Chapter 6Lesson 6.6

Probability6.6 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 E1 = first card is a spade and E2 = second card is a spade independent?NOP(E1 and E2) =

P(E1) × P(E2|E1) =

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 MultiplicationFor any two events E and F,

Ask yourself, “ Are these events independent?”

)( BAP

Yes No

)()( BPAP

)|()( ABPAP

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?

2.146

157

)G P(G 21

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 E1 = the first bulb is defective and E2 = 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 = .05

499/9999 = .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?

defective) eP(defectiv

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 D1 = first light bulb is defective

D2 = second light bulb is defective

= (.05)(.95) + (.95)(.05) = .095

2121defective) one exactly( DDDDPP CC

General Rule for Addition

For any two events E and F,

)()()()( FEPFPEPFEP

E F

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?

.4 + .3 -.1 = .6

)( FEP Ask yourself, “Are the events mutually exclusive?”

)()( FPEP

Yes No

)()()( FEPFPEP

If independent

)()( FPEP

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?

)()()()( BAPBPAPBAP

3611

61

61

61

61

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 B1 = event that brand 1 is purchased

B2 = event that brand 2 is purchased

E = event that extended warranty is purchased

P(B1) = .7 P(B2) = .3 P(E|B1) = .2 P(E|B2) = .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 playerOR

2) They purchased the extended warranty and Brand 2 DVD player

21 BEBEE

DVD Player Continued . . .

Let B1 = event that brand 1 is purchased

B2 = event that brand 2 is purchased

E = event that extended warranty is purchased

P(B1) = .7 P(B2) = .3 P(E|B1) = .2 P(E|B2) = .4

If a DVD customer is selected at random, what is the probability that they purchased the extended warranty? 21 BEBEE

These are disjoint events

21 BEPBEPEP

Use the General Multiplication Rule:

)(|)(| 2211 BPBEPBPBEPEP

DVD Player Continued . . .

Let B1 = event that brand 1 is purchased

B2 = event that brand 2 is purchased

E = event that extended warranty is purchased

P(B1) = .7 P(B2) = .3 P(E|B1) = .2 P(E|B2) = .4

If a DVD customer is selected at random, what is the probability that they purchased the extended warranty? )(|)(| 2211 BPBEPBPBEPEP

P(E) = (.2)(.7) + (.4)(.3) = .26

Practice with Homework

• Pg.372: #59-63odd, 64, 65, 76, 77