GATE CS 2021 Set 1 Q.64 (CS Q.54): Probability and Statistics

A sender (S) transmits a signal, which can be one of the two kinds: H and L with probabilities 0.1 and 0.9 respectively, to a receiver (R).
In the graph below, the weight of edge $(u, v)$ is the probability of receiving $v$ when $u$ is transmitted, where $u,v \in \{H, L\}$ . For example, the probability that the received signal is L given the transmitted signal was H, is 0.7.

Binary asymmetric communication channel diagram showing transition probabilities from sender states H and L to receiver states H and L.

If the received signal is H, the probability that the transmitted signal was H (rounded to 2 decimal places) is __________

Your answer

Official answer and solution

Official answer: 0.04
From: Official GATE 2021 answer key

Solution

Apply Bayes' theorem to find the posterior probability of transmitted signal H given received signal H.

Let $T_H$ and $T_L$ denote the events that signal H and L were transmitted respectively. We are given:

P(T_H) = 0.1, \quad P(T_L) = 0.9

From the channel transition graph:

For transmitted signal H: $P(R_L \mid T_H) = 0.7$ , so $P(R_H \mid T_H) = 1 - 0.7 = 0.3$ .
For transmitted signal L: $P(R_L \mid T_L) = 0.2$ , so $P(R_H \mid T_L) = 1 - 0.2 = 0.8$ .

By the law of total probability, the total probability of receiving signal H is:

P(R_H) = P(T_H) P(R_H \mid T_H) + P(T_L) P(R_H \mid T_L)

P(R_H) = (0.1 \times 0.3) + (0.9 \times 0.8) = 0.03 + 0.72 = 0.75

By Bayes' theorem, the probability that the transmitted signal was H given that the received signal is H is:

P(T_H \mid R_H) = \frac{P(T_H \cap R_H)}{P(R_H)} = \frac{0.03}{0.75} = \frac{3}{75} = \frac{1}{25} = 0.04

Final answer

0.04

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2021 Computer Science and Information Technology (CS, Set-1)
Question: 64 of 65, printed as CS Q.54
Section: Computer Science and Information Technology

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