Respuesta :
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
Answer:
P (suit) P(A) =0.22
P (shirt) P(B) =0.30
P (tie) P(C) = 0.28
P(AB) =0.11
P(AC) =0.14
P(BC) = 0.10
P(ABC) = 0.06
(i)
Probability that none of the items are bought = 1 - probability that all the items are bought
P(none) = 1 - P(A u B u C)
= P( A u B u C) =
P(A) + P(B) + P(C) - P(AB) - P(AB) - P(BC) + P(ABC)
= 0.22 + 0.30 + 0.28 - 0.11 - 0.14 - 0.10 + 0.06 = 0.51
P(A u B u C) = 0.51
Probability that he would buy none of the items = 1 - P( A u B u C)
1 - 0.51 = 0.49
(ii)
Probability of purchasing exactly one
= {(probability of buying a shirt) + (probability of purchasing a suit) + (probability of purchasing a tie)} - [(probability of purchasing a suit and a shirt) +(probability of purchasing a suit and a tie) + (probability of purchasing a shirt and a tie)]
= {(P(A) + P(B) + P(C)} - [P(AB) + P (AC) + P (BC)]
= {(0.22 + 0.30 + 0.28)} - [(0.11 + 14 + 0.10)]
= { 0.80 - 0.35}
= 0.45
Another way to calculate it would be:
1 - {(probability of purchasing none of the items ) + (probability of purchasing all the items)}
= 1 - {P (none) + P (all)}
= 1 - (0.49 + 0.06)
= 1- (0.55)
= 0.45
Step-by-step explanation:
Probability of purchasing a suit = 0.22, probability of purchasing a shirt =0.30 and probability of purchasing a tie = 0.28
both a suit and a shirt with probability .11: P(AB) =0.11
both a suit and a tie with probability .14, : P(AC) =0.14
and both a shirt and a tie with probability .10. : P(BC) = 0.10
all 3 items with probability .06. : P(ABC) = 0.06
So:
P (suit): P(A) =0.22
P (shirt): P(B) =0.30
P (tie): P(C) = 0.28
P (suit n shirt): P(AB) =0.11
P (suit and tie): P(AC) =0.14
P (shirt and tie): P(BC) = 0.10
Probability of purchasing all the items: P(ABC) = 0.06
Probability that none of the items are bought: P(none) = 1 - P(A u B u C)
P( A u B u C) = P(A) + P(B) + P(C) - P(AB) - P(AB) - P(BC) + P(ABC)
=0.22 + 0.30 + 0.28 - 0.11 - 0.14 - 0.10 + 0.06 = 0.51
P(A u B u C) = 0.51
Probability that he would buy all of the items = 1 - P( A u B u C)
1 - 0.51 = 0.49
(ii)
Probability of purchasing exactly one
={(probability of buying a shirt) + (probability of purchasing a suit) + ( probability of purchasing a tie)} {(probability of purchasing a suit and a shirt) +(probability of purchasing a suit and a tie) + (probability of purchasing a shirt and a tie)}
= {(P(A) + P(B) + P(C)) - ( P(AB) + P (AC) + P (BC)}
= {(0.22 + 0.30 + 0.28) - (0.11 + 14 + 0.10)
= { 0.80 - 0.35}
= 0.45
Another way to calculate it would be
1 {(probability of purchasing none of the items ) + (probability of purchasing all the items)}
= 1 - (0.49 + 0.06)
= 1- (0.55)
= 0.45
