Question

A company has two plants to manufacture scooters. Plant-1 manufactures 70% of the scooters and plant-2 manufactures 30%. At plant-1, 30% of the scooters are rated of standard quality and at plant-2, 90% of the scooters are rated as standard quality. A scooter is chosen at random and is found to be of standard quality. Find the probability that it has come from plant 2.

Answer 1

Hint: Here, we will use the concept of probability. First we will write the given probabilities of both the plants separately. Then we will apply the Baye’s theorem to get the probability of scooter chosen at random and is of standard quality from plant 2.

Complete step-by-step answer:
Let A be the manufactured scooters from plant 1 and B be the manufactured scooters from plant 2.
Let \[S\] be the manufactured scooter of standard quality.
It is given that Plant-1 manufactures 70% of the scooters and plant-2 manufactures 30%. Therefore, we get
\[\begin{array}{l}P\left( A \right) = \dfrac{{70}}{{100}} = \dfrac{7}{{10}}\\P\left( B \right) = \dfrac{{30}}{{100}} = \dfrac{3}{{10}}\end{array}\]
It is given in the question that at plant-1, 30% of the scooters are rated of standard quality.
Scooters rated of standard quality from plant 1, \[P\left( {\dfrac{A}{S}} \right) = \dfrac{{30}}{{100}} = \dfrac{3}{{10}}\].
Similarly it is given that at plant-2, 90% of the scooters are rated as standard quality.
\[ \Rightarrow \] Scooters rated of standard quality from plant 2, \[P\left( {\dfrac{B}{S}} \right) = \dfrac{{90}}{{100}} = \dfrac{9}{{10}}\].
Now we will find the probability of scooter chosen at random and is found to be of standard quality from plant 2 i.e. \[P\left( {\dfrac{S}{B}} \right)\]. By using the Baye’s theorem, we get
\[P\left( {\dfrac{S}{B}} \right) = \dfrac{{\left( {P\left( B \right).P\left( {\dfrac{B}{S}} \right)} \right)}}{{\left( {P\left( A \right).P\left( {\dfrac{A}{S}} \right)} \right) + \left( {P\left( B \right).P\left( {\dfrac{B}{S}} \right)} \right)}}\]
Now we will put the values of probabilities in the above equation. Therefore, we get
\[P\left( {\dfrac{S}{B}} \right) = \dfrac{{\left( {\dfrac{3}{{10}} \times \dfrac{9}{{10}}} \right)}}{{\left( {\dfrac{7}{{10}} \times \dfrac{3}{{10}}} \right) + \left( {\dfrac{3}{{10}} \times \dfrac{9}{{10}}} \right)}} = \dfrac{{\left( {\dfrac{{27}}{{100}}} \right)}}{{\left( {\dfrac{{21}}{{100}}} \right) + \left( {\dfrac{{27}}{{100}}} \right)}} = \dfrac{{27}}{{21 + 27}} = \dfrac{{27}}{{48}}\]
Hence, the probability of a scooter chosen at random and is found to be of standard quality from plant 2 is \[\dfrac{{27}}{{48}}\].

Note: We have used Baye’s theorem formula to solve the question. Baye’s theorem formula is generally used to calculate a probability when we know other certain probabilities also. Probability is the branch of mathematics, which gives the possibility of the event to occur. It is equal to the ratio of the number of favorable outcomes to the total number of outcomes.
Probability generally lies between the values of 0 to 1. Probability can never be above the value of 1 or less than 0. So if we got the value of probability greater than 1 then we need to check out the calculation to remove the error for calculating the probability.