Mathematics · Probability

JEE Advanced 2025 — Paper 2 — Question 27

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 \qquad

Answer: 0.3

Numerical answer — enter this value.

Step-by-step solution

figure

Now given probability P( Produced by M2 defective )=40100×14−x4020100=25\mathrm{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}

⇒14−x40=15\Rightarrow \frac{14-x}{40}=\frac{1}{5}

⇒14−x=8\Rightarrow 14-x=8

⇒x=6\Rightarrow x=6

So, the required probability

=620=0.3=\frac{6}{20}=0.3

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2025
Paper
Paper 2
Subject
Mathematics
Chapter
Probability
Topic
Conditional Probability and Multiplication Theorem