**The Meaning of and and or**

The words __and__ and __or__ are very important in many areas of mathematics. We use these words in several chapters in this book. including the chapter of Probabilities. The word __and__ is generally interpreted to mean **intersection**. whereas __or__ is generally interpreted to mean **union**. Suppose *A*={1,2,3,5,6,8} and *B*={1,3,4,7,9,10}. The elements that belong to set *A* __and__ set *B* are l and 3. These are the elements in the intersection of the sets. The elements that belong to set *A* __or__ set *B* are 1, 2, 3, 4, 5, 6, 7, 8, 9. and 10. These are the elements in the union of the sets.

**Addition rules**

If *A* and *B* be any two events, then the probability of the occurrence of either event *A* or event *B* is deﬁned by

(i) P(*A*∪*B*)=P(*A*)+P(*B*), when *A* and *B* are mutually exclusive.

(ii) P(*A*∪*B*)=P(*A*)+P(*B*)-P(*A*∩*B*), *A* and *B* are not mutually exclusive.

Note:

The above rules can be extended to more than two events.

**Compound events**

Before you start

You should be able to:

● add and multiply fractions.

Why do this?

Set notation can be used to describe the probability of two events occurring at the same time.

Objectives

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

Key Points

● Two events are mutually exclusive when they cannot occur at the same time. For mutually exclusive events *A* and *B*:

*A*∪

*B*)=P(

*A*)+P(

*B*)

● Two events are independent if one event does not affect the other event. For two independent events

*A*and

*B*:

*A*∩

*B*)=P(

*A*)×P(

*B*)

Review

● P(

*A*) represents the probability that the item is in set

*A*.

● P(Ā) represents the probability that the item is

__not__in set

*A*.

● P(Ā) =1-P(

*A*)

● P(

*A*∩

*B*) represents the probability that the 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.

● Two events are mutually exclusive when they cannot occur at the same time.

For mutually exclusive events

*A*and

*B*:

*A*∪

*B*)=P(

*A*)+P(

*B*)

● Two events are independent if one event does not affect the other event.

For two independent events

*A*and

*B*:

*A*∩

*B*)=P(

*A*)×P(

*B*)

**Mutually Exclusive Events (Disjoint Sets)**

Two events are mutually exclusive if

*A*∩

*B*=Ø

This means that

*A*and

*B*have no elements in common.

Draw a Venn Diagram that depicts two mutually exclusive events.

📌 Example 1. *M* and *N* are mutually exclusive events.

P(*M*)=4/9, P(*N*)=⅓

Work out P(*M∪N*).

✍

→[

*M*and

*N*are mutually exclusive events. So use P(

*M∪N*)=P(

*M*)+P(

*N*)]

⚠ Pay attention to every

underlinedsingular noun in Ex2 to Ex12 just on this page!

📌 Ex2. From aset of 15 cards whose faces are numbered 1 to 15, __one card__ is drawn at random. What is the probability that it is a multiple of 3 or 5 or both?

✍ Solution:

The sample space, if, is a set of 15 points.

*A*={3,6,9,12,15} and so n(*A*)=5

*B*={5,10,15} and so n(*B*)=3

There is one point in the intersection of *A* and B; i.e., there is one point that belongs to both *A* and *B*, namely the number 15.

The intersection of *A* and *B* is the set of all points belonging to both *A* and *B* andis denoted by *A*∩*B* and so n(*A*∩*B*)=1.

n(*A*∪*B*)=7, because the number 15 is a multiple of both 3 and 5, so the number 15 must be included only once, regarding it as a multiple of either 3 or 5. Thus

In this case,

*A*and

*B*are not mutually exclusive.

*A* and *B* are mutually exclusive, if the set of points in the intersection of *A* and *B* is a null set (an empty set). That is, *A*∩*B*=Ø={}, in which case P(*A*∩*B*)=0.

📌 Ex3. __A number__ is chosen at random from the ﬁrst 25 natural numbers. What is the probability that:

(i) it is multiple of 5 or 7, (ii) it is multiple of 3 or 7.

✍ Solution:

Here, *S*={1, 2, 3, …, 25}.

(i) Let *A*= the number which is a multiple of 5={5, 10, 15, 20, 25}. Then P(*A*)=5/25.

Let *B*= the number which is a multiple of 7={7, 14, 21}. Then P(*B*)=3/25.

Here, *A* and *B* are mutually exclusive because *A*∩*B*=Ø={},. Therefore,

(ii) Let

*C*= the number which is a multiple of 3

*C*)=8/25.

Let

*D*= the number which is a multiple of 7 ={7, 14, 21}. Then P(

*D*) =3/25.

Here,

*C*and

*D*are not mutually exclusive because

*C*∩

*D*={21}. Then P(

*)=1/25. Thus,*

*C*∩D💎 Probability: Play with numbers (part 1: The 10 easiest answered-questions)

📌 Ex4.

*A*dice and a coin are thrown.

Event F is getting a 5 on the dice. Event H is getting

__a head__on the coin.

Workout: a) P(

*F*), b) P(

*H*), c) P(

*F∩H*).

✍

Throwing a die and throwing a coin are independent events, since the outcome of one event does not affect the outcome of the other event. So use P(

A∩B)=P(A)×P(B).

a) P(*F*)=⅙, b) P(*H*)=½, c) P(*F∩H*)=⅙×½=1/12

💎 Probability to get anything on Rolling or Throwing a Die

📌 Ex5. __A card__ is drawn from a pack of 52 cards. What is the probability that it is either a spade or an ace?

