Chapter 6 Lesson 6.6 Probability 6.6 General Probability Rules

Example 1: Suppose I will pick two cards from a standard deck. This can be done two ways: 1)Pick a card at random, replace the card, then pick a second card 2) Pick a card at random, do NOT replace, then pick a second card. If I pick two cards from a standard deck without replacement, what is the probability that I select two spades? Are the events E 1 = first card is a spade and E 2 = second card is a spade independent? NO P(E 1 and E 2 ) = P(E 1 ) × P(E 2 |E 1 ) = Sampling with replacement – the events are typically independent events. Sampling without replacement – the events are typically dependent events. Probability of a spade given I drew a spade on the first card.

General Rule for Multiplication For any two events E and F,

Ask yourself, “ Are these events independent? ” Yes No Here is a process to use when calculating the intersection of two or more events.

There are seven girls and eight boys in a math class. The teacher selects two students at random to answer questions on the board. What is the probability that both students are girls? Are these events independent? NO

Example 2: Suppose the manufacturer of a certain brand of light bulbs made 10,000 of these bulbs and 500 are defective. You randomly pick a package of two such bulbs off the shelf of a store. What is the probability that both bulbs are defective? Are the events E 1 = the first bulb is defective and E 2 = the second bulb is defective independent? What would be the probability of selecting a defective light bulb? 500/10,000 =.05 To answer this question, let ’ s explore the probabilities of these two events?

Light Bulbs Continued... What would be the probability of selecting a defective light bulb? Having selected one defective bulb, what is the probability of selecting another without replacement? 500/10,000 = /9999 =.0499 These values are so close to each other that when rounded to three decimal places they are both.050. For all practical purposes, we can treat them as being independent.

Light Bulbs Continued... What is the probability that both bulbs are defective? Are the selections independent? We can assume independence. (0.05)(0.05) =.0025

Light Bulbs Revisited... A certain brand of light bulbs are defective five percent of the time. You randomly pick a package of two such bulbs off the shelf of a store. What is the probability that exactly one bulb is defective? Let D 1 = first light bulb is defective D 2 = second light bulb is defective = (.05)(.95) + (.95)(.05) =.095

General Rule for Addition For any two events E and F, EF Since the intersection is added in twice, we subtract out the intersection.

Musical styles other than rock and pop are becoming more popular. A survey of college students finds that the probability they like country music is.40. The probability that they liked jazz is.30 and that they liked both is.10. What is the probability that they like country or jazz? =.6

Ask yourself, “ Are the events mutually exclusive? ” Yes No If independent Here is a process to use when calculating the union of two or more events. In some problems, the intersection of the two events is given (see previous example). In some problems, the intersection of the two events is not given, but we know that the events are independent.

Suppose two six-sided dice are rolled (one white and one red). What is the probability that the white die lands on 6 or the red die lands on 1? Let A = white die landing on 6 B = red die landing on 1 Are A and B disjoint? NO How can you find the probability of A and B?

An electronics store sells DVD players made by one of two brands. Customers can also purchase extended warranties for the DVD player. The following probabilities are given: Let B 1 = event that brand 1 is purchased B 2 = event that brand 2 is purchased E = event that extended warranty is purchased P(B 1 ) =.7P(B 2 ) =.3 P(E|B 1 ) =.2 P(E|B 2 ) =.4 If a DVD customer is selected at random, what is the probability that they purchased the extended warranty? This can happen in one of two ways: 1)They purchased the extended warranty and Brand 1 DVD player OR 2) They purchased the extended warranty and Brand 2 DVD player

DVD Player Continued... Let B 1 = event that brand 1 is purchased B 2 = event that brand 2 is purchased E = event that extended warranty is purchased P(B 1 ) =.7P(B 2 ) =.3 P(E|B 1 ) =.2 P(E|B 2 ) =.4 If a DVD customer is selected at random, what is the probability that they purchased the extended warranty? These are disjoint events Use the General Multiplication Rule:

DVD Player Continued... Let B 1 = event that brand 1 is purchased B 2 = event that brand 2 is purchased E = event that extended warranty is purchased P(B 1 ) =.7P(B 2 ) =.3 P(E|B 1 ) =.2 P(E|B 2 ) =.4 If a DVD customer is selected at random, what is the probability that they purchased the extended warranty? P(E) = (.2)(.7) + (.4)(.3) =.26

Practice with Homework Pg.372: #59-63odd, 64, 65, 76, 77