ProbabilityDefinition:

Probability: the chance an event will happen.

# of ways a certain event can occur

# of possible events Probability =

Probability must be a value between 0 and 1.

The probability of the set of all possible outcomes of a trial must be 1

The probability of an event occurring is 1 minus the probability that it does not

occur. Complement of A(Ac ) A

Rule for Complementary Events:

P (A) + P(Ac)= 1

Example:

If the probability that a person lives in an industrialized country of the world is 1/5,

find the probability that a person does not live in an industrialized country.

Answer:

P (not living in an industrialized country) = 1- P (living in an industrialized

country)= 1 – 1/5= 4/5

Example 1:

What is the probability of rolling a 6 on a six sided die?

•A six sided die is labeled 1,2,3,4,5,6

•A 6 occurs only once in rolling a die

Know:

# of ways a certain outcome can occur

# of possible outcomes Probability =

1

6 Probability =

Simple Probability

Example 2

In a statistics class, 32 students of which 20 are females are selected to participate

in a study of eye color. It is discovered that 7 of the 32 students have blue eyes. It

is also noted that 5 out of the 20 females have blue eyes.

Males Females Total

Blue Eyes

No Blue Eyes

Total

Answer:

1. P(Blue eyes) = 7/32 = 0.219

2. P(Females) = 20/32 = 0.625

3. P(Males) = 12/32 = 0.375

4. P(No Blue eyes) = 25/32 = 0.781

Males Females Total

Blue Eyes 2 5 7

No Blue Eyes 10 15 25

Total 12 20 32

Example 4 :

In a sample of 50 people, 21 had type “O” blood, 22 had type “A”, 5 had type “B”

blood and 2 had type “AB” blood. Set up a frequency distribution and find the

following probabilities:

1.A person has type “O” blood.

2.A person has type “A” or type “B” blood.

3.A Person had neither “A” nor type “O” blood.

4.A person does not have type “AB” blood.

Example 4 :

In a sample of 50 people, 21 had type “O” blood, 22 had type “A”, 5 had type “B”

blood and 2 had type “AB” blood. Set up a frequency distribution and find the

following probabilities:

Group Frequency

Type “O” 21

Type “ A” 22

Type “B” 5

Type “AB” 2

Total 50

Answer:

Group Frequency

Type “O” 21

Type “ A” 22

Type “B” 5

Type “AB” 2

Total 50

1. A person has type “O” blood. P (X = “O”) = 21/50 = 0.42

2. A person has type “A” or type “B” blood. P (X = “A” or “B”) = 26/50 = 0.52

3. A Person had neither “A” nor type “O” blood.

P ( X = “B” or “ AB”) = 7/50 = 0.14

4. A person does not have type “AB” blood.

P ( X does not have type “AB”) = 48/50 = 0.96

Conditional ProbabilityFormula for Conditional Probability:

The probability that the second event B occurs given the first event A has occurred can be found by dividing the probability that both occurred by the probability that the first event has occurred.

( )( | )

( )

P A and BP B A

P A

A and B

Example 5 (Conditional Probability)

What is the probability that a student selected at random is a female given that the student has blue eyes?

P(Female | blue eyes) = 5/7 = 0.714

What is the probability that a student selected at random has blue eyes given that the student is male?

P(Blue eyes| Male) = 2/12 = 0.167

Males Females Total

Blue Eyes 2 5 7

No Blue Eyes 10 15 25

Total 12 20 32

Compound Probabilities

Events that occur in combination

• P(blue eyes and female) or in general:

P(A and B)

Events that occur as alternatives

• P(blue eyes or female) or in general:

P(A or B)

Multiplication (‘AND’) Law

Equation #1: If A and B are independent, then;

P (A and B) = P(A) x P(B)

Equation #2: If A and B are not independent i.e dependent, then;

P (A and B) = P (A | B) x P (B)

or

P (B | A) x P (A)

Test of Independent Events:

• Two events A and B are independent if the fact that A occurs does not

affect the probability of B occurring

•Two events A and B are independent events if

P(A | B) = P (A) or P(B | A) = P (B)

Note: If two events are not independent, they are dependent.

Example :

Males Females Total

Blue Eyes 2 5 7

No Blue Eyes 10 15 25

Total 12 20 32

In a statistics class, 32 students of which 20 are females are selected to

participate in a study of eye color. It is discovered that 7 of the 32 students have

blue eyes. It is also noted that 5 out of the 20 females have blue eyes.

Example :

What is the probability of being a female and having blue eyes?

Step 1: Is having blue eyes dependent on gender?

P(Blue eyes | Female) = P (Blue eye )

5/20 ≠ 7/32

Thus having blue eyes is dependent on gender.

Step 2: Use Equation #2:

P (Female and Blue eyes) = P (Blue eyes | Female) x P (Female)

= 5/20 x 20/32 = 5/32 = 0.156

Example :

A coin is flipped and a die is rolled. Find the probability of getting a head on the coin

and a 4 on the die.

Answer:

P (head and 4) = P (head). P (4)= 1/2 * 1/6 = 1/12 = 0.083

Addition (‘OR’) LawEquation #1: If A and B are mutually exclusive:

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

Equation #2: If A and B are not mutually exclusive:

P(A or B) = P(A) + P(B) – {P(A and B)}

P(A) P(B)(+)

P(A) P(B)

Note:

•Two events are mutually exclusive if they cannot occur at the same time (i.e. P(A and B) = 0)

A B

A and B

Example :

Males Females Total

Blue Eyes 2 5 7

No Blue Eyes 10 15 25

Total 12 20 32

In a statistics class, 32 students of which 20 are females are selected to

participate in a study of eye color. It is discovered that 7 of the 32 students have

blue eyes. It is also noted that 5 out of the 20 females have blue eyes.

Example :

What is the probability a student selected at random will be a female or has blue

eyes?

Answer:

Step 1: Is having blue eyes and gender mutually exclusive?

Since a given individual can be a female and have blue eyes, thus they are not

mutually exclusive.

Step 2: Equation #2:

P(Blue eyes or Female) = P(Blue Eyes) + P(Female) – [P(Blue eyes and Female)] =

7/32 + 20/32 – 5/32 = 22/32 = 0.688

Example :

A day of the week is selected at random. Find the probability that it is a

weekend day.

Answer:

P (Saturday or Sunday) = P ( Saturday) + P (Sunday)

= 1/7 + 1/7 = 2/7 = 0.286