JEE Advanced 2010

LEVELJEE Main

Animated Solution for Mathematics - Probability: A signal which can be green or red with probability $\frac{4}{5}$ and $\frac{1}{5}$ respectively, is received by station A and then transmitted to station B. The probability of each station receiving the signal correctly is $\frac{3}{4}$ . If the signal received at station B is green, then the probability that the original signal was green is

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Progressive Hints

Hint 1

The hidden trap in this problem is the 'Double Error' path. In a binary signal system, an error is essentially a bit-flip. If the signal is Green, it can reach Station B as Green in two distinct ways: either both stations transmit it correctly, or both stations make an error (flipping it to Red and then back to Green). Do not ignore the second path!

Hint 2

You are dealing with conditional probability, which points directly to Bayes' Theorem. You need to calculate

P (G ∣ B_{G}) = \frac{P ( B _{G} ∣ G ) P ( G )}{P ( B _{G} ∣ G ) P ( G ) + P ( B _{G} ∣ R ) P ( R )}

. Your goal is to find the total probability of receiving Green at Station B, which requires summing the probabilities of all paths that lead to a Green signal at the end.

Hint 3

Start by calculating the conditional probabilities for the two scenarios:

P (B_{G} ∣ G) = (\frac{3}{4} \times \frac{3}{4}) + (\frac{1}{4} \times \frac{1}{4})

and

P (B_{G} ∣ R) = (\frac{3}{4} \times \frac{1}{4}) + (\frac{1}{4} \times \frac{3}{4})

. Once you have these values, substitute them into the Bayes' formula to find the final probability.

Visualized Solution

The Transmission Setup

A signal is transmitted from Source to Station A, then to Station B.
The signal can be either Green ( $G$ ) or Red ( $R$ ).

Initial Probabilities

Probability of original signal being Green: $P (G) = \frac{4}{5}$
Probability of original signal being Red: $P (R) = \frac{1}{5}$

Transmission Accuracy

Probability of correct reception: $P (C) = \frac{3}{4}$
Probability of incorrect reception: $P (I) = 1 - \frac{3}{4} = \frac{1}{4}$

The Observed Event

Observed Event ( $B_{G}$ ): Station B receives a Green signal.
Goal: Find the probability that the original signal was Green, i.e., $P (G ∣ B_{G})$ .

Case 1: Original Green $\to$ B Receives Green

Assume the original signal was Green.
Path 1: Both stations receive correctly.
$G Correct G Correct G$
Probability = $\frac{3}{4} \times \frac{3}{4}$

Case 1: The Double Error Path

Path 2: Both stations receive incorrectly (Double Error).
$G Incorrect R Incorrect G$
Probability = $\frac{1}{4} \times \frac{1}{4}$

Total Probability for Case 1

$P (B_{G} ∣ G) = (\frac{3}{4} \times \frac{3}{4}) + (\frac{1}{4} \times \frac{1}{4})$
$P (B_{G} ∣ G) = \frac{9}{16} + \frac{1}{16} = \frac{10}{16} = \frac{5}{8}$

Case 2: Original Red $\to$ B Receives Green

Assume the original signal was Red.
Path 1: Station A is correct, Station B is incorrect.
$R Correct R Incorrect G$
Probability = $\frac{3}{4} \times \frac{1}{4}$

Case 2: The Alternate Error Path

Path 2: Station A is incorrect, Station B is correct.
$R Incorrect G Correct G$
Probability = $\frac{1}{4} \times \frac{3}{4}$

Total Probability for Case 2

$P (B_{G} ∣ R) = (\frac{3}{4} \times \frac{1}{4}) + (\frac{1}{4} \times \frac{3}{4})$
$P (B_{G} ∣ R) = \frac{3}{16} + \frac{3}{16} = \frac{6}{16} = \frac{3}{8}$

Applying Bayes' Theorem

We need $P (G ∣ B_{G})$ .
$P (G ∣ B_{G}) = \frac{P ( B _{G} ∣ G ) P ( G )}{P ( B _{G} ∣ G ) P ( G ) + P ( B _{G} ∣ R ) P ( R )}$

Substituting the Values

$P (G) = \frac{4}{5}$ , $P (B_{G} ∣ G) = \frac{5}{8}$
$P (R) = \frac{1}{5}$ , $P (B_{G} ∣ R) = \frac{3}{8}$
$P (G ∣ B_{G}) = \frac{\frac{5}{8} \times \frac{4}{5}}{( \frac{5}{8} \times \frac{4}{5} ) + ( \frac{3}{8} \times \frac{1}{5} )}$

Final Calculation

Numerator: $\frac{20}{40}$
Denominator: $\frac{20}{40} + \frac{3}{40} = \frac{23}{40}$
$P (G ∣ B_{G}) = \frac{\frac{20}{40}}{\frac{23}{40}}$

The Final Answer

$P (G ∣ B_{G}) = \frac{20}{23}$
The probability that the original signal was Green is $\frac{20}{23}$ .

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The Sigma Insight: Bayes' Theorem

Solution Diagram

The Signal in the Noise

A Journey Through Bayes' Theorem

Imagine you are an engineer working on a high-stakes communication relay. You have a source, a station A, and a station B. The signal is simple: Green or Red.

