MAE108 Homework 2 Solutions - Spring 2015

April 19, 2015

Problem 2.7

Define

• D1 = the event that the first weld is defective

• D2 = the event that the second weld is defective

• D3 = the event that the third weld is defective.

We know P (D1) = P (D2) = P (D3) = 0.10 and D1, D2, and D3 are independent of each other.

a)

P (D1D2D3) = P (D1)P (D2)P (D3) because D1, D2, D3 are independent

= (1− P (D1))(1− P (D2))(1− P (D3))

= (1− 0.9)(1− 0.9)(1− 0.9)

= 0.729.

b)

P (exactly 2 welds def.) = P (D1D2D3 ∪D1D2D3 ∪D1D2D3)

= P (D1)P (D2)(1− P (D3)) + P (D1)(1− P (D2))P (D3) + (1− P (D1))P (D2)P (D3)

= 0.1 ∗ 0.1 ∗ 0.9 + 0.1 ∗ 0.9 ∗ 0.1 + 0.9 ∗ 0.1 ∗ 0.1

= 0.027.

c)

P (D1D2D3) = P (D1)P (D2)P (D3)

= 0.1 ∗ 0.1 ∗ 0.1

= 0.001.

Problem 2.9

Define

• E1 = the event that tractor 1 is in good condition after 5 years

• E2 = the event that tractor 2 is in good condition after 5 years

• E3 = the event that tractor 3 is in good condition after 5 years.

1

a)

• A = E1E2E3

• B = E1E2E3 ∪ E1E2E3 ∪ E1E2E3

• C = E1 ∪ E2 ∪ E3

b)

We know P (E1) = 0.6, P (E2) = 0.6, P (E3) = 0.6, P (Ei|Ej) = 0.8 for every pair (i, j) of tractors, andP (Ei|EjEk) = 0.8 for each tuple (i, j, k) of tractors.

P (A) = P (E1E2E3)

= P (E1|E2E3)P (E2E3)

= (1− P (E1|E2E3))P (E2|E3)P (E3)

= (1− 0.8) ∗ 0.6 ∗ (1− 0.6)

= 0.048.

P (B) = P (E1E2E3 ∪ E1E2E3 ∪ E1E2E3) and because the intersection of any two events in the union is empty,

= P (E1E2E3) + P (E1E2E3) + P (E1E2E3) since P (E1E2E3) = P (E1E2E3) = P (E1E2E3)

= 3 ∗ (1− P (E1|E2E3))P (E2|E3)P (E3)

= 3 ∗ 0.048

= 0.144.

P (C) = P (E1 ∪ E2 ∪ E3) and applying De Morgan’s Rule,

= 1− P (E1E2E3)

= 1− P (E1|E2E3)P (E2E3)

= 1− P (E1|E2E3)P (E2|E3)P (E3)

= 1− 0.8 ∗ 0.6 ∗ (1− 0.6)

= 0.808.

Problem 2.11

Define

• X = the event of a waste leak from storage site X in 100 years

• SA = the event that there exists a continuous stream of sand from X to town A

• SB = the event that there exists a continuous stream of sand from X to town B

• A = the event that water in town A is contaminated

• B = the event that water in town B is contaminated.

We know P (X) = 0.01, P (SA) = 0.02, P (SB) = 0.03, P (SB |SA) = 0.2, and X is independent of SA, SB .

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a)

P (A) = P (XSA) and because X and SA are independent,

= P (X)P (SA)

= 0.01 ∗ 0.02

= 0.0002.

b)

P (A ∪B) = P (A) + P (B)− P (AB)

= P (XSA) + P (XSB)− P (XSASB)

= P (X)(P (SA) + P (SB)− P (SB |SA)P (SA))

= 0.01 ∗ (0.02 + 0.03− 0.2 ∗ 0.02)

= 0.00046.

Problem 2.13

Define

• G = the event of good weather

• L = the event of adequate labor supply

• M = the event of adequate materials supply

• C = the event of successful completion of the construction project.

We know P (L) = 0.7, P (G) = 0.6, P (M |G) = 1, P (M |G) = 0.5, and L is independent of G,M .

a)

C = G(L ∪M) ∪GLM

b)

P (C) = P (G(L ∪M)) + P (GLM)

= P (GL) + P (GM)− P (GLM) + P (LM |G)P (G)

= P (G)P (L) + P (M |G)P (G)− P (L)P (M |G)P (G) + P (L)P (M |G)(1− P (G))

= 0.6 ∗ 0.7 + 1 ∗ 0.6− 0.7 ∗ 1 ∗ 0.6 + 0.7 ∗ 0.5 ∗ (1− 0.6)

= 0.74.

c)

P (L|C) =P (LC)

P (C)

=P (LMG)

P (C)

=P (L)P (M |G)P (G)

P (C)

=0.3 ∗ 1 ∗ 0.6

0.74= 0.243.

