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

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: -

& \Rightarrow P\left( A \right)=\dfrac{9}{100} \\\ & \Rightarrow P\left( B \right)=\dfrac{5}{100} \\\ \end{aligned}$$ Here, we have to find the probability that the assembled product will not be defective. Since, the assembled product will be a combination of both parts X and Y, so for the assembled product to be non – defective we must have both the parts X and Y non – defective. Therefore, we have, $$\Rightarrow P\left( \overline{A} \right)$$ = probability of getting non – defective part X = $$1-P\left( A \right)$$ $$\begin{aligned} & \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{aligned}$$ $$\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, $$\begin{aligned} & \Rightarrow P\left( \overline{A}\cap \overline{B} \right)=\dfrac{91}{100}\times \dfrac{95}{100} \\\ & \Rightarrow P\left( \overline{A}\cap \overline{B} \right)=\dfrac{8645}{10000} \\\ & \Rightarrow P\left( \overline{A}\cap \overline{B} \right)=0.8645 \\\ \end{aligned}$$ **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, $$P\left( \overline{X} \right)=1-P\left( X \right)$$.