Probability using And,Or and Complements

Independent Events

• Two events are Independent if the occurrence of 1 has no effect on the occurrence of the other. (a coin tossed 2 times, the first toss has no effect on the 2nd toss)

• If A & B are independent events then the probability that both A & B occur is:

• P(A and B) = P(A) • P(B)

• A number cube is rolled and a coin is tossed. Find the probabilities:

1. P(5)

2. P(heads)

3. P(5 and heads)

1. P(5) = 1/6

2. P(heads) = 1/2

3. P(5 and heads) = P(5) P(heads)

=

=

1 1

6 2

1

12

Dependent Events

• Two events A and B are dependent events if the occurrence of one affects the occurrence of the other.

Dependent Events

• If A & B are dependent events, then the probability that both A & B occur is:

• P(A&B) = P(A) * P(B/A)

• The probability that B will occur given that A has occurred is called the conditional probability of B given A and is written P(B|A).

Comparing Dependent and Independent Events

• You randomly select two cards from a standard 52-card deck. What is the probability that the first card is not a face card (a king, queen, or jack) and the second card is a face card if

• (1) you replace the first card before selecting the second, and

• (2) you do not replace the first card?

• (1) If you replace the first card before selecting the second card, then A and B are independent events. So, the probability is:

• P(A and B) = P(A) • P(B) = 40 * 12 = 30 52 52 169

• ≈ 0.178• (2) If you do not replace the first card before

selecting the second card, then A and B are dependent events. So, the probability is:

• P(A and B) = P(A) • P(B|A) = 40*12 = 40 52 51 221

• ≈ .0181

Mutually Exclusive Events

Intersection of A & B

• To find P(A or B) you must consider what outcomes, if any, are in the intersection of A and B.

• If there are none, then A and B are mutually exclusive events and

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

• If A and B are not mutually exclusive, then the outcomes in the intersection (A and B) are counted twice when P(A) and P(B) are added.

• So P(A and B) must be subtracted once from the sum

EXAMPLE 1

• One six-sided die is rolled.

• What is the probability of rolling a multiple of 3 or 5?

• P(A or B) = P(A) + P(B) = 2/6 + 1/6 = 1/2

• 0.5

EXAMPLE 2

• One six-sided die is rolled. What is the probability of rolling a multiple of 3 or a multiple of 2?

• A = Mult 3 = 2 outcomes (3,6)

• B = mult 2 = 3 outcomes (2,4,6)

• P(A or B) = P(A) + P(B) – P(A and B)

• P(A or B) = 2/6 + 3/6 – 1/6 =

• 2/3 ≈ 0.67

EXAMPLE 3

• In a poll of high school juniors, 6 boys took French and 8 girls took french,11 boys took math class and 7 girls took math.

• How many juniors surveyed were either girl or took math?

• A = girl

• B = took math

• P(A) = 15/32, P(B) = 18/32

• P(A or B) = P(A) + P(B) – P(A and B)

• P(A or B) = 15/32 + 18/32 – 7/32

= 26/32

= 13/16

Using complements to find Probability

• The event A’, called the complement of event A, consists of all outcomes that are not in A.

• The notation A’ is read ‘A prime’.

Probability of the complement of an event

• The probability of the complement of A is :

• P(A’) = 1 - P(A)

EXAMPLE 4

• A card is randomly selected from a standard deck of 52 cards.

• Find the probability of the given event.

• a. The card is not a king. 1 – P(king) = 1 – 4/52

= 48/52 ≈ 0.923

• b. The card is not an ace or a jack.

• P(not ace or jack)

• 1 – P(ace or jack)= 1- P(4/52 + 4/52)

= 1- 8/52

= 44/52 ≈ 0.846

• In a survey of 200 pet owners, 103 owned dogs, 88 owned cats, 25 owned birds, and 18 owned reptiles.

• 1. None of the respondents owned both a cat and a bird.• What is the probability that they owned a cat or a bird?• 113/200 • = 0.565

• 2. Of the respondents, 52 owned both a cat and a dog. • What is the probability that a respondent owned a cat or a

dog?• 139/200• = 0.695

Assignment

Worksheet on Probability #’s 1-24