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?

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

Answer 1

The correct option is C

$0.84$

Explanation for correct option:

Option (C): $0.84$

Let P(A), P(B), and P(C) be the probability of parts A, B, and C working properly,

$\Rightarrow$ P(A) = $1-0.02$

= $0.98$

$\Rightarrow$ P(B) = $1-0.10$

= $0.90$

$\Rightarrow$ P(C) = $1-0.05$

= $0.95$

Probability of all parts working properly and not stopping,

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

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

= $0.98\times 0.90\times 0.95$

= $0.8379$

= $0.84$

Therefore, the probability of the machine working properly is $0.84$

Explanation for incorrect options:

Option (A): $0.06$

Here, P$(A\cap B\cap C)$ = $0.84$

Therefore, the probability of the machine working properly is not $0.06$

Option (B): $0.16$

Here, P$(A\cap B\cap C)$ = $0.84$

Therefore, the probability of the machine working properly

Option (D): $0.94$

Here, P$(A\cap B\cap C)$ = $0.84$

Therefore, the probability of the machine working properly is not $0.94$.

Hence, Option (C) is the correct answer.