JEE Challenger
More from Probability (Advanced)
MathMedAdv 2016 • Paper 1

Probability of non defective computer originating from plant T2

A computer producing factory has only two plants T1T_1 and T2T_2. Plant T1T_1 produces 20%20\% and plant T2T_2 produces 80%80\% of the total computers produced. 7%7\% of 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 T1)P(\text{computer turns out to be defective given that it is produced in plant } T_1) =10P(computer turns out to be defective given that it is produced in plant T2)= 10 P(\text{computer turns out to be defective given that it is produced in plant } T_2) where P(E)P(E) denotes the probability of an event EE. 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 T2T_2 is

Options

A

3673\frac{36}{73}

B

4779\frac{47}{79}

C

7893\frac{78}{93}

Correct
D

7583\frac{75}{83}

Topics & Concepts

Probability (Advanced)ProbabilityConditional ProbabilityBayes' Theorem

Step-by-Step Solution

To find the probability that a non-defective computer was produced in plant T2T_2, we define the following events:

  • Let T1T_1 be the event that a computer is produced in plant T1T_1.
  • Let T2T_2 be the event that a computer is produced in plant T2T_2.
  • Let DD be the event that a computer turns out to be defective.
  • Let DcD^c be the event that a computer does not turn out to be defective (i.e., non-defective).

From the problem statement, we are given the following probabilities: P(T1)=20%=0.20=15P(T_1) = 20\% = 0.20 = \frac{1}{5} P(T2)=80%=0.80=45P(T_2) = 80\% = 0.80 = \frac{4}{5} P(D)=7%=0.07=7100P(D) = 7\% = 0.07 = \frac{7}{100}

We are also given the conditional relationship: P(D∣T1)=10P(D∣T2)P(D | T_1) = 10 P(D | T_2)

Let P(D∣T2)=xP(D | T_2) = x. Then, P(D∣T1)=10xP(D | T_1) = 10x.

Using the Law of Total Probability for the event DD: P(D)=P(T1)⋅P(D∣T1)+P(T2)⋅P(D∣T2)P(D) = P(T_1) \cdot P(D | T_1) + P(T_2) \cdot P(D | T_2)

Substitute the given values into the equation: 0.07=(0.20)(10x)+(0.80)(x)0.07 = (0.20)(10x) + (0.80)(x) 0.07=2x+0.8x0.07 = 2x + 0.8x 0.07=2.8x0.07 = 2.8x x=0.072.8=7280=140x = \frac{0.07}{2.8} = \frac{7}{280} = \frac{1}{40}

Thus, we have: P(D∣T2)=140P(D | T_2) = \frac{1}{40}

The probability that a computer is non-defective given that it was produced in plant T2T_2 is: P(Dc∣T2)=1−P(D∣T2)=1−140=3940P(D^c | T_2) = 1 - P(D | T_2) = 1 - \frac{1}{40} = \frac{39}{40}

The overall probability that a computer is non-defective is: P(Dc)=1−P(D)=1−0.07=0.93=93100P(D^c) = 1 - P(D) = 1 - 0.07 = 0.93 = \frac{93}{100}

Now, we need to find P(T2∣Dc)P(T_2 | D^c), which is the probability that a randomly selected computer was produced in plant T2T_2 given that it is non-defective. Using Bayes' Theorem: P(T2∣Dc)=P(T2)⋅P(Dc∣T2)P(Dc)P(T_2 | D^c) = \frac{P(T_2) \cdot P(D^c | T_2)}{P(D^c)}

Substitute the known values into Bayes' formula: P(T2∣Dc)=45×394093100P(T_2 | D^c) = \frac{\frac{4}{5} \times \frac{39}{40}}{\frac{93}{100}}

Simplify the numerator: 45×3940=3950=78100\frac{4}{5} \times \frac{39}{40} = \frac{39}{50} = \frac{78}{100}

So, P(T2∣Dc)=7810093100=7893P(T_2 | D^c) = \frac{\frac{78}{100}}{\frac{93}{100}} = \frac{78}{93}

Hence, the correct option is (C).

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