Question

A machine has three parts, A, B, and C, whose chances of being defective are $0.02,0.01,$ 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?
1. $0.06$
2. $0.16$
3. $0.84$
4. $0.94$

Answer 1

Hint:In this question, given the probability of three parts which are defective. Using the complement rule $P\left( A \right) + P\left( {A'} \right) = 1$ find the probability of all the parts working properly. Now for the probability that the machine will not stop working. You have to calculate the probability that all parts will work properly.

Formula used:
Independent multiplication rule –
$P\left( {A \cap B \cap C} \right) = P\left( A \right) \times P\left( B \right) \times P\left( C \right)$
Probability complement rule –
$P\left( A \right) + P\left( {A'} \right) = 1$

Complete step by step solution: 
Let, Probability of parts A, B, and C are defective be $P\left( A \right),P\left( B \right),P\left( C \right)$respectively
And probability of parts A, B, and C are working properly be $P\left( {A'} \right),P\left( {B'} \right),P\left( {C'} \right)$
Given that,
$P\left( A \right) = 0.02,P\left( B \right) = 0.10,P\left( C \right) = 0.05$
Therefore,
$P\left( {A'} \right) = 1 - 0.02 = 0.98$
$P\left( {B'} \right) = 1 - 0.10 = 0.90$
$P\left( {C'} \right) = 1 - 0.05 = 0.95$
As we know, the probability that the machine will not stop working is equal to the probability that all parts will work properly.
So, probability that the machine will not stop working
$ = P\left( {A' \cap B' \cap C'} \right)$
$ = P\left( {A'} \right) \times P\left( {B'} \right) \times P\left( {C'} \right)$
$ = 0.98 \times 0.90 \times 0.95$
$ = 0.8379$
$ \approx 0.84$
Hence, Option (3) is the correct answer.

Note: To solve such questions, students must know about complements, union, and intersection. Basically, the union of two sets contains all the elements which are contained in either set (or both sets). The intersection of two sets contains only those elements that are in both sets and the complement of a set A contains everything that is not in the set A.