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Lesson 14-3 Probability of Compound Events. Transparency 3 Click the mouse button or press the Space Bar to display the answers.

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Presentation on theme: "Lesson 14-3 Probability of Compound Events. Transparency 3 Click the mouse button or press the Space Bar to display the answers."— Presentation transcript:

1 Lesson 14-3 Probability of Compound Events

2 Transparency 3 Click the mouse button or press the Space Bar to display the answers.

3 Transparency 3a

4 Objectives Find the probability of two independent events or dependent events Find the probability of two mutually exclusive or inclusive events

5 Vocabulary Simple event – Compound event – Independent events – Dependent events – Complements – Mutually exclusive – Inclusive –

6 Probability Part 1 Independent Events – one event does not affect the outcome of another Example: Rolling a 5 on a die twice or flipping a coin to get a head twice Dependent Events – one event effects the outcome of another Example: Drawing two cards from a normal deck of cards without replacing the first card and getting two aces P(A and B) = P(A)  P(B) P(A and B) = P(A)  P(B given A) P(5 and 5) = P(5)  P(5) = (1/6)  (1/6) = (1/36) AB P(A and A) = P(A)  P(A given 1 st A) = (4/52)  (3/51) = (12/2652)

7 Probability Part 2 Mutually Exclusive Events – events that cannot occur at the same time Example: Drawing a heart or a club from a normal deck of cards Inclusive Events – events overlap Example: Drawing an ace or a heart from a normal deck of cards Complement Events – all other events Example: flipping a coin to get a head or not a head (a tail) P(A or B) = P(A)+P(B) AB P(head) + P(not a head) = 1 AB P(A or B) = P(A)+P(B) - P(A and B) P(H or C) = P(H)+P(C) =(1/4)+(1/4) = (1/2) P(A or H) = P(A)+P(H) – P(A and H) =(4/52)+(13/52) - (1/52) = (16/52)

8 Example 1 Roberta is flying from Birmingham to Chicago to visit her grandmother. She has to fly from Birmingham to Houston on the first leg of her trip. In Houston she changes planes and heads on to Chicago. The airline reports that the flight from Birmingham to Houston has a 90% on time record, and the flight from Houston to Chicago has a 50% on time record. What is the probability that both flights will be on time?

9 Example 1 cont 0.90.5 Multiply. Answer:The probability that both flights will be on-time is 45%. Definition of independent events

10 Example 2 At the school carnival, winners in the ring-toss game are randomly given a prize from a bag that contains 4 sunglasses, 6 hairbrushes, and 5 key chains. Three prizes are randomly drawn from the bag and not replaced. A. Find P( sunglasses, hairbrush, key chain ). The selection of the first prize affects the selection of the next prize since there is one less prize from which to choose. So, the events are dependent. First prize: Second prize: Third prize:

11 Example 2 cont Multiply. Substitution Answer:The probability of drawing sunglasses, a hairbrush, and a key chain is 4/91 ≈ 0.044

12 Example 2b cont At the school carnival, winners in the ring-toss game are randomly given a prize from a bag that contains 4 sunglasses, 6 hairbrushes, and 5 key chains. Three prizes are randomly drawn from the bag and not replaced. B. Find P( hairbrush, hairbrush, key chain ). Notice that after selecting a hairbrush, not only is there one fewer prize from which to choose, there is also one fewer hairbrush. Multiply. Answer:The probability of drawing two hairbrushes and then a key chain is 5 out of 91 or 5/91 ≈ 0.055

13 Example 2c cont At the school carnival, winners in the ring-toss game are randomly given a prize from a bag that contains 4 sunglasses, 6 hairbrushes, and 5 key chains. Three prizes are randomly drawn from the bag and not replaced. C. Find P( sunglasses, hairbrush, not key chain ). Since the prize that is not a key chain is selected after the first two prizes, there are 10 – 2 or 8 prizes that are not key chains. Multiply. Answer:The probability of drawing sunglasses, a hairbrush, and not a key chain is 32/455 ≈ 0.07

14 Example 3 Alfred is going to the Lakeshore Animal Shelter to pick a new pet. Today, the shelter has 8 dogs, 7 cats, and 5 rabbits available for adoption. If Alfred randomly picks an animal to adopt, what is the probability that the animal would be a cat or a dog? Since a pet cannot be both a dog and a cat, the events are mutually exclusive.

15 Example 3 cont Definition of mutually exclusive events Substitution Add. Answer:The probability of randomly picking a cat or a dog is ¾ or 0.75

16 Example 4 A dog has just given birth to a litter of 9 puppies. There are 3 brown females, 2 brown males, 1 mixed-color female, and 3 mixed-color males. If you choose a puppy at random from the litter, what is the probability that the puppy will be male or mixed-color? Since three of the puppies are both mixed-colored and males, these events are inclusive. Definition of inclusive events

17 Example 4 cont Substitution LCD is 9. Simplify. Answer:The probability of a puppy being a male or mixed-color is 2/3 or about 67%.

18 Summary & Homework Summary: –For independent events, use P(A and B) = P(A)  P(B) –For dependent events, use P(A and B) = P(A)  P(B following A) –For mutually exclusive events, use P(A or B) = P(A) + P(B) –For inclusive events, use P(A or B) = P(A) + P(B) – P(A and B) Homework: –none


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