JEE Advanced 2005

LEVELJEE Main

Animated Solution for Mathematics - Probability: A person goes to office either by car, scooter, bus or train, the probability of which being $1/7, 3/7, 2/7$ and $1/7$ respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is $2/9, 1/9, 4/9$ and $1/9$ respectively. Given that he reached office in time, then what is the probability that he travelled by a car.

Progressive Hints

Hint 1

Visualize the sample space as a total probability rectangle. The problem gives you the probability of being 'Late', but the question asks about being 'In-Time'. Remember that the sum of probabilities for complementary events is always 1. How do you convert

P (L ∣ X)

P (\overset{ˉ}{L} ∣ X)

Hint 2

You are dealing with a classic 'inverse probability' scenario. We know the outcome (he arrived in time), and we want to trace back the cause (the mode of transport). This is the definition of Bayes' Theorem. Your toolkit must include the formula

P (A ∣ B) = \frac{P ( B ∣ A ) P ( A )}{P ( B )}

Hint 3

Start by writing the denominator, which is the total probability of arriving in time:

P (\overset{ˉ}{L}) = P (\overset{ˉ}{L} ∣ C) P (C) + P (\overset{ˉ}{L} ∣ S) P (S) + P (\overset{ˉ}{L} ∣ B) P (B) + P (\overset{ˉ}{L} ∣ T) P (T)

. Once you have this, the numerator is simply the specific branch for the car:

P (\overset{ˉ}{L} ∣ C) P (C)

Visualized Solution

Visualizing the Sample Space

Let $C$ , $S$ , $B$ , and $T$ represent the mutually exclusive and exhaustive events of choosing a Car, Scooter, Bus, or Train respectively.
The total area represents the entire sample space of transport choices.

Defining the Prior Probabilities

Prior probabilities of choosing each mode of transport:
$P (C) = \frac{1}{7}$
$P (S) = \frac{3}{7}$
$P (B) = \frac{2}{7}$
$P (T) = \frac{1}{7}$

Conditional Probabilities of Being Late

Let $L$ be the event that the person reaches office late.
Given conditional probabilities of being late:
$P (L ∣ C) = \frac{2}{9}$
$P (L ∣ S) = \frac{1}{9}$
$P (L ∣ B) = \frac{4}{9}$
$P (L ∣ T) = \frac{1}{9}$

Finding the 'In-Time' Probabilities

Let $\overset{ˉ}{L}$ be the event that the person reaches office in time.
Since $P (\overset{ˉ}{L} ∣ X) = 1 - P (L ∣ X)$ :
$P (\overset{ˉ}{L} ∣ C) = 1 - \frac{2}{9} = \frac{7}{9}$
$P (\overset{ˉ}{L} ∣ S) = 1 - \frac{1}{9} = \frac{8}{9}$
$P (\overset{ˉ}{L} ∣ B) = 1 - \frac{4}{9} = \frac{5}{9}$
$P (\overset{ˉ}{L} ∣ T) = 1 - \frac{1}{9} = \frac{8}{9}$

Applying Bayes' Theorem

We need to find the posterior probability: $P (C ∣ \overset{ˉ}{L})$
By Bayes' Theorem:
$P (C ∣ \overset{ˉ}{L}) = \frac{P ( L ˉ ∣ C ) P ( C )}{P ( L ˉ )}$

Expanding the Total Probability

The denominator $P (\overset{ˉ}{L})$ is the sum of in-time probabilities across all modes:
$P (\overset{ˉ}{L}) = P (\overset{ˉ}{L} ∣ C) P (C) + P (\overset{ˉ}{L} ∣ S) P (S) + P (\overset{ˉ}{L} ∣ B) P (B) + P (\overset{ˉ}{L} ∣ T) P (T)$

Substituting the Values

Substitute the known values into the formula:
Numerator: $\frac{7}{9} \cdot \frac{1}{7}$
Denominator: $(\frac{7}{9} \cdot \frac{1}{7}) + (\frac{8}{9} \cdot \frac{3}{7}) + (\frac{5}{9} \cdot \frac{2}{7}) + (\frac{8}{9} \cdot \frac{1}{7})$

Simplifying the Fractions

Calculate each term with a common denominator of $63$ :
Numerator: $\frac{7}{63}$
Denominator: $\frac{7}{63} + \frac{24}{63} + \frac{10}{63} + \frac{8}{63}$

Calculating the Final Ratio

Sum of denominator terms: $\frac{7 + 24 + 10 + 8}{63} = \frac{49}{63}$
Divide numerator by denominator:
$P (C ∣ \overset{ˉ}{L}) = \frac{\frac{7}{63}}{\frac{49}{63}} = \frac{7}{49}$

Final Answer and Key Takeaway

Simplifying the fraction:
$P (C ∣ \overset{ˉ}{L}) = \frac{1}{7}$
Key Takeaway: The posterior probability remains $\frac{1}{7}$ because the low prior probability balances the high in-time rate.

