Search any question & find its solution
Question:
Answered & Verified by Expert
A machine has three parts, A, B and $\mathrm{C}$, 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:
Solution:
2396 Upvotes
Verified Answer
The correct answer is:
$0.84$
Probability that machine stops working $=P(A \cup B \cup C)$
$\Rightarrow P(A \cup B \cup C)=P(A)+P(B)+P(C)$
$-P(A \cap B)-P(A \cap C)-P(B \cap C)$
$+P(A \cap B \cap C)$
$\begin{aligned} \Rightarrow P(A \cup B \cup C)=& P(A)+P(B)+P(C) \\ &-P(A) P(B)-P(A) P(C) \\ &-P(B) P(C)+P(A) P(B) P(C) \end{aligned}$
$(\because A, B \& C$ 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 $=1-P(A \cup B \cup C)=1-0.16=0.84$
$\Rightarrow P(A \cup B \cup C)=P(A)+P(B)+P(C)$
$-P(A \cap B)-P(A \cap C)-P(B \cap C)$
$+P(A \cap B \cap C)$
$\begin{aligned} \Rightarrow P(A \cup B \cup C)=& P(A)+P(B)+P(C) \\ &-P(A) P(B)-P(A) P(C) \\ &-P(B) P(C)+P(A) P(B) P(C) \end{aligned}$
$(\because A, B \& C$ 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 $=1-P(A \cup B \cup C)=1-0.16=0.84$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.