibilities in the sample space.

Example: When rolling a single die, the complement of rolling a 3 is the event of rolling a 1, 2, 4, 5 or 6.

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( ) ( ) ( )P T.O. and not in Chicago P T.O. P not inChicago= ⋅

0.2 0.4= ⋅0.08=

The probability of rain occurring Saturday inT.O. and not in Chicago is 8%.

If the chance of rain in Chicago is 60%,

then the chance of it not raining there is

40%.

Andrew is flying from Birmingham to Chicago. On the first leg of the trip he has to fly from

Birmingham to Houston. In Houston he’ll change planes and head to Chicago. Airline statistics

report that the Birmingham to Houston flight has a 90% on-time record and the flight from Houston to Chicago has a 50% on-time record. Assuming that one flight’s on-time status is independent of another, what’s the probability that both flights

will be on time?( ) ( ) ( )P plane1and plane 2 P plane1 P plane 2= ⋅

0.9 0.5= ⋅

0.45=

The probability of both flights being on time is 45%.

A B

CD

A die is rolled and a spinner like the one shown on the right is spun. Find each

probability:

124

=

( )P 4 and A

( ) ( )P 4 P A= ⋅

1 16 4

= ⋅

18

=

( )P even # and C

( ) ( )P even # P C= ⋅3 16 4

= ⋅

These are independent events. Multiply the probabilities together to find the probability of both occurring.

A B

CD

A die is rolled and a spinner like the one shown on the right is spun. Find each

probability:

12

=

( )P a number less than 5 and B, C, or D

( ) ( )P number less than5 P B,C, or D= ⋅

4 36 4

= ⋅

Find the probability that you’ll roll a 6 and then a five

when you roll a die twice.

136

=

( )First roll : P 6

( ) ( )P 6 P 5= ⋅

1 16 6

= ⋅

These are independent events. You will multiply the probabilities

together.16

= ( )Second roll : P 516

=

P(6 and 5)

1-15 11-25

Round plastic chips numbered 1-15 are placed in a box. Chips numbered 11-25 are placed in another box. A

chip is randomly drawn from the box on the left, then a second chip is randomly drawn from the box on the

right. These are independent

events. Multiply the probabilities together.

( )P both chips are greater than 14

( ) ( )P box 1 P box 2= ⋅

1 1115 15

= ⋅

11225

=

Probability of Dependent Events

If 2 events, A and B, are dependent, then the probability of both events occurring is the product of the probability of A and the

probability of B given A occurs. This is called a Conditional Probability.

( ) ( ) ( )P A andB P A P B f ollowingA= ⋅

When the outcome of one event affects the outcome of another event, the two events are

said to be dependent.

This can also be written using the formal notation for conditional probability: P(A and B)=P(A)•P(B|A)

€

P AandB( ) = P A( )P B A( )

Here’s an example of two events that are dependent.A bag contains 2 green, 9 brown, 7

yellow, and 4 blue marbles. Once a marble is selected, it is not replaced.

Find the probability of randomly drawing a brown, then a yellow marble.

1st marble : ( )P brown =number of brownmarbles9

22 total # of marbles2nd marble :

number of yellowmarbles721 total # of marbles remaining

9 722 21

= ⋅322

=

P(yellow following brown) =

P(brown and yellow)

= P(brown)•P(yellow following brown)

=P(brown)•P(yellow|brown)

So, putting all that all together, we see that:

Start again with a bag that contains 2 green, 9 brown, 7 yellow, and 4

blue marbles. Again, once a marble is selected, it is not replaced. Find the probability of randomly drawing a green marble, then a marble that’s

not blue.1st marble :

( )P green =number of greenmarbles2

22 total # of marbles

2nd marble :number of "not blue" marbles17

21 total # of marbles remaining

1 1711 21

= ⋅17231

=

111

=

P(green and not blue)

P(not blue|green) =

Find the probability of randomly drawing a yellow marble, a yellow

marble, and a blue marble. As before, marbles that are drawn are not

replaced.1st marble :

( )P yellow =number of yellowmarbles7

22 total # of marbles

2nd marble :number of yellowmarbles lef t6

21 total # of marbles remaining

7 2 122 7 5

= ⋅ ⋅1

55=

27

=

3rd marble :

P(yellow|yellow)=

P(blue|yellow and yellow)=

P(yellow and yellow and blue)

€

420

=15

number of blue marbles

total number of marbles remaining

At a school carnival, winners in a ring-toss game are randomly given a prize from a bag that

contains 4 sunglasses, 6 hairbrushes, and 5 key chains. Three prizes are randomly chosen from the bag and not replaced. Find the probability:

513⋅

491

=

415

=614⋅

3

3

7

P(sunglasses and hairbrush and key chain)

=P(sunglasses)•P(hairbrush|sunglasses)•P(key chain|sunglasses and hairbrush)

Day 2

Probability of Mutually Exclusive and

Inclusive Events

Mutually Exclusive Events

Mutually exclusive events, or disjoint events, are events that cannot occur

at the same time.

For example, consider the events of rolling a 1 or a 3 on a die? A die can’t show both a 1 and a 3 at the same

time, so the two events are considered mutually exclusive.

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

Probability of Mutually Exclusive Events

If 2 events, A and B, are mutually exclusive, then the probability that either

A or B occurs is the sum of their probabilities.

Find the probability of rolling a 1 or a 3

on a die.

13

=

( )P 1or 3

( ) ( )P 1 P 3= +

1 16 6

= +

These events are mutually exclusive. The die can’t

show two different numbers at the same time. This problem asks what the

probability is of one or the other number showing. In

other words, an outcome of 1 or an outcome of 3 is a

desired outcome. The two probabilities must be added together to show that either

roll is acceptable.

Janet is going to an animal shelter to choose a new pet. The shelter has 8 dogs, 7 cats, and 5 rabbits available for adoption. If she randomly picks an

animal to adopt, what is the probability it will be a cat or a dog?

15 320 4

= =

( )P cat or dog

( ) ( )P cat P dog= +

7 820 20

= +

The problem asks for a

cat or a dog, so either

outcome is desired. The

two probabilities

should be added

together.

Bart randomly drew one card from a standard

deck of cards. What is the probability that the card he drew was a club

or a diamond?

26 152 2

= =

( )P club or diamond

( ) ( )P club P diamond= +

13 1352 52

= +

Bart randomly drew one card from a standard

deck of cards. What is the probability that the card he drew was a face

card or a 10?

16 452 13

= =

( )P f ace card or 10

( ) ( )P f ace card P 10= +

12 452 52

= +

Mutually Inclusive Events

Mutually inclusive events are events that can occur at the same time.

For example, what is the probability of drawing either an ace or a spade randomly from a deck of cards?

It’s possible for a card to be both an ace and a spade at the same time. When we consider the probability of drawing either an ace or a spade there are 4 cards in the deck that are aces, 13 cards in the deck that are spades, and 1 card that’s both an ace and a spade at

the same time.That one card should not be

counted twice!

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

Probability of Mutually Inclusive Events

If 2 events, A and B, are mutually inclusive, then the probability that either A or B occurs is the sum of their probabilities, decreased by the

probability of both occurring.

Let’s go back to the earlier question:

What is the probability of

drawing either an ace or a spade? To figure this out you need to add the probability of

drawing an ace to the probability of drawing a spade, then subtract out the probability of drawing a card that’s both an ace and a spade at the same time.

( ) ( ) ( )P ace P spade P ace of spades+ −

4 13 152 52 52

= + −

1652

=413

=

P(ace or spade) P(ace) P(spade) P(ace and spade)

=

What is the probability of drawing either a red card or an

ace?

To figure this out you need to add the probability of drawing a red card to the probability of drawing an ace, then subtract out the probability of drawing a card that’s both red and an ace at the same time.

( ) ( ) ( )P red P ace P red ace+ −

26 4 252 52 52

= + −

2852

=

713

=

( ) ( ) ( )P f ace card P spade P spade f ace cards= + −

12 13 352 52 52

= + −

2252

=

1126

=

What is the probability of drawing a face card or a spade?

( )P f ace card or spade

Suppose your dog had 9 puppies!

• 3 are brown females• 2 are brown males• 1 is a mixed color female• 3 are mixed color males

If a puppy is randomly chosen from the litter, what is the probability that it will be

male or be mixed color?

( )( ) ( ) ( )

P male or mixed

P male P mixed P male and mixed= + −

5 4 39 9 9

= + −

6 29 3

= =

The puppies that are both male and mixed will be counted twice if not subtracted here.

In a bingo game, balls or tiles are numbered from 1 to 75. The numbers

correspond to columns on a bingo card.

B I N G O 1-15 16-30 31-45 46-60 61-75If a number is selected at random,

what is the probability it is

a multiple of 5 or in the “N”

column?

What multiples of 5 appear in the “N” column?

35, 40, and 45

( ) ( ) ( )P multiple of 5 P "N" column P multiple of 5 and"N" column= + −

( )P multiple of 5 or "N" column

The numbers that are both multiples of 5 and appear in the “N” column will be

counted twice if they’re not

subtracted out here.

15 15 375 75 75

= + −

925

=

2775

=

( ) ( ) ( )P even P "G" column P evens in"G" column= + −

( )P evenor "G" column

The numbers that are both even and

appear in the “G” column will be

counted twice if they’re not

subtracted out here.

37 15 875 75 75

= + −

4475

=

Students are selected at random from a group of 12 boys and 12 girls. In that group there are 4

boys and 4 girls from each of 6th, 7th, and 8th grades.

Find the probability:

23

=

( )P 6th grade or girl

( ) ( ) ( )P 6th grade P girl P 6th gradegirl= + −

8 12 424 24 24

= + −

1624

=

Students are selected at random from a group of 12 boys and 12 girls. In that group there are 4

boys and 4 girls from each of 6th, 7th, and 8th grades.

Find the probability:

56

=

( )P male or not 8th

( ) ( ) ( )P male P not 8th P males not in8th= + −

12 16 824 24 24

= + −

2024

=

To sum it all up for today:

Mutually exclusive events are events that cannot occur at the same time. You

will see the word “or” in the question.

Mutually inclusive events are events that can occur at the same time. You will see the word “or” in the question

here too.

Add the probabilities of mutually exclusive events together to consider the probability that either one or

the other will occur.

Add the probabilities of mutually inclusive events together to consider the probability that either one or

the other will occur, but remember to subtract out the events that overlap so they’re not counted twice.