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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Mathematics
Chapter
Probability

JEE Advanced 2025 · Paper 2 · Mathematics · Probability

A factory has a total of three manufacturing units, M_1, M_2 , and...

A factory has a total of three manufacturing units, M1,M2M_1, M_2, and M3M_3, which produce bulbs independent of each other. The units M1,M2M_1, M_2, and M3M_3 produce bulbs in the proportions of 2:2:1, respectively. It is known that 20%20 \% of the bulbs produced in the factory are defective. It is also known that, of all the bulbs produced by M1,15%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 M2M_2 is 25\frac{2}{5}. If a bulb is chosen randomly from the bulbs produced by M3M_3, then the probability that it is defective is ____.

Correct answer

0

Step-by-step solution

Now given probability P( Produced by M2 defective )=40100×14−x4020100=25⇒14−x40=15⇒14−x=8⇒x=6\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,therequiredprobability=the required probability =\frac{6}{20}=$0.3
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