Every time a signal passes through a station, there is a chance it gets corrupted. This is the fundamental challenge of information theory. We will solve this using the elegance of Bayes' Theorem.

Phase 1

The Anatomy of the Signal

We start with the source. The probability of the original signal being Green is

P (G) = \frac{4}{5}

, and the probability of it being Red is

P (R) = \frac{1}{5}

Each station has a transmission accuracy of

P (C) = \frac{3}{4}

. This implies an error probability of

P (I) = 1 - \frac{3}{4} = \frac{1}{4}

We are told that Station B received a Green signal, denoted as

B_{G}

. We need to find the probability that the source actually sent Green, given that B received Green, which is

P (G ∣ B_{G})

Phase 2

The Hidden Paths

To find

P (B_{G} ∣ G)

, we must consider all ways a Green signal can arrive at B as Green.

Path 1 (The 'Perfect' path): Source sends Green, A receives Green, B receives Green. The probability is:

\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}

Path 2 (The 'Double Error' path): Source sends Green, A receives Red (error), and B receives Green (error). The probability is:

\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}

Summing these, we get the total probability of B receiving Green, given the source sent Green:

P (B_{G} ∣ G) = \frac{9}{16} + \frac{1}{16} = \frac{10}{16} = \frac{5}{8}

Phase 3

The Counter-Intuitive Case

Now, we calculate

P (B_{G} ∣ R)

, the probability that B receives Green when the source sent Red. This occurs if there is exactly one error in the chain.

Path 1: Source sends Red, A receives Red (correct), but B receives Green (error). The probability is:

\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}

Path 2: Source sends Red, A receives Green (error), and B receives Green (correct). The probability is:

\frac{1}{4} \times \frac{3}{4} = \frac{3}{16}

Summing these gives the conditional probability:

P (B_{G} ∣ R) = \frac{3}{16} + \frac{3}{16} = \frac{6}{16} = \frac{3}{8}

Phase 4

The Final Synthesis

Bayes' Theorem provides the bridge to our final answer:

P (G ∣ B_{G}) = \frac{P ( B _{G} ∣ G ) P ( G )}{P ( B _{G} ∣ G ) P ( G ) + P ( B _{G} ∣ R ) P ( R )}

The numerator is:

(\frac{5}{8}) \times (\frac{4}{5}) = \frac{20}{40}

The denominator is the sum of our two cases:

(\frac{5}{8} \times \frac{4}{5}) + (\frac{3}{8} \times \frac{1}{5}) = \frac{20}{40} + \frac{3}{40} = \frac{23}{40}

Dividing these values, the denominators cancel out. The final probability is: $P (G ∣ B_{G}) = \frac{20}{23}$

Common Traps & FAQs

Wait, why do we count the double error case? If the signal is flipped twice, isn't that just 'wrong'?

That is a fantastic question! In a binary system, an error is a bit-flip. If you flip a Green signal to Red, and then flip that Red signal to Green, you have effectively restored the original signal. It is not 'wrong' in the final output; it is a valid path to the observed result. Ignoring this is the most common way students lose marks in communication theory problems!

Why can't we just use the probability of correct transmission directly?

Because you are looking for the probability of the *cause* given the *effect*. We know the effect (Station B received Green), but we don't know the cause (what the source sent). That is the very definition of Bayes' Theorem. We must weigh the likelihood of the cause by the probability of the observed outcome occurring under that cause.

Why is the denominator the sum of two cases?

The denominator represents the Law of Total Probability. It accounts for every possible way Station B could receive a Green signal. It could have come from an original Green signal, or it could have come from an original Red signal that got flipped. We must sum all these mutually exclusive possibilities to get the 'total' probability of the event occurring.

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Visualized Solution

The Transmission Setup

Initial Probabilities

Transmission Accuracy

The Observed Event

Case 1: Original Green $\to$ B Receives Green

Case 1: The Double Error Path

Total Probability for Case 1

Case 2: Original Red $\to$ B Receives Green

Case 2: The Alternate Error Path

Total Probability for Case 2

Applying Bayes' Theorem

Substituting the Values

Final Calculation

The Final Answer

The Sigma Insight: Bayes' Theorem

The Signal in the Noise

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is draw from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is:

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 :

Comprehension Passage

A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is 1/3, then the correct option(s) with the possible values of $n_{1}$ and $n_{2}$ is(are)

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 Transmission Setup

Initial Probabilities

Transmission Accuracy

The Observed Event

Case 1: Original Green .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } → B Receives Green

Case 1: The Double Error Path

Total Probability for Case 1

Case 2: Original Red .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } → B Receives Green

Case 2: The Alternate Error Path

Total Probability for Case 2

Applying Bayes' Theorem

Substituting the Values

Final Calculation

The Final Answer

The Sigma Insight: Bayes' Theorem

The Signal in the Noise

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is draw from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is:

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 :

Comprehension Passage

Case 1: Original Green $\to$ B Receives Green

Case 2: Original Red $\to$ B Receives Green