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The Sigma Insight: Bayes' Theorem

Analyzing the Setup

Let

C

S

B

, and

T

represent the mutually exclusive and exhaustive events of choosing a Car, Scooter, Bus, or Train, respectively. These events form our sample space.

The prior probabilities for each mode of transport are given as:

P (C) = 1/7

P (S) = 3/7

P (B) = 2/7

, and

P (T) = 1/7

These values represent our initial state of knowledge before any evidence is observed.

The 'In-Time' Trap

The problem provides the probability of being late,

P (L ∣ X)

. However, we are interested in the probability of being in time,

P (\overset{ˉ}{L} ∣ X)

Using the complement rule

P (\overset{ˉ}{L} ∣ X) = 1 - P (L ∣ X)

, we calculate the success rates for each mode:

For the car:

P (\overset{ˉ}{L} ∣ C) = 1 - 2/9 = 7/9

For the scooter:

P (\overset{ˉ}{L} ∣ S) = 1 - 1/9 = 8/9

For the bus:

P (\overset{ˉ}{L} ∣ B) = 1 - 4/9 = 5/9

For the train:

P (\overset{ˉ}{L} ∣ T) = 1 - 1/9 = 8/9

Applying Bayes' Theorem

We have observed the evidence that the person arrived in time (

\overset{ˉ}{L}

). We seek the posterior probability that they took the car, denoted as

P (C ∣ \overset{ˉ}{L})

According to Bayes' Theorem:

P (C ∣ \overset{ˉ}{L}) = \frac{P ( L ˉ ∣ C ) P ( C )}{P ( L ˉ )}

The denominator

P (\overset{ˉ}{L})

represents the total probability of arriving in time, calculated using the Law of Total Probability:

P (\overset{ˉ}{L}) = P (\overset{ˉ}{L} ∣ C) P (C) + P (\overset{ˉ}{L} ∣ S) P (S) + P (\overset{ˉ}{L} ∣ B) P (B) + P (\overset{ˉ}{L} ∣ T) P (T)

Final Calculation

Substituting the known values into the numerator and denominator:

The numerator is:

(7/9) \cdot (1/7) = 7/63

The denominator is the sum of all branches:

P (\overset{ˉ}{L}) = (\frac{7}{9} \cdot \frac{1}{7}) + (\frac{8}{9} \cdot \frac{3}{7}) + (\frac{5}{9} \cdot \frac{2}{7}) + (\frac{8}{9} \cdot \frac{1}{7})

P (\overset{ˉ}{L}) = \frac{7 + 24 + 10 + 8}{63} = \frac{49}{63}

Finally, we compute the ratio:

P (C ∣ \overset{ˉ}{L}) = \frac{7/63}{49/63} = \frac{7}{49} = \frac{1}{7}

The final probability that the person took the car, given they arrived in time, is $1/7$ .

Common Traps & FAQs

Wait, why do we subtract the given probabilities from 1? Can't we just use the 'Late' probabilities directly?

Great question! The problem asks for the probability of being *in time*. The given values are for being *late*. Since 'Late' and 'In-time' are mutually exclusive and exhaustive,

P (In-time) = 1 - P (Late)

. You must perform this conversion first, or your Bayes' calculation will be solving for the wrong event! Always check if the event in your conditional probability matches the event in your final question.

Why is the denominator so complicated? Do I really need to sum all those terms?

Yes, you do! The denominator represents the *total* probability of the event 'arriving in time' happening, regardless of the transport mode. Think of it as the 'normalization factor'. Without it, you aren't calculating a probability; you're just calculating a weighted likelihood. It ensures that the sum of all posterior probabilities equals 1. It is the 'universe' of all ways to arrive on time.

Is it a coincidence that the answer is

1/7

, which is the same as the prior probability

P (C)

It is a beautiful result, but not a coincidence! It happens because the 'In-time' rate for the car (

7/9

) is perfectly balanced against the weighted average of the 'In-time' rates of the other modes. It teaches us that sometimes, even when we gain new information, our initial belief remains unchanged if the evidence is perfectly neutral relative to our prior. It is a rare, elegant mathematical symmetry!

Visualized Solution

Visualizing the Sample Space

Defining the Prior Probabilities

Conditional Probabilities of Being Late

Finding the 'In-Time' Probabilities

Applying Bayes' Theorem

Expanding the Total Probability

Substituting the Values

Simplifying the Fractions

Calculating the Final Ratio

Final Answer and Key Takeaway

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

The 'In-Time' Trap

Applying Bayes' Theorem

Final Calculation

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