Question

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

Answer 1

Hint: Approach the solution by finding the probability of defective parts first. And then use the complement rule and product rule of probability.

Complete step-by-step answer:

Let us consider:
$A$ represents events for which part $X$ is defective while $B$ represents events for which Part $Y$ is defective.
Given that in part $X$, $9$ out of $100$ parts are defective
$\therefore$Probability of $A$, [ $P(A)$] = $\dfrac{9}{{100}}$
And also given that in part $Y$, $5$ out of $100$ parts are defective
$\therefore$Probability of $B$, [$P(B)$] = $\dfrac{5}{{100}}$
Here we have considered probability of defective parts as $P(A)$,$P(B)$
Then probability of non-defective parts will be $P\left( {\bar A} \right)$ and $P\left( {\bar B} \right)$
We have to find the probability of assembled products which are not defective (non-defective).
Required probability = Probability of assembled products which are non-defective
Using product rule of probability, we’ll get:
$\therefore$Required probability =$P\left( {\bar A} \right) \times P(\bar B)$
According to complement rule of probability, we know that:
$\Rightarrow$ $P(\bar A) = 1 - P(A)$
$\Rightarrow P(\bar B) = 1 - P(B)$
By using above condition we can write the required probability as
Required probability =$P\left( {\bar A} \right) \times P(\bar B)$
Required probability $\Rightarrow [1 - P(A)][1 - P(B)]$
$\Rightarrow \left( {1 - \dfrac{9}{{100}}} \right) \times \left( {1 - \dfrac{5}{{100}}} \right) \\\ \Rightarrow \dfrac{{91}}{{100}} \times \dfrac{{95}}{{100}} \\\ \Rightarrow 0.91 \times 0.95 \\\ \Rightarrow 0.8645 \\\$

Therefore the probability of assembled product of non-defective parts is $0.8645$

NOTE: Concentrate on converting the values of $P\left( {\bar A} \right)$ and $P\left( {\bar B} \right)$ when $P(A)$&$P(B)$ values are given.
If the probability of occurrence of two independent events are ${P_1}$ are ${P_2}$ respectively, then according to product rule, the probability of occurrence both the events simultaneously will be:
$\Rightarrow P = {P_1}{P_2}$