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Problem 2.15

Define

• F = the event that fuel cell technology is successful

• S = the event that solar technology is successful.

We know P (F ) = 0.7, P (S) = 0.85, and F and S are independent.

a)

P (F ∪ S) = P (F ) + P (S)− P (FS)

= P (F ) + P (S)− P (F )P (S)

= 0.7 + 0.85− 0.7 ∗ 0.85

= 0.955.

b)

P (exactly one is successful) = P (F ∪ S)− P (FS)

= 0.955− 0.7 ∗ 0.85

= 0.36.

Problem 2.17

Define

• A = the event that there is an open parking spot in lot A

• B = the event that there is an open parking spot in lot B

• C = the event that there is an open parking spot in lot C.

We know P (A) = 0.20, P (B) = 0.15, P (C) = 0.80, P (B|A) = 0.05, and P (C|AB) = 0.40.

a)

P (AB) = P (B|A)P (A)

= (1− P (B|A))(1− P (A))

= (1− 0.05)(1− 0.2)

= 0.76.

b)

P (A ∪B ∪ C) = 1− P (ABC)

= 1− P (C|AB)P (AB)

= 1− (1− P (C|AB))P (AB)

= 1− (1− 0.4) ∗ 0.76

= 0.544.

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c)

P (A ∪B|A ∪B ∪ C) =P (A ∪B)

P (A ∪B ∪ C)=

1− P (AB)

P (A ∪B ∪ C)

=1− 0.76

0.544= 0.441.

Problem 2.19

Define

• M = the event of master cylinder failure within 4 years or 50,000 miles

• W = the event of wheel cylinder failure within 4 years or 50,000 miles

• B = the event of brake pad failure within 4 years or 50,000 miles.

We know P (M) = 0.02, P (W ) = 0.05, P (B) = 0.50, P (MW ) = 0.01, and B is ind. of M,W .

a)

P (WM B) = P (B)P (M |W )P (W )

= P (B)(1− P (M |W ))P (W )

= (1− P (B))

(1− P (MW )

P (W )

)P (W )

= (1− 0.5)

(1− 0.01

0.05

)∗ 0.05

= 0.02.

b)

P (M ∪W ∪B) = P (M) + P (W ) + P (B)− P (MW )− P (MB)− P (WB) + P (MWB)

= P (M) + P (W ) + P (B)− P (MW )− P (M)P (B)− P (W )P (B) + P (MW )P (B)

= 0.02 + 0.05 + 0.50− 0.01− 0.02 ∗ 0.5 ∗ 0.05 ∗ 0.5 + 0.5 ∗ 0.01

= 0.53.

c)

P (exactly one fails, if total failure) = P (MW B|M ∪W ∪B) + P (MWB|M ∪W ∪B) + P (M WB|M ∪W ∪B)

=P (MW B) + P (MWB) + P (M W )P (B)

P (M ∪W ∪B)

=P (MW B) + (1− P (B))

(1− P (MW )

W

)P (W ) + P (1 + P (MW )− P (M)− P (W ))P (B)

P (M ∪W ∪B)

=0.02 + (1− 0.5)

(1− 0.01

0.05

)∗ 0.05 + 0.5 ∗ (1 + 0.01− 0.02− 0.05)

0.53= 0.962.

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Problem 2.21

Define

• E1 = the event that summer 1 is hot

• E2 = the event that summer 2 is hot

• E3 = the event that summer 3 is hot.

We know P (E1) = P (E2) = P (E3) = 0.2, P (E2|E1) = 0.4, P (E3|E2) = 0.4, and E1 and E3 are indepen-dent.

a)

P (E1E2E3) = P (E3|E1E2)P (E1E2) and because E1 does not effect P (E3),

= P (E3|E2)P (E2|E1)P (E1)

= 0.4 ∗ 0.4 ∗ 0.2

= 0.032.

b)

P (E2|E1) =P (E2E1)

P (E1)

=1 + P (E1E2)− P (E1)− P (E2)

1− P (E1)

=1− 0.2− 0.2 + 0.4 ∗ 0.2

1− 0.2

= 0.85.

c)

P (E1 ∪ E2 ∪ E3) = 1− P (E1E2E3)

= 1− P (E3|E2E1)P (E2E1)

= 1− P (E3|E2)P (E2|E1)P (E1)

= 1− 0.85 ∗ 0.85 ∗ 0.8

= 0.422.

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