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 computers produced in the factory turn out to be defective. It is known that $P(\text{computer turns out to be defective given that it is produced in plant } T_1) = 10P(\text{computer turns out to be defective given that it is produced in plant } T_2)$, where $P(E)$ denotes the probability of an event $E$. 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 Options: (A) 36/73 (B) 47/79 (C) 78/93 (D) 75/83

### Analyzing the Factory Setup The factory is divided into two plants, $T_1$ and $T_2$. The production distribution is given as: $P(T_1) = 0.2$ $P(T_2) = 0.8$ These values represent the foundation of our probability tree, tracing the origin of every computer produced. ### The Defect Mystery Let $D$ represent the event that a computer is defective. We are given that the defect rate at $T_1$ is ten times that of $T_2$. Let $P(D|T_2) = x$. Consequently, the defect rate at $T_1$ is: $P(D|T_1) = 10x$ The overall defect rate of the factory is $P(D) = 0.07$. By the **Law of Total Probability**, we have: $$P(D) = P(T_1)P(D|T_1) + P(T_2)P(D|T_2)$$ Substituting the known values into the equation: $$0.07 = (0.2)(10x) + (0.8)(x)$$ $$0.07 = 2x + 0.8x$$ $$0.07 = 2.8x$$ Solving for $x$: $$x = \frac{0.07}{2.8} = 0.025$$ Thus, the defect rate for plant $T_2$ is $2.5\%$, and for plant $T_1$, it is $25\%$. ### The Non-Defective Path We are interested in the event $D^c$, representing a non-defective computer. The total probability of a non-defective computer is: $$P(D^c) = 1 - P(D) = 1 - 0.07 = 0.93$$ For plant $T_2$, the non-defective rate is: $$P(D^c|T_2) = 1 - P(D|T_2) = 1 - 0.025 = 0.975$$ ### The Final Reveal: Bayes' Theorem We seek the probability that a non-defective computer originated from plant $T_2$. Applying **Bayes' Theorem**: $$P(T_2 | D^c) = \frac{P(D^c | T_2)P(T_2)}{P(D^c)}$$ Substituting our calculated values: $$P(T_2 | D^c) = \frac{0.975 \times 0.8}{0.93}$$ Calculating the numerator: $$0.975 \times 0.8 = 0.78$$ The final probability is: $$P(T_2 | D^c) = \frac{0.78}{0.93} = \frac{78}{93} = \frac{26}{31}$$ The final probability that the non-defective computer came from plant $T_2$ is **$\frac{26}{31}$**.

JEE Advanced 2016

LEVELJEE Main

Animated Solution for Mathematics - Probability: 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 computers produced in the factory turn out to be defective. It is known that $P (computer turns out to be defective given that it is produced in plant T_{1}) = 10 P (computer turns out to be defective given that it is produced in plant T_{2})$ , where $P (E)$ denotes the probability of an event $E$ . 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

Select Answer:

Visualized Solution

The Factory Setup

Let $T_{1}$ and $T_{2}$ be the events that a computer is produced in plant 1 and plant 2 respectively.
The factory splits its production between these two plants.

Production Probabilities

$P (T_{1}) = 20% = 0.2$
$P (T_{2}) = 80% = 0.8$

The Defective Reality

Let $D$ be the event that a computer is defective.
Both plants produce some defective computers.

Defective Rates Relationship

Given: $P (D ∣ T_{1}) = 10 \times P (D ∣ T_{2})$
Let $P (D ∣ T_{2}) = x$
Then $P (D ∣ T_{1}) = 10 x$

Law of Total Probability

Overall defective rate is $7%$ .
$P (D) = 0.07$
By Law of Total Probability:
$P (D) = P (T_{1}) P (D ∣ T_{1}) + P (T_{2}) P (D ∣ T_{2})$

Substituting Known Values

$0.07 = (0.2) (10 x) + (0.8) (x)$

Simplifying the Equation

$0.07 = 2 x + 0.8 x$
$0.07 = 2.8 x$

Solving for $x$

$x = \frac{0.07}{2.8}$
$x = \frac{7}{280} = \frac{1}{40} = 0.025$
So, $P (D ∣ T_{2}) = 0.025$

The Non-Defective Computers

The selected computer is not defective ( $D^{c}$ ).
We need to focus on the non-defective branches.

