Question

james surveyed people at school and asked whether they bring their lunch to school or buy their lunch at school more often. the results are shown below.

bring lunch: 46 males, 254 females
buy lunch: 176 males, 264 females

the events \male\ and \buys lunch\ are not independent because

\\(p(\text{buys lunch} \mid \text{male}) = p(\text{male}) = 0.4\\).
\\(p(\text{male} \mid \text{buys lunch}) = p(\text{male}) = 0.3\\).
\\(p(\text{buys lunch} \mid \text{male}) = 0.3\\) and \\(p(\text{male}) = 0.4\\).
\\(p(\text{male} \mid \text{buys lunch}) = 0.4\\) and \\(p(\text{male}) = 0.3\\).

Explanation:

Construct a two-way frequency table

Using the Two-Way Frequency Tables knowledge point, we organize the survey data:

	Male	Female	Total
Buy lunch	176	264	440
Total	222	518	740

Calculate the probability of being male

Using the Conditional Probability Calculation and Independent Events knowledge points

$$ P(\text{male}) = \frac{222}{740} = 0.3 $$

Calculate the conditional probability of being male given they buy lunch

Using the Conditional Probability Calculation and Independent Events knowledge points

$$ P(\text{male} \mid \text{buys lunch}) = \frac{176}{440} = 0.4 $$

Determine independence condition

Using the Independent Events knowledge point

$$ P(\text{male} \mid \text{buys lunch}) eq P(\text{male}) \implies 0.4 eq 0.3 $$

Answer:

• (A) \(P(\text{buys lunch} \mid \text{male}) = P(\text{male}) = 0.4\).
• (B) \(P(\text{male} \mid \text{buys lunch}) = P(\text{male}) = 0.3\).
• (C) \(P(\text{buys lunch} \mid \text{male}) = 0.3\) and \(P(\text{male}) = 0.4\).
• (D) \(P(\text{male} \mid \text{buys lunch}) = 0.4\) and \(P(\text{male}) = 0.3\). (Correct answer)