⚠ Pay attention to every

underlinedsingular noun just on this page!

**SETS AND PROBABILITY**

It is far easier to alk about probability using the language of sets. The set of all outcomes is called the **sample space**, a subset of the sample space is called an event. Thus when we throw three coirs, we can take the sample space as the set

*S*={HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

and the event ‘tlerowing at least one head and at least one tail’ is then the subset

*E*= {HHT, HTH, HTT, THH, THT, TTH}

Since each outcome is equally likely,

P(at least one head and at least one tail)=

^{|E|}⁄

_{|S|}=¾

The event space of the complem ntary event ‘throwing all heads or all tails’ is the complement of the event space in the sample space, which we take as the universal set, so

*E*

^{∁}={HHH,TTT}

Since |

*E*| +|

*E*

^{∁}|=|

*S*|, it follows after dividing by |

*S*| that P(

*E*

^{∁})=1-P(

*E*), so

P(throwing all head or all tails)=1-¾=¼.

Let

*F*be the event ‘throwing at least two heads’. Then

*F*= {HHH, HHT, HTH, THH}

A Venn diagram is the best way to sort out the relationship between the two events

*E*and

*F*. We can then conclude that

*E*and

*F*|=3 and |

*E*or

*F*|=7

**Union of 2 Events A and B**

**General Addition Rule**

The general addition rule is a way of finding the **probability of a union of 2 events**. It is

*A*∪

*B*)=P(

*A*)+P(

*B*)-P(

*A*∩

*B*)

**Additive rule of probability**

Given events *A* and *B*, the probability of the union of events *A* and *B* is the sum of the probability of events *A* and *B* minus the probability of the intersection of events *A* and *B*

*A*∪

*B*)=P(

*A*)+P(

*B*)-P(

*A*∩

*B*)

📌 Example 1.

If *A*=band members and *B*=club members, find the probability of *A*∪*B* in the school.

The number of members is n(U).

n(

*A*∪

*B*)=160+35+530=725,

n(

*A*)=160+35=195,

n(

*B*)=530+35=565,

n(

*A*∩

*B*)=35

📌 Ex2. Data for the students enrolled in a local high school are shown in the Venn diagram below.

If

__a student__from the high school is selected at random, what is the probability that the student is a sophomore given that the student is enrolled in

*A*lgebra II?

📌 Ex3. In a class of 28 students, 20 study French and 15 study Chemistry. Each student in the class studies either French or Chemistry.

a) Represent this information on a Venn diagram.

b) __One student__ is selected at random from the group. What is the probability that the student studies:

b.i) both French and Chemistry (ii) Chemistry but not French?

✍ Solution:

a) Let *F* be the event ‘student studies French’ and *C* be the event ‘student studies Chemistry’. Twenty students study French and fifteen study Chemistry in a class of 28.

*F*∪

*C*|=|

*F*|+|

*C*|-|

*F*∩

*C*|

28=15+20—|

*F*∩

*C*|

Therefore |

*F*∩

*C*|=7.

Then 20-7=13 students study French but not Chemistry, and 15-7=8 students study Chemistry but not French.

b.i) Using the Venn diagram, P(

*F*∩

*C*)=7/28=¼,

ii) Using the Venn diagram, P(

*F*

^{∁}∩

*C*)=8/28=2/7.

📌 Ex4. In a group of 50 learners, 35 take Mathematics and 30 take History. 12 learners do not take Mathematics or History.

a) Draw a Venn diagram to represent this information.

b) If __a learner__ is chosen at random from this group, what is the probability that he takes both Mathematics and History?

✍ Solution:

a) Use *M* for Mathematics and *H* for History.

Draw the sample space and sets for the events

*M*and

*H*.

We do not know yet how many learners (outcomes) are in the intersection of

*M*and

*H*.

So let

*M*∩

*H*=

*x*

We do know that 12 learners are not in

*M*or

*H*.

*x*+

*x*+30-

*x*+12=50

–

*x*=-27

*x*=27

So place ‘27’ in the intersection of

*M*and

*H*.

*M*=35-27=8

*H*=30-27=3

b) P(

*M*and

*H*)=27/50

📌 Ex5. A survey was conducted at Mutende Primary School to establish how many of the 650 learners buy vetkoek and how many buy sweets during break. The following was found:

● 50 learners bought nothing

● 400 learners bought vetkoek

● 300 learners bought sweets

a) Represent this information with a Venn diagram.

✍ Solution:

The following Venn diagram represents the given information. However we can calculate more information from this that will help us answer the second part of this question.

We note the following information:

● 400 learners bought vetkoek, some of these also bought sweets.

● 300 learners bought sweets, some of these also bought vetkoek.

● Of the total number of learners, 50 did not buy anything, so 650-50=600 bought either vetkoek or sweets or both.

Let the number of learners who bought vetkoek only be v, the number of learners who bought sweets only be s and the number of learners who bought both be b. Now we note the following:

*v+s+b*

But

*v+b*=400

∴600=400+

*s→s*=200

Also

*s+b*=300

∴

*b*=100

∴

*v*=600-

*s-b*

∴

*v*=300

We can fill this in on the Venn diagram:

b) If

__a learner__is chosen randomly, calculate the probability that this learner buys:

i. sweets only ii. vetkoek only iii. neither vetkoek nor sweets

iv. vetkoek and sweets v. vetkoek or sweets

✍ Solution:

📌 Ex6.

