Question

Suppose we have the following facts about customers at a supermarket:

60% buy bread (B).

55% buy milk (M).

45% buy eggs (E).

78% buy bread or eggs.

38% buy bread and milk.

30% buy milk and eggs.

19% buy all three items.

. If we randomly select a customer, what is the probability the customer buys none of the

three items?

(A) 0.12

(B) 0.13

(C) 0.14

(D) 0.15

(E) 0.16

Answer 1

Probability the customer buys none of the three items = 1 - P[ customer buy at least one of the three items ]

60% buy bread (B).

55% buy milk (M).

45% buy eggs (E).

78% buy bread or eggs.

38% buy bread and milk.

30% buy milk and eggs.

19% buy all three items.

P[ customer buy at least one of the three items ] = P[ buy bread ] + P[ buy milk ] + P[ buy egg ] - P[ buy milk and eggs ] - P[ buy bread and milk ] - P[ buy bread and eggs ] + P[ buy all three items ]

P[ buy bread or eggs ] = P[ buy bread ] + P[ buy egg ] - P[ buy bread and eggs ]

P[ buy bread and eggs ] = P[ buy bread ] + P[ buy egg ] - P[ buy bread or eggs ]

P[ buy bread and eggs ] = 60% + 45% - 78%

P[ buy bread and eggs ] = 27%

P[ customer buy at least one of the three items ] = P[ buy bread ] + P[ buy milk ] + P[ buy egg ] - P[ buy milk and eggs ] + P[ buy bread and milk ] + P[ buy bread and eggs ] - P[ buy all three items ]

P[ customer buy at least one of the three items ] = 60% + 55% + 45% - 30% - 38% - 27% + 19%

P[ customer buy at least one of the three items ] = 179 - 95

P[ customer buy at least one of the three items ] = 84%

Probability the customer buys none of the three items = 1 - P[ customer buy at least one of the three items ]

Probability the customer buys none of the three items = 1 - 0.84

Probability the customer buys none of the three items = 0.16