Probability using And Or and Complements Independent Events

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 2 nd 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)

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