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  • An item is manufactured by three machines A, B and C. Out of the total number of items manufactured during a specified period, 50% are manufactured on A, 30% on B and 20% on C. 2% of the items produced on A and 2% of items produced on B are defective, and

An item is manufactured by three machines A, B and C. Out of the total number of items manufactured during a specified period, 50% are manufactured on A, 30% on B and 20% on C. 2% of the items produced on A and 2% of items produced on B are defective, and 3% of these produced on C are defective. All the items are stored at one go down. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine A?

Answers (1)

Solution

Let E1, E2, E3 be the following events:

E1 -event that item is manufactured by machine A
E2- event that item is manufactured by machine B
E3- event that item is manufactured by machine C

Since, E1, E2 and E3 are mutually exclusive and exhaustive events. Hence, they represent a partition of sample space.

As we know,
Items manufactured on machine A = 50%
Items manufactured on machine B = 30%
Items manufactured on machine C = 20%

Therefore,
\\P\left(E_{1}\right)=50 \% =\frac{50}{100}=\frac{1}{2}\\\\P\left(E_{2}\right)=30 \%=\frac{30}{100}=\frac{3}{10}\\\\P\left(E_{3}\right)=20 \%=\frac{20}{100}=\frac{1}{5}

Let E be the event that ‘an item is defective’.

Therefore, P(E|E1) is the probability that the item drawn is defective given that it is manufactured on machine A = 2%
P(E|E2) is the probability that the item drawn is defective given that it is manufactured on machine B = 2%
P(E|E3) is the probability that the item drawn is defective given that it is manufactured on machine C = 3%

Therefore,

P\left(E \mid E_{1}\right)=\frac{2}{100}, P\left(E \mid E_{2}\right)=\frac{2}{100}, P\left(E \mid E_{3}\right)=\frac{3}{100}

To find- the probability that the item that is picked up is defective, it was manufactured on machine A

Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

\therefore{P}(A \mid B)=\frac{\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B} \mid \mathrm{A})}{P(B)}

\mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{E}\right)  is the probability that the item drawn is defective and it was manufactured on machine A

\\ \therefore P\left(E_{1} \mid E\right)=\frac{P\left(E_{1}\right) \times P\left(E \mid E_{1}\right)}{P\left(E_{1}\right) \times P\left(E \mid E_{1}\right)+P\left(E_{2}\right) \times P\left(E \mid E_{2}\right)+P\left(E_{3}\right) \times P\left(E \mid E_{3}\right)} \\\\=\frac{\frac{1}{2} \times \frac{2}{100}}{\frac{1}{2} \times \frac{2}{100}+\frac{3}{10} \times \frac{2}{100}+\frac{1}{5} \times \frac{3}{100}} \\\\=\frac{\frac{1}{100}}{\frac{1}{100}+\frac{3}{5 \times 100} \times \frac{3}{5 \times 100}} \\\\=\frac{1}{1+\frac{3}{5}+\frac{3}{5}} \\\\=\frac{5}{5+6} \\\\=\frac{5}{11}

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