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**Unformatted text preview: **3.24 a. Let H = Hyundai Elantra, T: Toyota Prius, and S = Subaru Forrester. All possible rankings are as
follows, where the first car listed is ranked ﬁrst, the second car listed is ranked second, and the third.
car listed is ranked third: HES H61?" 3,1353" SEE EH3 IEH b. If each set of rankings is equally likely, then each has a probability of 1.36. The probability that the Toyota Prius is ranked ﬁrst
=P(T,H,S)+P(T,S,Hj=1.-"6+l.r'6 =2;‘6=1:‘3 The probability that the Hyundai Elantra is ranked third
= P(S,T,H}+P(T,S,H) =1.r6+1.r'6 = 2.-‘6 =1.-'3_ The probability that the Toyota Prius is ranked first and the Subaru Forrester is ranked second
= PCT,S,H) =1.r6_ 3.30 The experiment consists of rolling a pair of fair dice. The sample points are: Since each die is fair, each sample point is equally likely. The probability of each sample point is 1;"35.
a. A: {(1, 6), (2, 5), (3, 4), (4, 3),(5,2),(6,1)} BI {(134)114);(3:4)=(434);(5,4):(5,4);(4; 1), (4, 2): (4; 3), (43 5), (4, 5)} A03: {(3, 4), (4, 3)} AUBI {(1: 4): (2, 4), (3} 4): (4, 4), (5; 4)} (5, 4), (4, 1); (4: 21501331391: 5):
(4= 5), (1, 61(2, 5): C5: 2), (5, 13'} A“: {(1} 11'2 (1, 231,“; 310, 4L0; 5); (23 115(13): (2? 3); (3, 410; (5)76: 1), (3; 3); (3, 3), (3; 5);
(151(4, 13,01; 332(4; 4): (4, 5); (4: 6); (5, 13; (52 3); (5: 43L (5;. 5): (5, 6), C6; 3): (5: 3): (6; 4):
(6: 511:5, 6)} i\ =£=l P(a)=11[i]=ﬂ P(AHB)=2[-Hi\=i=i 36, 36 6 36 36 I36} 36 Is b P(A)—6[ W. F - r
Pmoaj=15£i =1—3—i PKAE)=30| £133
30; 35 12 k30,73 1 11 l 6+‘ll—2 15 S
c- PIZAUB)=P(A)+P{B)—P[AﬂB}=—+———=—=—=—
6 36 18 36 36 12 d. A andB are not mutually exclusive- To be mutually exclusive. P(.«1 rs 3) must be 0. Here, 1
Fame =—.
( ) 18 3.113 Deﬁne the following events:
3: {System shuts down}
F1: {Hardware failure}
F3: {Software failure}
F3: {Power failure}
From the Exercise, we know: P(F1) = .01, P(F2)=.05 ,and P(F3)=.ﬂ2. Also; P(S|F1)=.73, P(S|F1)=.12,andP(S|P;)=.38. The probability that the current shutdown is due to a hardware failure is: HF I S) 2 as n3) : P(S|F1)P(F1)
1 Po“) PIS I FIJHFJ + P(3 I Poms) +PcSI mes)
_ 03001) _ .00r3 _ .00s3 _ 3362
03001) + .12(.05)+.88(.l]2) .00?3+.005+.01?5 .0309 'L
The probability that the current shutdown is due to a software failure is:
HF 3)_P(F2FISJ _ P(S|F:)P(F2)
‘ Po) P(S|F1)P(EJ+P(SI was) + P(S| Fame)
.12(.05) .005 _ .006 21942 _ .7‘3(.01)+.12(.05) +.33(.02) _ .00r3 +.005+ .0115 _ .0309 The prebahility that the current shutdown is due to a pewer failure is: HF IS) _P(F3ﬂSJ _ PESIEFGD
3 _ Pia) _PIES|E)P(E)+P(S|EJP(F2)+P(SIF;)P(P;)
.33(_02) Lane __{}1T6 : _ = — _ — = .5596
_?3(_:]1)+_12(_e5)+_33(_02) Janna +_ne5+_m?e .3313? ...

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