A computer producing factory has only two plants $T_{1}$ and $T_{2}$ . Plant $T_{1}$ produces 20% and plant $T_{2}$ produces 80% of the total computers produced. 7% of the computers produced in the factory turn out to be defective. It is known that P(computer turns out to be defective given that it is produced in plant $T_{1}$ ) = 10 P(computer turns out to be defective given that it is produced in plant $T_{2}$ ). A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant $T_{2}$ is

Answer (Detailed Solution Below)View answer ▾

Answer: Option C — $\frac{78}{93}$

Detailed Solution

Let $P (D ∣ T_{2}) = x$ , so $P (D ∣ T_{1}) = 10 x$ . Total: $0.2 (10 x) + 0.8 x = 2.8 x = 0.07$ , so $x = 0.025$ .

$P (\hat{D} ∣ T_{1}) = 1 - 0.25 = 0.75$ and $P (\hat{D} ∣ T_{2}) = 1 - 0.025 = 0.975$ .

$P (T_{2} ∣ \hat{D}) = \frac{0.8 \times 0.975}{0.2 \times 0.75 + 0.8 \times 0.975} = \frac{0.78}{0.93} = \frac{78}{93}$ .

✓ Check

P (\hat{D}) = 0.15 + 0.78 = 0.93 = 1 - 0.07

, as it should be.

⚠ Common trap

Using

P (D ∣ T_{i})

instead of

P (\hat{D} ∣ T_{i})

gives

P (T_{2} ∣ D) = \frac{0.02}{0.07} = \frac{2}{7}

, the answer for a defective computer, which is not what is asked.

Quick tip

$0.2 (10 x) + 0.8 x = 0.07 \Rightarrow x = 0.025$ ; $P (T_{2} ∣ \overset{ˉ}{D}) = \frac{0.8 ( 0.975 )}{0.2 ( 0.75 ) + 0.8 ( 0.975 )} = \frac{78}{93}$

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Probability — Preparation

This question tests your understanding of Probability, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

Probability measures how likely an event is, as a number between 0 and 1. It equals the number of favourable outcomes divided by the total number of possible outcomes, when all outcomes are equally likely. This topic covers coins, dice, playing cards, drawing balls from a bag, complementary events, independent events, and conditional probability. The most useful tools are the complement rule (P(at least one) = 1 - P(none)) and the multiplication rule for independent events. For engineering and MCA entrance exams, the topic also includes the addition theorem, conditional probability, Bayes' theorem, and probability distributions. A clear sample space is the key to avoiding mistakes, so always list or count all outcomes carefully. Practise the probability questions below. Each comes with a step-by-step solution that shows how the sample space and favourable outcomes are counted.

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