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MathMedAdv 2025 • Paper 2

Defective Bulbs Probability in Manufacturing Units

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 2:2:12 : 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 M1M_1, 15%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 __________.

Official Numerical Answer0.27 to 0.33

Topics & Concepts

Probability (Advanced)ProbabilityBayes' TheoremConditional Probability

Step-by-Step Solution

Let M1,M2M_1, M_2, and M3M_3 denote the events that a randomly chosen bulb was manufactured by unit M1,M2M_1, M_2, and M3M_3, respectively. Let DD denote the event that a randomly chosen bulb is defective.

From the given proportions of production 2:2:12 : 2 : 1, the probabilities of selecting a bulb from each unit are: P(M1)=22+2+1=25P(M_1) = \frac{2}{2 + 2 + 1} = \frac{2}{5} P(M2)=22+2+1=25P(M_2) = \frac{2}{2 + 2 + 1} = \frac{2}{5} P(M3)=12+2+1=15P(M_3) = \frac{1}{2 + 2 + 1} = \frac{1}{5}

We are given the overall probability of a defective bulb in the factory: P(D)=20%=15=0.20P(D) = 20\% = \frac{1}{5} = 0.20

We are also given the conditional probability of a defective bulb given it is from M1M_1: P(D∣M1)=15%=15100=320P(D \mid M_1) = 15\% = \frac{15}{100} = \frac{3}{20}

It is given that if a bulb chosen at random is defective, the probability that it was produced by M2M_2 is 25\frac{2}{5}: P(M2∣D)=25P(M_2 \mid D) = \frac{2}{5}

Using Bayes' Theorem: P(M2∣D)=P(D∣M2)⋅P(M2)P(D)P(M_2 \mid D) = \frac{P(D \mid M_2) \cdot P(M_2)}{P(D)}

Substitute the known values into the equation: 25=P(D∣M2)⋅2515\frac{2}{5} = \frac{P(D \mid M_2) \cdot \frac{2}{5}}{\frac{1}{5}}

Simplifying the right-hand side: 25=2⋅P(D∣M2)  ⟹  P(D∣M2)=15=0.20\frac{2}{5} = 2 \cdot P(D \mid M_2) \implies P(D \mid M_2) = \frac{1}{5} = 0.20

Now, applying the Law of Total Probability for P(D)P(D): P(D)=P(D∣M1)⋅P(M1)+P(D∣M2)⋅P(M2)+P(D∣M3)⋅P(M3)P(D) = P(D \mid M_1) \cdot P(M_1) + P(D \mid M_2) \cdot P(M_2) + P(D \mid M_3) \cdot P(M_3)

Substituting the known probabilities: 15=(320)⋅(25)+(15)⋅(25)+P(D∣M3)⋅(15)\frac{1}{5} = \left(\frac{3}{20}\right) \cdot \left(\frac{2}{5}\right) + \left(\frac{1}{5}\right) \cdot \left(\frac{2}{5}\right) + P(D \mid M_3) \cdot \left(\frac{1}{5}\right)

15=6100+225+15P(D∣M3)\frac{1}{5} = \frac{6}{100} + \frac{2}{25} + \frac{1}{5} P(D \mid M_3)

Convert all fractions on the right side to a common denominator of 100100: 20100=6100+8100+15P(D∣M3)\frac{20}{100} = \frac{6}{100} + \frac{8}{100} + \frac{1}{5} P(D \mid M_3)

20100=14100+15P(D∣M3)\frac{20}{100} = \frac{14}{100} + \frac{1}{5} P(D \mid M_3)

Subtracting 14100\frac{14}{100} from both sides: 6100=15P(D∣M3)\frac{6}{100} = \frac{1}{5} P(D \mid M_3)

Multiply both sides by 55: P(D∣M3)=30100=0.3P(D \mid M_3) = \frac{30}{100} = 0.3

Thus, if a bulb is chosen randomly from the bulbs produced by M3M_3, the probability that it is defective is 0.30.3 (or 0.300.30).

Defective Bulbs Probability in Manufacturing Units | Mathematics PYQ Solution - JEE Challenger