__One card__is drawn from a set of cards numbered 1 to 10. Find the probability of drawing out an odd number or a multiple of 3.

✍ Solution:

Odd cards are 1, 3, 5, 7 and 9

Multiples of 3 are 3, 6 and 9

The numbers 3 and 9 are both odd and multiples of 3, If there are not too many numbers, simply count them.

i.e. odd numbers or multiples of 3 are 1, 3, 5, 6, land 9

i.e. P(odd or multiple of 3)=6/10=⅗.

A Venn diagram can be used for this type of question.

There are 6 numbers inside the circles and 10 numbers altogether.

Count all the numbers inside the two circles. From the Venn diagram: Probability is 6/10=⅗.

**Probability and Venn diagrams**

Before you start

You should be able to:

● draw and interpret Venn diagrams

● find the probability that an event will occur.

Why do this?

Venn diagrams can be used to help work out probabilities.

Objective

● You will be able to use set notation to describe events.

● You will be able to use Venn diagrams to find probabilities

When working out probabilities from a Venn diagram:

● P(*A*) represents the probability that the item is in set *A*

● P(Ā) represents the probability that the item is notin set *A*

● P(Ā)=1-P(*A*)

● P(*A*∩*B*) represents the probability thatthe item is in both set *A* and set *B*

● P(*A*∪*B*) represents the probability that the item is in set *A* or in set *B* or in both sets.

📌 Ex7. The Venn diagram shows the integers from 4 to 12.

A number is taken at random from those shown on the Venn diagram.

Find: a) P(

*A*), b) P(Ā), c) P(

*A*∩

*B*).

✍

There are 9 numbers in total in the Venn diagram so 9 goes on the bottom of the fraction.

a) P(

*A*)=5/9 ←[There are 5 numbers altogether in set

*A*so 5 goes on the top of the fraction.]

b) P(Ā)=1-5/9 ←[There are 4 numbers that are not in set

*A*. Alternatively, work out 1-5/9]

c) P(

*A*∩

*B*)=2/9 ←[There are 2 numbers in both

*A*and

*B*.]

📌 Ex8. The Venn diagram shows information about the students in Year 12.

*B*={students who take Biology}

*C*={students who take Chemistry}

If

__a student__is chosen at random work out:

a) P(

*B*), b) P(

*B*∩

*C*), c) P(

*C*∩B̄), d) P(

*B*∪

*C*)

✍

a) P(

*B*)=51/117 ←[49+44+17+7=117 thereare 117 students in Year 12, so the bottom number of each fraction will be 117. 44+7=51 so 51 students study Biology.]

b) P(

*B*∩

*C*)=7/117 ←[7 is in the intersection. This shows that 7 students study Biology and Chemistry.]

c) P(

*C*∩B̄)=17/117 ←[There are 17 students who study Chemistry and not Biology.]

d) P(

*B*∪

*C*)=68/117 ←[17+7+44=68 so 68 students study Biology or Chemistry (or both).]

**Venn diagrams**

In this section we introduce the ideas of Venn diagrams and probability. A Venn diagram is a way of representing information visually.

Consider two events *A* and *B*,

*A*∩

*B*= This section contains those numbers which are in both

*A*and

*B*.

*A*∩B̄=

*A*–

*B*= This section contains those numbers which are in

*A*but

**not**in

*B*.

Ā∩

*B*=

*B*–

*A*= This section contains those numbers which are in

*B*but not

**in**

N̄∩

(

Check: 8+7+11+4=30

*A*.📌 Ex9. In a group of 30 friends, 15 play netball, 18 play badminton, 7 play netball and badminton.

(a) Represent this information on a Venn diagram.

(b) Work out the probability that __a friend__ plays netball only.

(c) Work out the probability that __a friend__ plays badminton only.

(d) Work out the probability that __a friend__ plays netball or badminton but not both.

(e) Work out the probability that __a friend__ does not play netball or badminton.

✍

(a)

*N*∩*B*= This section contains those friends who play netball__and__badminton*N*∩B̄=*N*–*B*=This section contains those friends who play netball but not badminton.N̄∩

*B*=*B*–*N*= This section contains those friends who play badminton but not netball.(

*N*∪*B*)= This section contains those friends who do not play either sport.Check: 8+7+11+4=30

NOTE: Always start from the overlap and then work your way outwards to complete the Venn diagram. Always label your circles.

(b) netball only=15-7=8 ←[15 play netball but 7 play both]

probability netball only=8/30 ←[8 play netball only out of 30 friends]

(c) badminton only=18-7=11 ←[18 play netball but 7 play both]

probability netball only=11/30 ←[Write your answer as a fraction]

(d) netball only or badminton only=8+11=19 ←[Do not include 7 as 7 play both]

probability netball only or badminton only=19/30

(e) not netball or badminton=30-(8+11+7)=4 ←[4 is not in any of the circles]

probability not netball or badminton=4/30.

💪 How can we Calculate Probabilities with a 2-Set Venn Diagram? 👑

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