JEE Advanced · Mathematics · 32. Probability
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 ____.
- A 0.4
- B 0.5
- C 0.3
- D 0.2
Answer & Solution
Correct Answer
(C) 0.3
Step-by-step Solution
Detailed explanation
Now given probability
\(\begin{aligned}
& P\left(\frac{\text { Produced by } \mathrm{M}_2}{\text { defective }}\right)=\frac{\frac{40}{100} \times \frac{14-\mathrm{x}}{40}}{\frac{20}{100}}=\frac{2}{5} \\
& \Rightarrow \frac{14-\mathrm{x}}{40}=\frac{1}{5} \\
& \Rightarrow 14-\mathrm{x}=8 \\
& \Rightarrow \mathrm{x}=6
\end{aligned}\)
So, the required probability
\(=\frac{6}{20}=0.3\)
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