The Practice of Statistics, 5th Edition

Starnes, Tabor, Yates, Moore

Bedford Freeman Worth Publishers

CHAPTER 5Probability: What Are the Chances?

5.3

Conditional Probability and Independence

The Practice of Statistics, 5th Edition 2

5.3 Reading Quiz: Matching

1. formula for calculating conditional probability

2. general multiplication rule

3. formula for determining independent events

4. multiplication rule for independent events

5. Can disjoint events be independent?

a) 𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 ∗ 𝑃(𝐵)

b) 𝑃 𝐴 𝐵 =𝑃(𝐴∩𝐵)

𝑃(𝐵)

c) 𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 ∗ 𝑃 𝐴 𝐵

d) 𝑃 𝐴 𝐵 = 𝑃(𝐴)

e) 𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 ∗ 𝑃(𝐵|𝐴)

f) yes

g) no

The Practice of Statistics, 5th Edition 7

Try this

A recent census at a major university revealed that 40% of its students

mainly used Macintosh computers (Macs). The rest mainly used PCs. At the

time of the census, 67% of the school’s students were undergraduates. The

rest were graduate students. In the census, 23% of the respondents were

graduate students who said that they used PCs as their primary computers.

Suppose we select a student at random from among those who were part of

the census and learn that the student mainly uses a PC. Find the probability

that this person is a graduate student. Show your work.

𝑃 𝐺 𝑃𝐶 =𝑃(𝐺∩𝑃𝐶)

𝑃(𝑃𝐶)=

0.23

0.60= 0.383. Given that a student mainly uses a PC,

there is a 0.383 probability that the person is a graduate student.

0.10 0.370.23

0.30

Mac PC Total

Graduate Student

0.10 0.23 0.33

Undergraduate Student

0.30 0.37 0.67

Total 0.40 0.60 1.0

Learning Objectives

After this section, you should be able to:

The Practice of Statistics, 5th Edition 8

✓ USE the general multiplication rule to CALCULATE probabilities.

✓ USE tree diagrams to MODEL a chance process and CALCULATE

probabilities involving two or more events.

Conditional Probability and Independence

The Practice of Statistics, 5th Edition 9

The General Multiplication Rule

The probability that events A and B both occur can be found

using the general multiplication rule

P(A ∩ B) = P(A) • P(B | A)

where P(B | A) is the conditional probability that event B occurs

given that event A has already occurred.

General Multiplication Rule

In words, this rule says that for both of two events to occur, first one

must occur, and then given that the first event has occurred, the

second must occur.

The Practice of Statistics, 5th Edition 10

Playing in the NCAA

About 55% of high school students participate in a school athletic team

at some level, and about 5% of these athletes go on to play on a

college team in the NCAA

(http://www.washingtonpost.com/wp-

dyn/content/article/2009/09/23/AR2009092301947.html,

http://www.collegesportsscholarships.com/percentage-high-school-

athletes-ncaa-college.htm).

Problem: What percent of high school students play a sport in high

school and go on to play a sport in the NCAA?

Solution: We know P(high school sport) = 0.55 and P(NCAA sport |

high school sport) = 0.05, so P(high school sport and NCAA sport) =

P(high school sport) x P(NCAA sport | high school sport) = (0.55)(0.05)

= 0.0275. Almost 3% of high school students will play a sport in high

school and in the NCAA.

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Tree Diagrams

The general multiplication rule is especially useful when a chance

process involves a sequence of outcomes. In such cases, we can use a

tree diagram to display the sample space.

Consider flipping a coin

twice.

What is the probability of

getting two heads?

Sample Space:

HH HT TH TT

So, P(two heads) = P(HH) = 1/4

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Example: Tree Diagrams

 

P(online) = 0.93

P(profile | online) = 0.55

 

P(online and have profile) = P(online) × P(profile |online)

 

= (0.93)(0.55)

= 0.5115

51.15% of teens are online and have posted

a profile.

The Pew Internet and American Life Project finds that 93% of

teenagers (ages 12 to 17) use the Internet, and that 55% of online

teens have posted a profile on a social-networking site.

What percent of teens are online and have posted a profile?

The Practice of Statistics, 5th Edition 13

Late for school

Shannon hits the snooze bar on her alarm clock on 60% of school days. If she

doesn’t hit the snooze bar, there is a 0.90 probability that she makes it to class

on time. However, if she hits the snooze bar, there is only a 0.70 probability

that she makes it to class on time. On a randomly chosen day, what is the

probability that Shannon is late for class?

The probability that she hits the snooze bar and is late is (0.60)(0.30) = 0.18. The

probability that she doesn’t hit the snooze bar and is late is (0.40)(0.10) = 0.04.

Because these outcomes are mutually exclusive, P(late) = 0.18 + 0.04 = 0.22.

Shannon is late on about 22% of school days.

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Late for school

Suppose that Shannon is late for school. What is the probability that

she hit the snooze bar that morning?

𝑃(ℎ𝑖𝑡𝑠 𝑠𝑛𝑜𝑜𝑧𝑒 𝑏𝑎𝑟 | 𝑙𝑎𝑡𝑒 𝑓𝑜𝑟 𝑠𝑐ℎ𝑜𝑜𝑙) =𝑃(ℎ𝑖𝑡𝑠 𝑠𝑛𝑜𝑜𝑧𝑒 𝑏𝑎𝑟 𝑎𝑛𝑑 𝑙𝑎𝑡𝑒 𝑓𝑜𝑟 𝑠𝑐ℎ𝑜𝑜𝑙)

𝑃(𝑙𝑎𝑡𝑒 𝑓𝑜𝑟 𝑠𝑐ℎ𝑜𝑜𝑙)

𝑃(ℎ𝑖𝑡𝑠 𝑠𝑛𝑜𝑜𝑧𝑒 𝑏𝑎𝑟 | 𝑙𝑎𝑡𝑒 𝑓𝑜𝑟 𝑠𝑐ℎ𝑜𝑜𝑙) =0.18

0.18 + 0.04= 0.818

When Shannon is late, there is a 0.818 probability that she hit the snooze bar that

morning.

Section Summary

In this section, we learned how to…

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✓ USE the general multiplication rule to CALCULATE probabilities.

✓ USE tree diagrams to MODEL a chance process and CALCULATE

probabilities involving two or more events.

Conditional Probabilities and Independence

The Practice of Statistics, 5th Edition 16

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Homework