Probability Question 356
Question: A machine has three parts, A, B and C, whose chances of being defective are 0.02, 0.10 and 0.05 respectively. The machine stops working if any one of the parts becomes defective. What is the probability that the machine will not stop working?
Options:
A) 0.06
B) 0.16
C) 0.84
D) 0.94
Show Answer
Answer:
Correct Answer: C
Solution:
Probability that machine stops working $ =P(A\cup B\cup C) $
$ \Rightarrow P(A\cup B\cup C)=P(A)+P(B)+P(C) $ $ -P(A\cap B)-P(A\cap C)-P(B\cap C) $
$ +P(A\cap B\cap C) $
$ \Rightarrow P(A\cup B\cup C)=P(A)+P(B)+P(C) $ $ -P(A)P(B)-P(A)P(C) $ $ -P(B)P(C)+P(A)P(B)P(C) $ ( $ \because $ A, B & C are independent events)
$ \Rightarrow P(A\cup B\cup C)=0.02+0.1+0.05-(0.02\times 0.1) $ $ -(0.02\times 0.05)-(0.1\times 0.05) $ $ +(0.02\times 0.05\times 0.1) $
$ \Rightarrow P(A\cup B\cup C)=0.16 $
$ \therefore $ Probability that the machine will not stop working $ =1-P(A\cup B\cup C)=1-0.16=0.84 $
BETA
