Question

In: Statistics and Probability

Please answer with method or formula used. Select a person at random at a local baseball...

Please answer with method or formula used.

Select a person at random at a local baseball game. Consider these events: C = person buys soda, B = person buys beer, N = person buys peanuts

Consider these probabilities: a) The probability the person buys none of the things. b) The probability the person buys exactly one of the things c) The probability the person buys all three, given that he buys at least 2. d) The probability the person buys peanut, given that he doesn’t buy beer

A) Calculate (a)-(d) assuming that ?(?) = 0.40, ?(?) = 0.60, ?(?) = 0.30, ?(? ∩ ? ∩ ?) = 0.08

B) ?(? ∪ ?) = 0.83, ?(? ∪ ?) = 0.50, ?(? ∩ ?) = 0.15 , ,

C) Calculate (a)-(d) assuming that ?(?) = 0.40, ?(?) = 0.60, ?(?) = 0.30 , and that C, B, N are independent events.

D) A pair of 6-sided die is tossed. If the sum of the two die is at least 10, you win $10. sum of the die is a number between 4 and 9, you win $5. less than 4, you win $0.
1) What are the possible winnings? 2) What is the probability for each possibility, assuming that the die are fair? 3) What is the expected winnings? 4) What is the variance of winnings?

E) Let X be the payout from the previous problem. Compute a) ?(1/?) b) c) ?(2.3? − 1.5) d) ?(2.3 − 1.5)

Expert Solution

A)

a)

Probability that a person buys none of these things

= P(C' B' N')

= P[(CBN)']

= 1 - P(CBN)

= 1 - [P(C)+P(B)+P(N)-P(CB)-(CN)-P(BN)+P(CBN)]

= 1 - [0.40 + 0.60 + 0.30 - {P(C)+P(B)-P(CB)} - {P(C)+P(N)-P(CN)} - 0.15 + 0.08]

= 1 - [1.23 - {0.40+0.60-0.83} - {0.40+0.30-0.50}] = 1 - [1.23-0.17-0.20] = 0.14

b)

P(C' B' N') = 0.14

P(C' B') = P[(CB)'] = 1 - P(CB) = 1 - 0.83 = 0.17

P(C' N') = P[(CN)'] = 1 - P(CN) = 1 - 0.50 = 0.50

P(B' N') = P[(BN)'] = 1 - P(BN) = 1 - [P(B) + P(N) - P(BN)] = 1 - [0.60+0.30-0.15] = 0.25

P(C B' N') = Probability that a person buys only soda = P(B' N') - P(C' B' N') = 0.25 - 0.14 = 0.11

P(C' B N') = Probability that a person buys only beer = P(C' N') - P(C' B' N') = 0.50 - 0.14 = 0.36

P(C' B' N) = Probability that a person buys only peanuts = P(C' B') - P(C' B' N') = 0.17 - 0.14 = 0.03

Probability that a person buys exactly one of the things = P(C B' N') + P(C' B N') + P(C' B' N)

= 0.11+0.36+0.03 = 0.50

c)

Probability that a person buys all three, given that he buys at least two

P(a person buys all three things) = P(CBN) = 0.08

P(C B N') = P(a person buys soda and beer and does not buy peanuts)

= P(CB) - P(CBN) = P(C) + P(B) - P(CB) - P(CBN) = 0.40+0.60-0.83-0.08 = 0.09

P(C B' N) = P(a person buys soda and peanuts and does not buy beer)

= P(CN) - P(CBN) = P(C) + P(N) - P(CN) - P(CBN) = 0.40+0.30-0.50-0.08 = 0.12

P(C' B N) = P(a person buys beer and peanuts and does not buy soda)

= P(BN) - P(CBN) = 0.15-0.08 = 0.07

P(a person buys exactly two things) = P(C B N') + P(C B' N) + P(C' B N) = 0.09+0.12+0.07 = 0.28

P(a person buys at least two things)

= P(a person buys exactly two things) + P(a person buys all three things) = 0.28+0.08 = 0.36

Probability that a person buys all three, given that he buys at least two

d)

Probability that a person buys peanut, given that he doesn’t buy beer

.

