Practice 12-2 Conditional Probability 1. The table | Chegg.com - Free Printable

Let’s solve each problem step by step. We’ll be careful with the numbers and make sure we understand what each probability is asking for.

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Problem 1: Education and Salary of Employees

Total employees = 1205 (given)

We’ll use the table to find counts for each part.

First, let’s add up all the values in the table to confirm total:

Row 1 (Less than high school): 69 + 36 + 2 = 107
Row 2 (High school): 112 + 98 + 14 = 224
Row 3 (Some college): 102 + 193 + 143 = 438
Row 4 (College degree): 13 + 178 + 245 = 436

Total = 107 + 224 + 438 + 436 = 1205 ✔

Now, let’s do each part:

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a. P(employee has less than a high school education)

That’s row 1 total: 107

P = 107 / 1205 ≈ ?

Calculate: 107 ÷ 1205 = 0.08879... → 8.9%

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b. P(employee earns under $20,000)

Column 1 total: 69 + 112 + 102 + 13 = 296

P = 296 / 1205 ≈ 0.2456 → 24.6%

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c. P(employee earns over $30,000 AND has less than a high school education)

Look at intersection: “Less than high school” row and “Over $30,000” column → 2

P = 2 / 1205 ≈ 0.00166 → 0.2%

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d. P(employee earns under $20,000 AND has a college degree)

Intersection: “College degree” row and “Under $20,000” column → 13

P = 13 / 1205 ≈ 0.01079 → 1.1%

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e. Given that the employee has only a high school education, probability they earn over $30,000

This is conditional probability.

“Only high school education” = row 2 total = 224

Of those, how many earn over $30,000? → 14

So P = 14 / 224 = 0.0625 → 6.3%

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f. Given that the employee earns over $30,000, probability they have only a high school education or less

First, total who earn over $30,000: column 3 total = 2 + 14 + 143 + 245 = 404

Now, among these, how many have “only high school education or less”? That means:

- Less than high school: 2
- High school: 14

Total = 2 + 14 = 16

P = 16 / 404 ≈ 0.0396 → 4.0%

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Problem 2: Favorite Leisure Activities

Table:

Female: Sports=39, Hiking=48, Reading=85, Phonng=62, Shopping=71, Other=29
Male: Sports=67, Hiking=58, Reading=76, Phonng=54, Shopping=68, Other=39

First, find totals:

Female total = 39+48+85+62+71+29 = let’s compute:

39+48=87; 87+85=172; 172+62=234; 234+71=305; 305+29=334

Male total = 67+58+76+54+68+39 =

67+58=125; 125+76=201; 201+54=255; 255+68=323; 323+39=362

Grand total = 334 + 362 = 696

Now, each part:

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a. P(sports | female)

Given female, so denominator = 334

Sports among females = 39

P = 39 / 334 ≈ 0.11677 → 11.7%

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b. P(female | sports)

Given sports, so first find total sports players: 39 (female) + 67 (male) = 106

Of those, female = 39

P = 39 / 106 ≈ 0.3679 → 36.8%

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c. P(reading | male)

Given male, denominator = 362

Reading among males = 76

P = 76 / 362 ≈ 0.2099 → 21.0%

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d. P(male | reading)

Total readers: 85 (female) + 76 (male) = 161

Males who read = 76

P = 76 / 161 ≈ 0.4720 → 47.2%

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e. P(hiking | female)

Hiking among females = 48

Denominator = 334

P = 48 / 334 ≈ 0.1437 → 14.4%

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f. P(hiking | male)

Hiking among males = 58

Denominator = 362

P = 58 / 362 ≈ 0.1602 → 16.0%

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g. P(male | shopping)

Total shoppers: 71 (female) + 68 (male) = 139

Males who shop = 68

P = 68 / 139 ≈ 0.4892 → 48.9%

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h. P(female | shopping)

Same total shoppers = 139

Females who shop = 71

P = 71 / 139 ≈ 0.5108 → 51.1%

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Problem 3: Senior Class Probability

Given:

- 55% are female → P(Female) = 0.55
- 32% are females who play competitive sport → P(Female AND Sport) = 0.32

We want: P(Sport | Female) = ?

Formula: P(A|B) = P(A and B) / P(B)

So P(Sport | Female) = P(Female and Sport) / P(Female) = 0.32 / 0.55 ≈ 0.5818 → 58.2%

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Problem 4: Softball Game Cancellation

We’re told:

- If light drizzle → 80% chance canceled
- If heavy fog (and no drizzle) → 30% chance canceled
- P(heavy fog) = 70%
- P(light drizzle) = 30%

Note: It says “when there is no drizzle” for heavy fog — so we assume heavy fog and light drizzle are mutually exclusive? The problem doesn’t say they can happen together, so we’ll treat them as separate cases.

Also, note: 70% heavy fog + 30% light drizzle = 100%, so no other weather? Okay.

Draw tree diagram mentally:

Root: Weather

Branch 1: Light Drizzle (30%) → then Cancel (80%), Not Cancel (20%)

Branch 2: Heavy Fog (70%) → then Cancel (30%), Not Cancel (70%)

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a. Find probability game will be canceled

Use law of total probability:

P(Cancel) = P(Cancel | Drizzle) * P(Drizzle) + P(Cancel | Fog) * P(Fog)

= (0.80)(0.30) + (0.30)(0.70) = 0.24 + 0.21 = 0.45 → 45%

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b. Probability there will be light drizzle AND game will NOT be canceled

P(Drizzle and Not Cancel) = P(Not Cancel | Drizzle) * P(Drizzle) = (0.20)(0.30) = 0.06 → 6%

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Problem 5: High School Students Attending Concerts

Given:

- 51% males → P(Male) = 0.51 → so P(Female) = 1 - 0.51 = 0.49
- 45% of males attend concerts → P(Concert | Male) = 0.45
- 49% of females attend concerts → P(Concert | Female) = 0.49

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a. Find probability that a student attends concerts

Use total probability:

P(Concert) = P(Concert | Male)*P(Male) + P(Concert | Female)*P(Female)

= (0.45)(0.51) + (0.49)(0.49)

Compute:

0.45 * 0.51 = 0.2295

0.49 * 0.49 = 0.2401

Sum = 0.2295 + 0.2401 = 0.4696 → 47.0%

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b. Find probability that a student is female AND does not attend concerts

P(Female and Not Concert) = P(Not Concert | Female) * P(Female)

P(Not Concert | Female) = 1 - 0.49 = 0.51

So: 0.51 * 0.49 = 0.2499 → 25.0%

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Final Answer:

1.
a. 8.9%
b. 24.6%
c. 0.2%
d. 1.1%
e. 6.3%
f. 4.0%

2.
a. 11.7%
b. 36.8%
c. 21.0%
d. 47.2%
e. 14.4%
f. 16.0%
g. 48.9%
h. 51.1%

3. 58.2%

4.
a. 45%
b. 6%

5.
a. 47.0%
b. 25.0%