MATH SOLVE

5 months ago

Q:
# 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 femalesBuy lunch: 176 males, 264 femalesThe events "male” and "buys lunch” are not independent becauseP(buys lunch | male) = P(male) = 0.4.P(male | buys lunch) = P(male) = 0.3.P(buys lunch | male) = 0.3 and P(male) = 0.4.P(male | buys lunch) = 0.4 and P(male) = 0.3.

Accepted Solution

A:

Answer:The correct option is 4.Step-by-step explanation:Given information:Bring lunch : 46 males, 254 femalesBuy lunch : 176 males, 264 femalesTotal number of peoples is[tex]46+254+176+264=740[/tex]Total number of males is[tex]46+176=222[/tex]The probability of male is[tex]P(Male)=\frac{Males}{Total}=\frac{222}{740} =0.3[/tex]Since probability of males is 0.3, therefore options A and C are incorrect.Total number of persons who buys lunch is[tex]176+264=440[/tex]The probability of persons who buys lunch is[tex]P(\text{Buys lunch})=\frac{\text{Buys lunch}}{Total}=\frac{440}{740} =\frac{22}{37}[/tex]We need to find the probability of P(male | buys lunch).According to the conditional probability, we get[tex]P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}[/tex]P(male | buys lunch)[tex]=\frac{P(\text{male }\cap \text{ buys lunch})}{P(\text{buys lunch})}[/tex]P(male | buys lunch)[tex]=\frac{\frac{176}{740}}{\frac{22}{37}}[/tex]P(male | buys lunch)[tex]=\frac{2}{5}=0.4[/tex]Therefore the correct option is 4.