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: it’s D

Explanation:

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Answer 2

The correct option is 4.

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)=(Males)/(Total)=(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}=(440)/(740) =(22)/(37)`

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

According to the conditional probability, we get

`P((A)/(B))=(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)
`=((176)/(740))/((22)/(37))`

P(male | buys lunch)
`=(2)/(5)=0.4`

Therefore the correct option is 4.