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 D 1 , D2 , and D 3 are independent of each other.
a) P (D 1 D2 D 3 ) = P (D1 )P (D2 )P (D 3 )
bec because ause D1 , D2 , D3 are independent = (1 − (1 − P (D1 ))(1 − ))(1 − P (D2 ))(1 − ))(1 − P (D3 )) = (1 − (1 − 0 0.9)(1 − 9)(1 − 0 0.9)(1 − 9)(1 − 0 0.9) = 0.729.
b) P (exactly 2 welds def.) = P (D1 D2 D 3 ∪ D1 D2 D3 ∪ D 1 D2 D3 )
= P (D1 )P (D2 )(1 − )(1 − P (D3 )) + P (D1 )(1 − )(1 − P (D2 ))P (D3 ) + (1 − (1 − P (D1 ))P (D2 )P (D3 ) = 0.1 ∗ 0 ∗ 0.1 ∗ 0 ∗ 0.9 + 0 .1 ∗ 0 ∗ 0.9 ∗ 0 ∗ 0.1 + 0 .9 ∗ 0 ∗ 0.1 ∗ 0 ∗ 0.1 = 0.027.
c) P (D1 D2 D3 ) = P (D1 )P (D2 )P (D3 )
= 0.1 ∗ 0 ∗ 0.1 ∗ 0 ∗ 0.1 = 0.001.
Problem 2.9 Define • E 1 = the event that tractor 1 is in good condition after 5 years • E 2 = the event that tractor 2 is in good condition after 5 years • E 3 = the event that tractor 3 is in good condition after 5 years.
1
a) • A = E 1 E 2 E 3 • B = E 1 E 2 E 3 ∪ E 1 E 2 E 3 ∪ E 1 E 2 E 3 • C = E 1 ∪ E 2 ∪ E 3
b) We know P (E 1 ) = 0.6, P (E 2 ) = 0.6, P (E 3 ) = 0.6, P (E i |E j ) = 0.8 for every pair ( i, j ) of tractors, and P (E i |E j E k ) = 0.8 for each tuple ( i,j,k ) of tractors. P (A) = P (E 1 E 2 E 3 )
= P (E 1 |E 2 E 3 )P (E 2 E 3 ) = (1 − P (E 1 |E 2 E 3 ))P (E 2 |E 3 )P (E 3 ) = (1 − 0.8) ∗ 0.6 ∗ (1 − 0.6) = 0.048. P (B ) = P (E 1 E 2 E 3 ∪ E 1 E 2 E 3 ∪ E 1 E 2 E 3 )
and because the intersection of any two events in the union is empty,
= P (E 1 E 2 E 3 ) + P (E 1 E 2 E 3 ) + P (E 1 E 2 E 3 )
since P (E 1 E 2 E 3 ) = P (E 1 E 2 E 3 ) = P (E 1 E 2 E 3 )
= 3 ∗ (1 − P (E 1 |E 2 E 3 ))P (E 2 |E 3 )P (E 3 ) = 3 ∗ 0.048 = 0.144. P (C ) = P (E 1 ∪ E 2 ∪ E 3 )
and applying De Morgan’s Rule,
= 1 − P (E 1 E 2 E 3 ) = 1 − P (E 1 |E 2 E 3 )P (E 2 E 3 ) = 1 − P (E 1 |E 2 E 3 )P (E 2 |E 3 )P (E 3 ) = 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 • S A = the event that there exists a continuous stream of sand from X to town A • S B = 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 (S A ) = 0.02, P (S B ) = 0.03, P (S B |S A ) = 0 .2, and X is independent of S A , S B .
Page 2
a) P (A) = P (XS A )
and because X and S A are independent, = P (X )P (S A ) = 0.01 ∗ 0.02 = 0.0002.
b) P (A ∪ B ) = P (A) + P (B ) − P (AB )
= P (XS A ) + P (XS B ) − P (XS A S B ) = P (X )(P (S A ) + P (S B ) − P (S B |S A )P (S A )) = 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 (LM G) P (C )
P (L)P (M |G)P (G) P (C ) 0.3 ∗ 1 ∗ 0.6 = 0.74 = 0 .243.
=
Page 3
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 (F S )
= 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 (F S )
= 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 |A B ) = 0 .40.
a) P (A B ) = 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 (A B C )
= 1 − P (C |A B )P (A B ) = 1 − (1 − P (C |A B ))P (A B ) = 1 − (1 − 0.4) ∗ 0.76 = 0.544.
Page 4
c) P (A ∪ B ) P (A ∪ B ∪ C ) 1 − 0.76 = 0.544 = 0.441.
P (A ∪ B |A ∪ B ∪ C ) =
=
1 − P (A B ) P (A ∪ B ∪ C )
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 (M W ) = 0.01, and B is ind. of M, W .
a) P (W M B ) = P (B )P (M |W )P (W )
= P (B )(1 − P (M |W ))P (W ) P (M W ) = (1 − P (B )) 1 − P (W )
0 .01 = (1 − 0.5) 1 − 0.05 = 0.02.
P (W )
∗ 0.05
b) P (M ∪ W ∪ B ) = P (M ) + P (W ) + P (B ) − P (M W ) − P (M B ) − P (W B ) + P (M W B )
= P (M ) + P (W ) + P (B ) − P (M W ) − P (M )P (B ) − P (W )P (B ) + P (M W )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 (M W B |M ∪ W ∪ B ) + P (M W B |M ∪ W ∪ B ) + P (M W B |M ∪ W ∪ B )
=
P (M W B ) + P (M W B ) + P (M W )P (B ) P (M ∪ W ∪ B ) P (M W B ) + (1 − P (B
= =
)) 1 − ( ) + (1 + ( ) − ( ∪ ∪ ) ∗ 0 05 + 0 5 ∗ (1 + 0 01 − 0 02 − 0 05) P (MW ) W
P W
P M
0 02 + (1 − 0 5) 1 − .
.
0.01 0.05
.
.
0.53
= 0.962.
Page 5
P
W
.
P M W
B
.
.
P (M ) − P (W ))P (B )
Problem 2.21 Define • E 1 = the event that summer 1 is hot • E 2 = the event that summer 2 is hot • E 3 = the event that summer 3 is hot. We know P (E 1 ) = P (E 2 ) = P (E 3 ) = 0.2, P (E 2 |E 1 ) = 0.4, P (E 3 |E 2 ) = 0.4, and E 1 and E 3 are independent.
a) P (E 1 E 2 E 3 ) = P (E 3 |E 1 E 2 )P (E 1 E 2 )
and because E 1 does not effect P (E 3 ), = P (E 3 |E 2 )P (E 2 |E 1 )P (E 1 ) = 0.4 ∗ 0.4 ∗ 0.2 = 0.032.
b) P (E 2 |E 1 ) =
P (E 2 E 1 )
P (E 1 ) 1 + P (E 1 E 2 ) − P (E 1 ) − P (E 2 ) = 1 − P (E 1 ) 1 − 0.2 − 0.2 + 0 .4 ∗ 0.2 = 1 − 0.2 = 0.85.
c) P (E 1 ∪ E 2 ∪ E 3 ) = 1 − P (E 1 E 2 E 3 )
= 1 − P (E 3 |E 2 E 1 )P (E 2 E 1 ) = 1 − P (E 3 |E 2 )P (E 2 |E 1 )P (E 1 ) = 1 − 0.85 ∗ 0.85 ∗ 0.8 = 0 .422.
Page 6