C)

C,B,N are independent events C',B',N' are independent events

a)

Probability that a person buys none of these things

= P(C' B' N') = P(C')P(B')P(N') = [1-P(C)][1-P(B)][1-P(N)] = (1-0.40)(1-0.60)(1-0.30) = 0.168

b)

P(C B' N') = Probability that a person buys only soda

= P(C)P(B')P(N') = P(C)[1-P(B)][1-P(N)] = 0.40(1-0.60)(1-0.30) = 0.112

P(C' B N') = Probability that a person buys only beer

= P(C')P(B)P(N') = [1-P(C)]P(B)[1-P(N)] = (1-0.40)0.60(1-0.30) = 0.252

P(C' B' N) = Probability that a person buys only peanuts

= P(C')P(B')P(N) = [1-P(C)][1-P(B)]P(N) = (1-0.40)(1-0.60)0.30 = 0.072

Probability that a person buys exactly one of the things = P(C B' N') + P(C' B N') + P(C' B' N)

= 0.112+0.252+0.072 = 0.436

c)

Probability that a person buys all three, given that he buys at least two

P(a person buys all three things) = P(CBN) = P(C)P(B)P(N) = 0.400.600.30 = 0.072

P(C B N') = P(a person buys soda and beer and does not buy peanuts)

= P(C)P(B)P(N') = P(C)P(B)[1-P(N)] = 0.400.60(1-0.30) = 0.168

P(C B' N) = P(a person buys soda and peanuts and does not buy beer)

= P(C)P(B')P(N) = P(C)[1-P(B)]P(N) = 0.40(1-0.60)0.30 = 0.048

P(C' B N) = P(a person buys beer and peanuts and does not buy soda)

= P(C')P(B)P(N) = [1-P(C)]P(B)P(N) = (1-0.40)0.600.30 = 0.108

P(a person buys exactly two things) = P(C B N') + P(C B' N) + P(C' B N) = 0.168+0.048+0.108 = 0.324

P(a person buys at least two things)

= P(a person buys exactly two things) + P(a person buys all three things) = 0.324+0.168 = 0.492

Probability that a person buys all three, given that he buys at least two

d)

Probability that a person buys peanut, given that he doesn’t buy beer

.

D)

1) Possible earnings are $10 , $5 and $0.

2)

Probability of winning $10 = P(sum of the two die is at least 10)

= P(4,6) + P(5,5) + P(5,6) + P(6,4) + P(6,5) + P(6,6) = 6(1/6)² = 1/6

Probability of winning $5

= P(sum of the two die is at 4 to 9)

= P(1,3) + P(1,4) + P(1,5) + P(1,6) + P(2,2) + P(2,3) + P(2,4) + P(2,5) + P(2,6) + P(3,1) + P(3,2) + P(3,3) + P(3,4) + P(3,5) + P(3,6) + P(4,1) + P(4,2) + P(4,3) + P(4,4) + P(4,5) + P(5,1) + P(5,2) + P(5,3) + P(5,4) + P(6,1) + P(6,2) + P(6,3)

= 27(1/6)² = 3/4

Probability of winning $0 = P(sum of the two die is less than 4) = P(1,1) + P(1,2) + P(2,1) = 3(1/6)² = 1/12

3)

Expected winnings = $10(1/6) + $5(3/4) + $0(1/12) = $(65/12) = $5.42

4)

Variance of winnings = ($10)²(1/6) + ($5)²(3/4) + ($0)²(1/12) - ($65/12)² = $²(875/144) = $² 6.0764

.

E)

a) E(1/X) = (1/$10)(1/9) + (1/$5)(13/18) + (1/$0)(1/12) = . So, E(1/X) does not exist.