PROBLEM SET 12-2 (Conditional Probability) 1. A ... - Frankumstein - Free Printable

Here are the step-by-step solutions for the problems in Problem Set 12-2.

Problem 1

Question: A student says that if $P(A) = P(A | B)$, then $A$ and $B$ must be independent events. Is the student correct? Explain.

Step-by-step Reasoning:
1. Definition of Independence: Two events, $A$ and $B$, are considered independent if the occurrence of one does not affect the probability of the other occurring. Mathematically, this is defined as $P(A | B) = P(A)$.
2. Analyzing the Statement: The equation $P(A) = P(A | B)$ literally means "The probability of A happening is the same as the probability of A happening given that B has already happened."
3. Conclusion: Since this equation is the exact definition of statistical independence, the student's statement is correct. If knowing that $B$ occurred doesn't change the likelihood of $A$, the events are independent.

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Problems 2–5 (Education and Salary Table)

First, we need to calculate the Grand Total of all employees to use as the denominator for our probabilities. We sum every number in the table:

* Row 1 (Less than HS): $69 + 36 + 2 = 107$
* Row 2 (High School): $112 + 98 + 14 = 224$
* Row 3 (Some College): $102 + 193 + 143 = 438$
* Row 4 (College): $13 + 173 + 245 = 431$

Total Employees = $107 + 224 + 438 + 431 = \mathbf{1200}$

We also need specific column totals for later questions:
* Total earning Over \$30,000: $2 + 14 + 143 + 245 = \mathbf{404}$
* Total with Less than High School education: $\mathbf{107}$ (calculated above)

2. $P(\text{has less than high school education})$
* This asks for the total number of people with "Less than high school" divided by the grand total.
* Calculation: $\frac{107}{1200}$
* Decimal: $\approx 0.0892$

3. $P(\text{earns over \$30,000 and has less than high school education})$
* This is an "AND" probability (intersection). We look for the single cell where the row "Less than high school" meets the column "Over \$30,000".
* Value in table: $2$
* Calculation: $\frac{2}{1200}$
* Simplified: $\frac{1}{600} \approx 0.0017$

4. $P(\text{earns over \$30,000 } | \text{ has only high school education})$
* The vertical bar $|$ means "given that". We restrict our view to only the "High School" row.
* Total people in High School row = $224$.
* People in that row who earn over \$30,000 = $14$.
* Calculation: $\frac{14}{224}$
* Simplified: $\frac{1}{16} = 0.0625$

5. $P(\text{has less than high school education } | \text{ earns over \$30,000})$
* Here, the condition is "earns over \$30,000". We restrict our view to the "Over \$30,000" column.
* Total people in that column = $404$.
* People in that column who have "Less than high school" education = $2$.
* Calculation: $\frac{2}{404}$
* Simplified: $\frac{1}{202} \approx 0.0050$

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Problems 6–8 (Gender and College Majors Table)

First, let's find the totals for this table.

* Male Row Total: $40 + 16 + 35 = 91$
* Female Row Total: $15 + 24 + 20 = 59$
* Grand Total Students: $91 + 59 = \mathbf{150}$

We also need column totals:
* Physics Total: $16 (\text{Male}) + 24 (\text{Female}) = \mathbf{40}$
* Chemistry Total: $35 (\text{Male}) + 20 (\text{Female}) = \mathbf{55}$

6. $P(\text{male})$
* Total males divided by total students.
* Calculation: $\frac{91}{150}$
* Decimal: $\approx 0.6067$

7. $P(\text{male or majors in Chemistry})$
* Formula for OR: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$.
* $P(\text{Male}) = \frac{91}{150}$
* $P(\text{Chemistry}) = \frac{55}{150}$
* $P(\text{Male AND Chemistry})$: Look at the intersection of Male row and Chemistry column = $35$. So, $\frac{35}{150}$.
* Calculation: $\frac{91 + 55 - 35}{150} = \frac{111}{150}$
* Simplified: $\frac{37}{50} = 0.74$

8. $P(\text{majors in Physics } | \text{ male})$
* The condition is "male". We look only at the Male row.
* Total Males = $91$.
* Males majoring in Physics = $16$.
* Calculation: $\frac{16}{91}$
* Decimal: $\approx 0.1758$

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Final Answer:

1. Yes, the student is correct. By definition, two events are independent if the probability of one event occurring is unaffected by the occurrence of the other ($P(A) = P(A|B)$).

2. $\frac{107}{1200}$ (or approx. $0.0892$)

3. $\frac{1}{600}$ (or approx. $0.0017$)

4. $\frac{1}{16}$ (or $0.0625$)

5. $\frac{1}{202}$ (or approx. $0.0050$)

6. $\frac{91}{150}$ (or approx. $0.6067$)

7. $\frac{37}{50}$ (or $0.74$)

8. $\frac{16}{91}$ (or approx. $0.1758$)