✍ Solution:

Let *A*= the spade card and *B*=the ace card.

Then

Therefore,

P(the card is either a spade or an ace)

📌 Ex6. __A card__ is drawn from a Well—shuffled deck of 52 playing cards. What is the probability that it is a queen or a heart?

✍

*Q*=Queen and *H*=Heart

💎 Probability to get a certain card, Card Withdrawal Chance, featured by 20 examples and more

📌 Ex7. The probability that a student belongs to a club is P(

*C*)=0.4. The probability that

__a student__works part time is P(PT)=0.5. The probability that a student belongs to a club AND works part time is P(C and PT)=0.05. What is the probability that

__a student__belongs to a club OR works part time?

✍ Answer:

*C*∪PT)=P(

*C*)+P(PT)-P(

*C*∩PT)

=0.4+0.5-0.05

=0.85

📌 Ex8.

A=owns a car |
B=has a pet |

P(A)=0.87 |
P(B)=0.57 |

*A*and

*B*)=0.53

What is the probability that a student owns a car OR has a pet??

✍ Answer:

*A*OR

*B*)=P(

*A*)+P(

*B*)-P(

*A*AND

*B*)

=0.87+0.57-0.53=0.91

📌 Ex9. A survey finds that 56% of people are married. They ask the same group of people, and 67% have at least one child. If there are 41% that are married and have at least one child, what is the probability that __a person__ in the survey is married OR has a child?

✍ Answer:

💎 Probability that Either Event A or B, or Both will happen

**Complement Rule**

The complement rule is a way to calculate a probability based on the probability of its complement.

❶ By the definition of complement

*A*∪

*A*

^{∁}=U=U

❷ Apply the probability operator

*A*∪

*A*

^{∁})=P(U)=1

❸ Since

*A*and

*A*

^{∁}are mutually exclusive

*A*∪

*A*

^{∁})=P(

*A*)+P(

*A*

^{∁})

❹ Hence we get P(

*A*)=1-P(

*A*

^{∁})

Suppose we rolled a fair, six-sided die 10 times. Let *T* be the event that we roll at least 1 three. If one were to calculate *T* you would need to find the probability of 1 three, 2 threes, …, and 10 threes and add them all up. However, you can use the complement rule to calculate P(*T*).

Solution.

Let *X* be the times that we rolled a 3, then

If we apply the complement rule

*T*)=1-P(

*T*

^{∁})=1-P(

*X*=0)

📌 Ex10. You are given that the P(*A*)=0.4. Calculate P(*A*^{∁}).

✍ P(*A*^{∁})=1-0.4=0.6

📌 Ex11. If P(*A*)=0.6,P(*B*)=0.5 and P(*A*∪*B*=0.4), find:

(a) P(*A*∩*B*), (b) P(Ā), (c) P(B̄), (d) P(Ā∩B̄).

✍ Solution:

(a) For any two events *A* and *B* we know that

*A*∪

*B*)=P(

*A*)+P(

*B*)-P(

*A*∩

*B*).

Therefore,

*A*∩

*B*)=P(

*A*)+P(

*B*)-P(

*A*∪

*B*)

=0.6+0.5-0.4=0.7.

(b) P(Ā)=1-P(

*A*)=1-0.6=0.4.

(c) P(B̄)=1-P(

*B*)=1-0.5=0.5.

(d) P(Ā+B̄)=1-P(Ā+B̄)

*A*∩

*B*)

=1-0.7=0.3.

📌 Ex12. Two friends, David and Colin, frequently play golf and tennis with each other. In the long run, it has been found that David wins 3 rounds of golf out of every 5, and one game of tennis out of every 4. If they play __one round__ of golf and __one game__ of tennis, find the probability that David

(a) wins both,

(b) loses both,

(c) wins the round of golf only,

(d) wins either the golf or the tennis but not both.

✍ Solution:

We assume that no game ends in a draw.

Let *A* be the event “David wins the golf”, and so Ᾱ is the event “David loses the golf”.

*A*)+P(Ᾱ)=1

P(

*A*)=⅗ and P(Ᾱ)=⅖

Let

*B*be the event “David wins the tennis”, and so B̄ is the event “David loses the tennis”.

*B*)=¼ and P(B̄)=¾

On the assumption that

*A*and Bare independent events, then

(a) P(

*A*∩

*B*)=P(

*A*)⋅P(

*B*)=⅗⋅¼=3/20

(b) P(Ᾱ∩B̄)=P(Ᾱ)⋅P(B̄)=⅖⋅¾=3/10

(c) P(

*A*∩B̄)=P(

*A*)⋅P(B̄)=⅗⋅¾=9/20

(d) The statement “David wins either the golf or the tennis but not both” implies that

(i) David wins the golf and loses the tennis, or

(ii) David loses the golf and wins the tennis.

(i) and (ii) are mutually exclusive.

(i) P(

*A*∩B̄)=P(

*A*)⋅P(B̄)=⅗⋅¾=9/20

(ii) P(Ᾱ∩

*B*)=P(Ᾱ)⋅P(

*B*)=⅖⋅¼=2/20

Hence the required probability

💎 Compound Events: Probability of Complement of An Event