james surveyed people at school and asked whether they bring their lunch to school or buy their lunch at…

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\np(buys lunch | male) = p(male) = 0.4.\np(male | buys lunch) = p(male) = 0.3.\np(buys lunch | male) = 0.3 and p(male) = 0.4.\np(male | buys lunch) = 0.4 and p(male) = 0.3.

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\np(buys lunch | male) = p(male) = 0.4.\np(male | buys lunch) = p(male) = 0.3.\np(buys lunch | male) = 0.3 and p(male) = 0.4.\np(male | buys lunch) = 0.4 and p(male) = 0.3.

Answer

Answer:

P(buys lunch | male) = 0.3 and P(male) = 0.4.

Explanation:

Step1: Calculate total number of people

Total = (46 + 254)+(176 + 264)=740

Step2: Calculate P(male)

Number of males = 46+176 = 222. So (P(male)=\frac{222}{740}=0.3)

Step3: Calculate P(buys lunch and male)

Number of males who buy lunch = 176. So (P(buys\ lunch\ and\ male)=\frac{176}{740})

Step4: Calculate P(buys lunch | male)

By the formula (P(A|B)=\frac{P(A\cap B)}{P(B)}), (P(buys\ lunch | male)=\frac{P(buys\ lunch\ and\ male)}{P(male)}=\frac{\frac{176}{740}}{\frac{222}{740}}=\frac{176}{222}\approx0.3) Two events A and B are independent if (P(A|B) = P(A)). Here (P(buys\ lunch | male)\neq P(male)), so the events "male" and "buys lunch" are not independent.