Chapter 5: Probability What are the Chances?

Section 5-3: Conditional Probability and Independence

2

5

2

5 P(A)⋅ P(B)=

2

5⋅2

5=

4

25

= 0.16

2

5

1

4 P(A)⋅ P(B)=

2

5⋅1

4=

2

20

= 0.10

conditional

P(B|A)

P(B|A)

Was one way more difficult to win?

independent

P(B)

P(B) & P(A|B) = P(A)

dependent

P(A) ⋅ P(B|A)

P(B) P(A) ⋅ P(B)

Mutually Exclusive Events: Independent or Dependent?

Choose a U.S. adult at random.

Define Event A: the person is male Event B: the person is pregnant

Are events A & B mutually exclusive (disjoint)? If you know that event A has occurred, does this affect the probability of event B happens?

So are events A & B independent or dependent?

Yes, they can’t happen together.

YES! If we know the person is male, then the chance that that person is pregnant is 0.

Dependent! P(B|A) ≠ P(B)

Two mutually exclusive events are NEVER independent.

P(HS) = 310

500

= .62

P(HS | owns home) = 221

340 = 0.65

Being a high school grad and owning a home are dependent events because 0.62 ≠ 0.65.

P(owns home |HS) = 221310 = 0.71

They are completely different conditional probabilities. Order Matters!!!

Go to the page with all the Check Your Understanding problems. You will now work on the first and second sets.

Students at the University of New Harmony received 10,000 course grades last semester. The two-way table breaks down these grades by which school of the university taught the course. The schools are Liberal Arts, Engineering and Physical Sciences, and Health and Human Services. College grades tend to be lower in engineering and the physical sciences (EPS) than in liberal arts and social sciences (which includes Health and Human Services). Consider the two events E: the grade comes from an EPS course, and L: the grade is lower than a B.

1) Find P(L). Interpret this probability in context.

2) Find P(E | L) and P(L | E). Which of these conditional probabilities tells you whether this college’s EPS students tend to earn lower grades than students in liberal arts and social sciences? Explain.

1) Find P(L). Interpret this probability in context.

P(L) = 2268 + 800 + 588

10000 =

3656

10000= 0.3656

Total # of course grades: 10,000

L: the grade is lower than a B

36.56% of grades are lower than a B.

2) Find P(E | L) and P(L | E). Which of these conditional probabilities tells you whether this college’s EPS students tend to earn lower grades than students in liberal arts and social sciences? Explain.

𝑃 𝐸 𝐿) =

𝑃 𝐿 𝐸) =

E: the grade comes from an EPS course L: the grade is lower than a B

3656

1600

800

2268 + 800 + 588=

800

3656= 0.219

800

368 + 432 + 800=

800

1600= 0.50

𝑃 𝐸 𝐿) is the probability of someone being an EPS student, given they have a low grade. 𝑃 𝐿 𝐸) is the probability of someone having a low grade, given they are an EPS student. We would want to compare this to the probability of getting a lower grade if the student is in liberal arts and social sciences.

For each chance process below, determine whether the events are independent. Justify your answer. 1) Shuffle a standard deck of cards, and turn over the top card. Put it back in the deck, shuffle again, and turn over the top card. Define events A: first card is a heart, and B: second card is a heart. 2) Shuffle a standard deck of cards, and turn over the top two cards, one at a time. Define events A: first card is a heart, and B: second card is a heart. 3) The 28 students in Mr. Tabor’s AP Statistics class completed a brief survey. One of the questions asked whether each student was right or left-handed. The two-way table summarizes the class data. Choose a student from the class at random. The events of interest are “female” and “right-handed.”

1) Shuffle a standard deck of cards, and turn over the top card. Put it back in the deck, shuffle again, and turn over the top card. Define events A: first card is a heart, and B: second card is a heart.

Independent. Since we are putting the first card back and then reshuffling the cards before drawing the second card, knowing what the first card was will not tell us anything about what the second card will be. 2) Shuffle a standard deck of cards, and turn over the top two cards, one at a time. Define events A: first card is a heart, and B: second card is a heart.

Dependent. Once we know the suit of the first card, then the probability of getting a heart on the second card will change depending on what the first card was.

3) The 28 students in Mr. Tabor’s AP Statistics class completed a brief survey. One of the questions asked whether each student was right or left-handed. The two-way table summarizes the class data. Choose a student from the class at random. The events of interest are “female” and “right-handed.”

Define Event F: chosen student is female Event R: chosen student is right-handed

P(F) = P(F|R) =

We want to use the definition: If events A and B are independent, then P(A|B) = P(A) and P(B|A) = P(B).

P(R) = P(R|F) =

21

28=

3

4

18

24=

3

4

same

24

28=

6

7

18

21=

6

7

same

Since P(F|R) = P(F) and P(R|F) = P(R), the events of a student being female and right-handed are independent. Once we know that the chosen person is female, this does not tell us anything more about whether she is right-handed or not.

Using a tree diagram is another way to model chance behavior that involves a sequence of outcomes.

Here is a basic tree diagram example:

If you toss a fair coin twice, what’s the probability you get two tails?

Toss a coin

Heads

Tails

Heads

Heads

Tails

Tails

12

12

12

12

12

12

P(Tails 1𝑠𝑡and 2𝑛𝑑) = P(Tails 1𝑠𝑡) ⋅ P(Tails 2𝑛𝑑|Tails 1𝑠𝑡) = 1

2

1

2=

1

4

P(Tails 1𝑠𝑡) = 1

2

P(Tails 2𝑛𝑑|Tails 1𝑠𝑡) = 1

2

Remember!! P(A and B) = P(A) ⋅ P(B|A)

Internet Users

Young

Older

P(Young)

. 85

.15

P(Older)

Use YouTube

Use YouTube

No YouTube

No YouTube

P(Use YT|Young)

. 75

P(Use YT|Older)

. 55

.25 P(No YT|Young)

.45 P(No YT|Older)

.85 .75 = .6375

.85 .25 = .2125

.15 .55 = .0825

.15 .45 = .0675

1

P(Use YT ∩ Young)

P(No YT ∩ Young)

P(Use YT ∩ Older)

P(No YT ∩ Older)

Internet Users

Young

Older

Use YouTube

Use YouTube

No YouTube

No YouTube

.85 .75 = .6375

.85 .25 = .2125

.15 .55 = .0825

.15 .45 = .0675

.75

.25

.55

.45

.85

.15

P(Older ∩ Use YouTube) = .0825 = 8.25%

P(Use YouTube) = .6375 + .0825 = 0.72 = 72%

P(Young|Use YouTube) =

Recall: P(A ∩ B) = P(A) ⋅ P(B|A) P(B|A) = P A ∩ B

P A

P Young ∩ Use YouTube

P Use YouTube =

.6375

.72= .89 = 89%

A computer company makes desktop and laptop computers at factories in three states—California, Texas, and New York. The California factory produces 40% of the company’s computers, the Texas factory makes 25%, and the remaining 35% are manufactured in New York. Of the computers made in California, 75% are laptops. Of those made in Texas and New York, 70% and 50%, respectively, are laptops. All computers are first shipped to a distribution center in Missouri before being sent out to stores. Suppose we select a computer at random from the distribution center. 1) Construct a tree diagram to represent this situation. 2) Find the probability that the computer is a laptop. Show your work. 3) Find the probability that a computer was made in California given that it’s a laptop.

1) Construct a tree diagram to represent this situation.

2) Find the probability that the computer is a laptop. Show your work.

P(laptop) = 0.3 + 0.175 + 0.175 = 0.65

3) Find the probability that a computer was made in California given that it’s a laptop.

P(Made in Cal | laptop) =0.30

0.65= 0.46

P(A) ⋅ P(B|A)

P A and B

P A

P(A) ⋅ P(B)

P(wins all 5) = P(wins 1𝑠𝑡 and wins 2𝑛𝑑 and W 3𝑟𝑑 and W 4𝑡ℎ and W 5𝑡ℎ)

= P(wins 1𝑠𝑡) ⋅ P(wins 2𝑛𝑑) ⋅ P(wins 3𝑟𝑑) ⋅ P(wins 4𝑡ℎ) ⋅ P(wins 5𝑡ℎ)

By multiplication rule of independent events:

=1

6

1

6

1

6

1

6

1

6 =

1

6

5

= 0.000219

P(at least 1 win) = P(1 win)+P(2 wins)+P(3 wins)+P(4 wins)+P(5 wins) Is there any easier way? YES

P(at least 1 win) = 1 − P(0 wins)

= 1 − 5

6

5

6

5

6

5

6

5

6 = 1 − .402 = 0.598

Whenever you have to find P(at least one), do 1 − P(none)!!

S

L C

0.60 0.18 0.20

0.02

Start with both (middle)

Using Venn Diagram: Using P(B|A) formula and written information:

P(cell | landline) = 0.60

0.60 + 0.18=

P(cell | landline) = P(cell ∩ landline)

P(landline)=

0.60

0.78= 0.77

0.60

0.78= 0.77

Internet Users

Young

Older

Use YouTube

Use YouTube

No YouTube

No YouTube

.85 .75 = .6375

.85 .25 = .2125

.15 .55 = .0825

.15 .45 = .0675

.75

.25

.55

.45

.85

.15

P(Older | Use YouTube) =P Older ∩ Use YouTube

P Use YouTube

=.0825

.6375 + .0825 =

.0825

.72= 0.115