Pleas show all steps to the solution. Legible handwriting is desirable. PROBLEM
ID: 3353545 • Letter: P
Question
Pleas show all steps to the solution. Legible handwriting is desirable.
PROBLEM 3.2 (pg 87, #80-see diagram below) Consider the system of components in the accompanying picture. Components 3 and 4 are connected in series (call this subsystem 3-4). Subsystem 3-4 will work only if both comppnents 3 and 4 work. In order for the entire system to function, it must be the case that component 1 functions (A1) or component 2 functions (A2) or that subsystem 3-4 functions (A34). Suppose that each individual component functions independently of all other components and each component functions with probability 0.6. Find the probability that the entire system functions or P(AUUA4) Hint: determine P(A34). It isn t 0.60 2 3 4Explanation / Answer
Probability that the lowest branch functions = Probability that both 3 and 4 function = 0.6*0.6 = 0.36
Probability that all the three branches function = 0.6*0.6*0.36 = 0.1296
Probability that upper branch functions and lower two don't = 0.6*0.4*0.64 = 0.1536
Probability that upper and middle branches function and lower doesn't = 0.6*0.6*0.64 = 0.2304
Probability that upper and lower branches function and middle doesn't = 0.6*0.4*0.36 = 0.0864
Probability that middle branch functions and other two don't = 0.4*0.6*0.64 = 0.1536
Probability that lower and middle branches function and upper doesn't = 0.4*0.6*0.36 = 0.0864
Probability that lower branch functions and other two don't = 0.4*0.4*0.36 = 0.0576
So the probability that the whole system functions = 0.1296+0.1536+0.2304+0.0864+0.1536+0.0864+0.0576 = 0.8976
Hope this helps !