Answer:
a) The probability that he doesnt but any items is 0.49
b) He buys exactly 1 of those items with probability 0.28
Step-by-step explanation:
lets call su the event that the customer purchases a suit, sh the event that teh customer purchases a shirt and t the event that the customer purchases a tie.
Remembe that for events A, B and C we have that
P(A U B) = P(A) + P(B) - P(A ∩ B)
P(A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)
Also, we are given that
P(su) = 0.22
P(sh) = 0.3
p(t) = 0.28
p(su ∩ sh) = 0.11
P(su ∩ t) = 0.14
P(sh ∩ t) = 0.1
P(sh ∩ t ∩ su) = 0.06
The event that he doesnt buy any item has as complementary event su ∪ sh ∪ t, therefore
P( he doesnt but any items) = 1-P(su U sh U t) =
1-( P(su) + p(sh) + p(t) - P(su ∩ sh) - p(su∩t) - p(sh∩t) + p(su∩sh∩t) ) =
1-(0.22+0.30+0.28-0.11-0.14-0.1+0.06) = 1-0.51 = 0.49
b) The probability that he buys at least 2 items is equal to
p(su ∩ t) + p(su ∩ sh) + p(sh ∩ t) -2 p(su ∩ t ∩ sh) (because we are counting the triple intersection 3 times, so we need to remove it twice)
This number is
0.14+0.11+0.1-2*0.06 = 0.23
Thus, the probability that he buys exactly one item can be computed by substracting from one the probability of the complementary event : she buys 2 or more or non items
P(he buys exactly one item) = 1- ( p(he buys none items) + p(he buys at least 2) ) = 1- 0.49-0.23 = 0.28
