Question

An article manufactured by a company consists of two parts X and y. In the process of manufacturing the part X, 9 out of 100 parts may be effective. Similarly, 5 out of 100 are likely to be defective in part Y. Calculate the probability that the assembled product will not be defective.

Answer 1

Hint: Assume ‘A’ as the event of getting defective part X and ‘B’ as the event of getting defective part Y. Find their probability by using the given information. Now, use the formula: - \[P\left( \overline{X} \right)=1-P\left( X \right)\], to get the values of \[P\left( \overline{A} \right)\] and \[P\left( \overline{B} \right)\], which are the probabilities of getting non – defective part X and Y respectively. Take the product of \[P\left( \overline{A} \right)\] and \[P\left( \overline{B} \right)\] to calculate the probability that the assembled part will not be defective.

Complete step-by-step solution:
Here, let us assume the following things: -
A = Event of getting defective part X.
B = Event of getting defective part Y.
Now, it is given in the question that we get 9 out of 100 parts are defective in X and 5 out of 100 parts are defective in Y. Therefore, probabilities of getting defective part in X and Y can be given as: -
\[\begin{align}
  & \Rightarrow P\left( A \right)=\dfrac{9}{100} \\
 & \Rightarrow P\left( B \right)=\dfrac{5}{100} \\
\end{align}\]
Here, we have to find the probability that the assembled product will not be defective. Since, the assembled product will be a combination of both parts X and Y, so for the assembled product to be non – defective we must have both the parts X and Y non – defective. Therefore, we have,
 \[\Rightarrow P\left( \overline{A} \right)\] = probability of getting non – defective part X = \[1-P\left( A \right)\]
\[\begin{align}
  & \Rightarrow P\left( \overline{A} \right)=1-P\left( A \right) \\
 & \Rightarrow P\left( \overline{A} \right)=1-\dfrac{9}{100} \\
 & \Rightarrow P\left( \overline{A} \right)=\dfrac{91}{100} \\
\end{align}\]
\[\Rightarrow \]\[P\left( \overline{B} \right)\] = probability of getting non – defective part Y = \[1-P\left( B \right)\]
\[\Rightarrow P\left( \overline{B} \right)=1-\dfrac{5}{100}\]
\[\Rightarrow P\left( \overline{B} \right)=\dfrac{95}{100}\]
Now, the event that both the parts will be non – defective will be represented as \[\overline{A}\cap \overline{B}\]. Therefore, the probability be given as \[P\left( \overline{A}\cap \overline{B} \right)\]. Since, both \[\overline{A}\] and \[\overline{B}\] are independent of each other, therefore, we have,
\[\Rightarrow P\left( \overline{A}\cap \overline{B} \right)=P\left( \overline{A} \right)\times P\left( \overline{B} \right)\]
Substituting the values of \[P\left( \overline{A} \right)\] and \[P\left( \overline{B} \right)\], we get,
\[\begin{align}
  & \Rightarrow P\left( \overline{A}\cap \overline{B} \right)=\dfrac{91}{100}\times \dfrac{95}{100} \\
 & \Rightarrow P\left( \overline{A}\cap \overline{B} \right)=\dfrac{8645}{10000} \\
 & \Rightarrow P\left( \overline{A}\cap \overline{B} \right)=0.8645 \\
\end{align}\]
Hence, the probability that the assembled product will be non – defective is 0.8645.

Note: One may note that for a product to be non – defective, all its part must be non – defective and that is why we have taken the intersection of the events \[\overline{A}\] and \[\overline{B}\]. Note that the formula, \[P\left( \overline{A}\cap \overline{B} \right)=P\left( \overline{A} \right)\times P\left( \overline{B} \right)\] is used above because the events \[\overline{A}\] and \[\overline{B}\] does not depend on each other. One of the most important formulas that must be remembered to solve the above question is, \[P\left( \overline{X} \right)=1-P\left( X \right)\].