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(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.

Answer 1

Answer:

The correct option is 4.

Step-by-step explanation:

Given information:

Bring lunch : 46 males, 254 females

Buy lunch   : 176 males, 264 females

Total number of peoples is

$46+254+176+264=740$

Total number of males is

$46+176=222$

The probability of male is

$P(Male)=\frac{Males}{Total}=\frac{222}{740} =0.3$

Since probability of males is 0.3, therefore options A and C are incorrect.

Total number of persons who buys lunch is

$176+264=440$

The probability of persons who buys lunch is

$P(\text{Buys lunch})=\frac{\text{Buys lunch}}{Total}=\frac{440}{740} =\frac{22}{37}$

We need to find the probability of P(male | buys lunch).

According to the conditional probability, we get

$P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}$

P(male | buys lunch)$=\frac{P(\text{male }\cap \text{ buys lunch})}{P(\text{buys lunch})}$

P(male | buys lunch)$=\frac{\frac{176}{740}}{\frac{22}{37}}$

P(male | buys lunch)$=\frac{2}{5}=0.4$

Therefore the correct option is 4.

Answer 2

Answer: it’s D

Step-by-step explanation:

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