A customer visiting the suit department of a certain store will purchase a suit with probability .22, a shirt with probability .30, and a tie with probability .28. The customer will purchase both a suit and a shirt with probability .11, both a suit and a tie with probability .14, and both a shirt and a tie with probability .10. A customer will purchase all 3 items with probability .06. What is the probability that a customer purchases

(a) none of these items?

(b) exactly 1 of these items?

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

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

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb