A factory has a total of three manufacturing units, M 1 , M 2 , and M 3 , which produce bulbs independent of each other. The units M 1 , M 2 , and M 3 produce bulbs in the proportions of 2

Question

A factory has a total of three manufacturing units, $M_{1}, M_{2}$ , and $M_{3}$ , which produce bulbs independent of each other. The units $M_{1}, M_{2}$ , and $M_{3}$ produce bulbs in the proportions of 2:2:1, respectively. It is known that $20%$ of the bulbs produced in the factory are defective. It is also known that, of all the bulbs produced by $M_{1}, 15%$ are defective. Suppose that, if a randomly chosen bulb produced in the factory is found to be defective, the probability that it was produced by $M_{2}$ is $\frac{2}{5}$ . If a bulb is chosen randomly from the bulbs produced by $M_{3}$ , then the probability that it is defective is ____.

Answer:

0.30

Solution:

Now given probability $P (\frac{Produced by M _{2}}{defective}) = \frac{\frac{40}{100} \times \frac{14 - x}{40}}{\frac{20}{100}} = \frac{2}{5} \Rightarrow \frac{14 - x}{40} = \frac{1}{5} \Rightarrow 14 - x = 8 \Rightarrow x = 6$ So, the required probability $= \frac{6}{20} = 0.3$

TestHub · Accepted Answer

Now given probability ​ P ( defective Produced by M 2 ​ ​ ) = 100 20 ​ 100 40 ​ × 40 14 − x ​ ​ = 5 2 ​ ⇒ 40 14 − x ​ = 5 1 ​ ⇒ 14 − x = 8 ⇒ x = 6 ​ So, the required probability = 20 6 ​ = 0.3

Mathematics - Probability Question with Solution | TestHub

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