Question

A customer visiting the suit department of a certain store will purchase a suit with probability 0.22, a shirt with probability 0.30, and a tile with probability 0.28. The customer will purchase both a suit and a shirt with probability 0.11, both a suit and a tie with probability 0.14, and both a shirt and a tie with probability 0.10. A customer will purchase all 3 items with probability 0.06. What’s the probability that a customer purchase: (a) none of these items? (b) exactly 1 of these items?

Answer 1

a. The probability that a customer purchase none of these items is 0.49

b. The probability that a customer purchase exactly 1 of these items would be of 0.28

Explanation:

a. In order to calculate the probability that a customer purchase none of these items we would have to make the following:

let A represents suit

B represents shirt

C represents tie

P(A) = 0.22

P(B) = 0.30

P(C) = 0.28

P(A∩B) = 0.11

P(C∩B) = 0.10

P(A∩C) = 0.14

P(A∩B∩C) = 0.06

Therefore, the probability that a customer purchase none of these items we would have to calculate the following:

1 - P(A∪B∪C)

P(A∪B∪C) =P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C)

= 0.22+0.28+0.30-0.11-0.10-0.14+0.06

= 0.51

Hence, 1 - P(A∪B∪C) = 1-0.51 = 0.49

The probability that a customer purchase none of these items is 0.49

b.To calculate the probability that a customer purchase exactly 1 of these items we would have to make the following calculation:

= P(A∪B∪C) - ( P(A∩B) +P(C∩B) +P(A∩C) - 2 P(A ∩ B ∩ C))

=0.51 -0.23 = 0.28

The probability that a customer purchase exactly 1 of these items would be of 0.28