identifying two independent events

brown law firm collected data on the transportation choices of its employees for their morning commute. the table shows the percentages of the type of transportation of the male and female employees.

\begin{tabular}{|c|c|c|c|c|}
\hline
& public & own & other & total \\
\hline
male & 12 & 20 & 4 & 36 \\
\hline
female & 8 & 10 & 6 & 24 \\
\hline
total & 20 & 30 & 10 & 60 \\
\hline
\end{tabular}

consider the following events:
a: the employee is male.
b: the employee is female.
c: the employee takes public transportation.
d: the employee takes his/her own transportation.
e: the employee takes some other method of transportation.

which two events are independent?

a and c
a and d
b and d
b and e

Question

identifying two independent events

brown law firm collected data on the transportation choices of its employees for their morning commute. the table shows the percentages of the type of transportation of the male and female employees.

\begin{tabular}{|c|c|c|c|c|}
\hline
& public & own & other & total \\
\hline
male & 12 & 20 & 4 & 36 \\
\hline
female & 8 & 10 & 6 & 24 \\
\hline
total & 20 & 30 & 10 & 60 \\
\hline
\end{tabular}

consider the following events:
a: the employee is male.
b: the employee is female.
c: the employee takes public transportation.
d: the employee takes his/her own transportation.
e: the employee takes some other method of transportation.

which two events are independent?

a and c
a and d
b and d
b and e

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Independent Events", "Two-Way Frequency Tables", "Conditional Probability Calculation" ], "new_concepts": [], "current_concepts": [ "Independent Events", "Two-Way Frequency Tables", "Conditional Probability Calculation" ] } ## Calculate individual probabilities Using the *Two-Way Frequency Tables* knowledge point $$ \begin{aligned} &P(A) = \frac{36}{60} = 0.60\ &P(B) = \frac{24}{60} = 0.40\ &P(C) = \frac{20}{60} = \frac{1}{3} \approx 0.333\ &P(D) = \frac{30}{60} = 0.50\ &P(E) = \frac{10}{60} = \frac{1}{6} \approx 0.167 \end{aligned} $$ ## Calculate joint probabilities Using the *Two-Way Frequency Tables* knowledge point $$ \begin{aligned} &P(A \cap C) = \frac{12}{60} = 0.20\ &P(A \cap D) = \frac{20}{60} = \frac{1}{3} \approx 0.333\ &P(B \cap D) = \frac{10}{60} = \frac{1}{6} \approx 0.167\ &P(B \cap E) = \frac{6}{60} = 0.10 \end{aligned} $$ ## Test for independence Using the *Independent Events* knowledge point $$ \begin{aligned} &\text{Check A and C: } P(A) \times P(C) = 0.60 \times \frac{1}{3} = 0.20 = P(A \cap C)\ &\text{Check A and D: } P(A) \times P(D) = 0.60 \times 0.50 = 0.30 eq P(A \cap D)\ &\text{Check B and D: } P(B) \times P(D) = 0.40 \times 0.50 = 0.20 eq P(B \cap D)\ &\text{Check B and E: } P(B) \times P(E) = 0.40 \times \frac{1}{6} \approx 0.067 eq P(B \cap E) \end{aligned} $$

A and C

A and D

B and D

B and E

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Statistics", "Independent Events" ] }

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QUESTION IMAGE

Question

Explanation:

Calculate individual probabilities

Calculate joint probabilities

Test for independence

Answer: