A computer producing factory has only two plants T 1 and T 2 . Plant T 1 produces 20 % and plant T 2 produces 80 % of the total computers produced. 7 % of computers produced in the factory turn out to

Question

A computer producing factory has only two plants $T_{1}$ and $T_{2}$ . Plant $T_{1}$ produces $20 %$ and plant $T_{2}$ produces $80 %$ of the total computers produced. $7 %$ of computers produced in the factory turn out to be defective. It is known that,

$P$ (computer turns out to be defective given that it is produced in plant $T_{1}$ )

$= 10 P$ (computer turns out to be defective given that it is produced in plant $T_{2}$ ).

Where, $P (E)$ denotes the probability of an event $E .$ A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant $T_{2}$ is

TestHub · Accepted Answer

P T 1 = 20 100 P T 2 = 80 100 Let, P D T 2 = x (where, D represents defective units) P D T 1 = 10 x P D = 7 100 (given) P T 1 P D T 1 + P T 2 P D T 2 = 7 100 20 100 × 10 x + 80 100 × x = 7 100 x = 1 40 P D T 2 = 1 40 ⇒ P D - T 2 = 39 40 = Probability of not defective, given that it is produced in plant T 2 P D T 1 = 10 40 ⇒ P D - T 1 = 30 40 Now using Bayes' theorem Probability of computer from T 2 given that it is not defective : P T 2 D - = 80 100 × 39 40 20 100 × 30 40 + 80 100 × 39 40 = 78 93

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