A machine has three independent components. The components are grouped together

Question

A machine has three independent components. The components are grouped together into two groups. The first group contains a single component, which has failed with probability q1. The second group contains the other two components, connected in parallel. These components have failed with probabilities q2 and q3. The two groups are connected in series to form a machine.

A)Draw a picture of the second group

B)What is the probability that the second group has failed.

C)Draw a picture of the entire machine and produce a formula for the probability that the entire machine has failed. (For practice, test your general formula: if q1=.15, q2= 0.1, and q3= 0.05, then group 2 has failed with probability 0.005 and the entire machine works with probability 0.84575).

Explanation / Answer

(A) second group: .......-----(machine 2)-----........ ...../......................... ...... --------- ..... ........................./...... .......-----(machine 3)-----........ (B) Second group fails if both machines fail therefore this is an "AND" case P(machine 2 and machine 3 both fail) = P(machine 2 fails) x P(machine 3 fails) Therefore P(group 2 fails) = q2 times q3 If q2= 0.1 and q3 = 0.05, then P(group 2 fails) = 0.1 x 0.05 which is 0.005 (C) Full machine (first and second group): .........................-----(machine 2)----- ......................../........................ -----(machine 1)--------- ................................................./ ..........................-----(machine 3)----- The whole machine fails IF (machine 1 fails) OR [ (macine 1 works) and (Both machine 2 and machine 3 faIL)]. or IS A plus, SO: P(Global failure) = P (machine 1 fails) + P(machine 1 does not fail, and both machines in group 2 fail) P(Global failure) = q1 + (1-q1)x(q2.q3) If q1 = 0.15, q2= 0.1 and q3 = 0.05, Then P(Global failure) = 0.15 + 0.85x(0.1 x 0.05) P(Global failure) = 0.15 + 0.85x0.005 which is 0.15425 P (No failure) 1 -(Failure) P (No failure) = 1 - [ q1 + (1-q1).(q2.q3) ] So: P (No failure) = 1 - 0.15425 P (No failure) = 0.84575

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