Probabilities of Non-Defective

Overall non-defective: $P (D^{c}) = 1 - P (D) = 1 - 0.07 = 0.93$
Non-defective from $T_{2}$ : $P (D^{c} ∣ T_{2}) = 1 - P (D ∣ T_{2}) = 1 - 0.025 = 0.975$

Bayes' Theorem Setup

We need to find the probability that it came from $T_{2}$ , given it is not defective.
Find $P (T_{2} ∣ D^{c})$ .
By Bayes' Theorem:
$P (T_{2} ∣ D^{c}) = \frac{P ( D ^{c} ∣ T _{2} ) P ( T _{2} )}{P ( D ^{c} )}$

Substituting into Bayes' Theorem

$P (T_{2} ∣ D^{c}) = \frac{0.975 \times 0.8}{0.93}$

Calculating the Numerator

Numerator: $0.975 \times 0.8 = 0.78$
$P (T_{2} ∣ D^{c}) = \frac{0.78}{0.93}$

Final Probability

Multiply numerator and denominator by 100:
$P (T_{2} ∣ D^{c}) = \frac{78}{93}$
Correct Option: $\frac{78}{93}$

00:00 / 00:00

The Sigma Insight: Bayes' Theorem

Solution Diagram

Analyzing the Factory Setup

The factory is divided into two plants,

T_{1}

and

T_{2}

. The production distribution is given as:

P (T_{1}) = 0.2

P (T_{2}) = 0.8

These values represent the foundation of our probability tree, tracing the origin of every computer produced.

The Defect Mystery

Let

D

represent the event that a computer is defective. We are given that the defect rate at

T_{1}

is ten times that of

T_{2}

Let

P (D ∣ T_{2}) = x

. Consequently, the defect rate at

T_{1}

is:

P (D ∣ T_{1}) = 10 x

The overall defect rate of the factory is

P (D) = 0.07

. By the Law of Total Probability, we have:

P (D) = P (T_{1}) P (D ∣ T_{1}) + P (T_{2}) P (D ∣ T_{2})

Substituting the known values into the equation:

0.07 = (0.2) (10 x) + (0.8) (x)

0.07 = 2 x + 0.8 x

0.07 = 2.8 x

Solving for

x

x = \frac{0.07}{2.8} = 0.025

Thus, the defect rate for plant

T_{2}

2.5%

, and for plant

T_{1}

, it is

25%

The Non-Defective Path

We are interested in the event

D^{c}

, representing a non-defective computer. The total probability of a non-defective computer is:

P (D^{c}) = 1 - P (D) = 1 - 0.07 = 0.93

For plant

T_{2}

, the non-defective rate is:

P (D^{c} ∣ T_{2}) = 1 - P (D ∣ T_{2}) = 1 - 0.025 = 0.975

The Final Reveal

Bayes' Theorem

We seek the probability that a non-defective computer originated from plant

T_{2}

. Applying Bayes' Theorem:

P (T_{2} ∣ D^{c}) = \frac{P ( D ^{c} ∣ T _{2} ) P ( T _{2} )}{P ( D ^{c} )}

Substituting our calculated values:

P (T_{2} ∣ D^{c}) = \frac{0.975 \times 0.8}{0.93}

Calculating the numerator:

0.975 \times 0.8 = 0.78

The final probability is:

P (T_{2} ∣ D^{c}) = \frac{0.78}{0.93} = \frac{78}{93} = \frac{26}{31}

The final probability that the non-defective computer came from plant

T_{2}

is $\frac{26}{31}$ .

Common Traps & FAQs

Why do we use

1 - P (D ∣ T_{i})

? Can't we just use

P (D ∣ T_{i})

directly?

Great question! The problem explicitly states that the selected computer is NOT defective. Therefore, we are working in the space of non-defective computers (

D^{c}

). We must convert our defective rates into non-defective rates using the complement rule:

P (D^{c} ∣ T_{i}) = 1 - P (D ∣ T_{i})

Why is the denominator

0.93

in the final step?

That is the total probability of selecting a non-defective computer,

P (D^{c})

. Since

7%

of all computers are defective,

100% - 7% = 93%

must be non-defective. This is the total probability space we are conditioning upon.

Is it possible to solve this without Bayes' Theorem?

You could use a frequency table approach (assuming a total of 1000 computers), but Bayes' Theorem is the most elegant and standard way to handle conditional probability reversals in JEE problems. It is a vital tool to master.

Select Answer:

Visualized Solution

The Factory Setup

Production Probabilities

The Defective Reality

Defective Rates Relationship

Law of Total Probability

Substituting Known Values

Simplifying the Equation

Solving for $x$

The Non-Defective Computers

Probabilities of Non-Defective

Bayes' Theorem Setup

Substituting into Bayes' Theorem

Calculating the Numerator

Final Probability

The Sigma Insight: Bayes' Theorem

Analyzing the Factory Setup

The Defect Mystery

The Non-Defective Path

The Final Reveal

Similar Questions

Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn in white, is :

Bag $B_{1}$ contains 6 white and 4 blue balls, Bag $B_{2}$ contains 4 white and 6 blue balls, and Bag $B_{3}$ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from Bag $B_{2}$ is:

Select Answer:

Visualized Solution

The Factory Setup

Production Probabilities

The Defective Reality

Defective Rates Relationship

Law of Total Probability

Substituting Known Values

Simplifying the Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

The Non-Defective Computers

Probabilities of Non-Defective

Bayes' Theorem Setup

Substituting into Bayes' Theorem

Calculating the Numerator

Final Probability

The Sigma Insight: Bayes' Theorem

Analyzing the Factory Setup

The Defect Mystery

The Non-Defective Path

The Final Reveal

Similar Questions

Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn in white, is :

Solving for $x$