211 A factory has two assembly lines each of which is shut d

2.11 A factory has two assembly lines, each of which is shut down (S), at partial capacity (P), or at full capacity (F) The sample space is given in Figure 1.25, where, for example, (S, P) denotes that the first assembly line is shut down and the second one is operating at partial capacity. What is the probability that (a) both assembly lines are shut down? (b) neither assembly line is shut down? (c) at least one assembly line is at full capacity? (d) exactly one assembly line is at full capacity? What is the complement of the event in part (b)? What is the complement of the event in part (c)? (This problem is continued in Problem 1.4.10.) (S, S) (S, P) (S, F) 0.06 0.02 0.05 OP S) (P P) (P F) 0.07 0.14 0.20 (F, S) (F, P) (F, F) 0.21 0.06 0.19 GURE 1.25

Solution

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A factory has two assembly lines, each of which is shut down (S), at partial capcity (P), or at full capacity (F).

Probability values for assembly line operations:

(S,S)=.02    (S,P)=.06     (S,F)=.05

(P,S)=.07     (P,P)=.14     (P,F)=.20

(F,S)=.06     (F,P)=.21     (F,F)=.19

Evaluate the following scenarios:

(a) Both lines are at full capacity conditional on neither line being shut down.

(b) At least one line is at full capacity conditional on neither line being shut down.

(c) One line is at full capacity conditional on exactly one line being shut down.

(d) Neither line is at full capacity conditional on at least one line operating at partial capacity.

Ans

(a) neither line being shut down, so ignore all above pairs with S in it
left with (P,P), (P,F), (F,P) (F,F)

Probability = (F,F) /[(P,P)+(P,F)+ (F,P) +(F,F)] = 0.19/(0.14+0.2+0.21+0.19) = 0.26

(b) Probability = (P,F) + (F,P) + (F,F) /((P,P)+ (P,F) + (F,P) + (F,F)) = (0.2+0.21+0.19)/(0.14+0.20+0.21+0.19) = 0.81

(c) look for all pairs with exact one S in it:
(S,P) (S,F) (P,S) (F,S)

Probability = (S,F) + (F,S) / [(S,P) + (S,F) + (P,S) + (F,S)] = (0.05+0.06)/(0.06+0.05+0.07+0.06) = 0.46

(d) look for all pairs with at least one P in it

(S,P) (P,S) (P,P) (P,F) (F,P)

Probability = (S,P)+ (P,S) +(P,P) /[(S,P) + (P,S)+ (P,P)+ (P,F)+(F,P)] = (0.06+0.07+0.14)/(0.06+0.07+0.14+0.20+0.21)

= 0.40