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 arts becomes defective. What is the probability that the machine will not stop working?

• A. $0.06$
• B. $0.16$
• C. $0.84$
• D. $0.94$

Answer 1

The correct option is C $0.84$

Let $P\left(A\right),P\left(B\right),P\left(C\right)$ be the probability of parts A,B and C working properly.

$\therefore P\left(A\right)=1-0.02=0.98$

$\therefore P\left(B\right)=1-0.1=0.9$

$\therefore P\left(C\right)=1-0.05=0.95$

Probaility that machine will not stop working $=$ Probaility that all parts are working properly

$=P(A\cap B\cap C)$

$=P\left(A\right)P\left(B\right)P\left(C\right)$ ...( A,B and C are independent events)

$=\left(0.98\right)\left(0.9\right)\left(0.95\right)$

$=0.8379=0.84$

Answer is option (C)