# 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 $100$are likely to be defective in part $Y$.Calculate the probability that the assembled product will not be defective.

Answer

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Hint: Equate the part X that is not defective to a variable and do the same thing with Y as well and then find the intersection of the two cases.

Let $A = Part X$ is not defective.

Probability of $A$ is $P\left( A \right) = \dfrac{{91}}{{100}}$

Similarly for $B$,

$B = Part Y $is not defective

Probability of $B$ is $P\left( B \right) = \dfrac{{95}}{{100}}$

Required Probability is,

$P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right)$

$P\left( {A \cap B} \right) = \dfrac{{91}}{{100}} \times \dfrac{{95}}{{100}}$

$P\left( {A \cap B} \right) = \dfrac{{8645}}{{10000}}$

Note: We took 2 variables A and B as the non defective ones and then found the probability by taking the intersection of these 2 cases.

Let $A = Part X$ is not defective.

Probability of $A$ is $P\left( A \right) = \dfrac{{91}}{{100}}$

Similarly for $B$,

$B = Part Y $is not defective

Probability of $B$ is $P\left( B \right) = \dfrac{{95}}{{100}}$

Required Probability is,

$P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right)$

$P\left( {A \cap B} \right) = \dfrac{{91}}{{100}} \times \dfrac{{95}}{{100}}$

$P\left( {A \cap B} \right) = \dfrac{{8645}}{{10000}}$

Note: We took 2 variables A and B as the non defective ones and then found the probability by taking the intersection of these 2 cases.

Last updated date: 01st Jun 2023

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