GATE CSE 2021 Set 1 | Question: 54

Question

A sender $(\textsf{S})$ 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 $(\textsf{R})$.

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 $0.7$.

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

Comments

we have to find the probability of transmitted signal was H when given that received signal is H

we can do like this ->total no. of favourable outcomes/total possible outcomes

in this case the favourable outcomes is when transmitted signal is H and received signal also H=>0.1*0.3

total possible way of receiving signal H  =>

when transmitted signal is L and received signal H+when transmitted signal H and received signal H=>0.9*0.8+0.1*0.3

so 0.1*0.3/0.9*0.8+0.1*0.3 so by solving this we will get 0.04 as the answer.

Answer 1

 

Given that:

$P(H_s) = 0.1, P(L_s)=0.9$

From the diagram we get, 

$P(H_r\mid H_s) = 0.3, \quad P(L_r\mid H_s) = 0.7,$

$P(H_r\mid L_s) = 0.8, \quad P(L_r\mid L_s) = 0.2$

We have to find: $P(H_s\mid H_r)$

By Bayes  theorem,

Probability of sending signal ‘H’ given that signal received is ‘H’,

$P(H_s\mid H_r) = \dfrac{P(H_s \cap H_r)}{P(H_r)}$

$\qquad  = \dfrac{P(H_r \mid H_s).P(H_s)}{P(H_r \mid H_s).P(H_s)+ P(H_r \mid L_s).P(L_s)}$

$\qquad  = \dfrac{0.3 \times 0.1}{0.3 \times 0.1+ 0.8\times 0.9} = 0.04$

Comments

nice explanation

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• @Akash 15  @Hira ThakurPlease explain the 2nd part of bays theorem denominator

It should be $P(Hr)* P(Hs/Hr)$ why it is written as

$P(Ls)*P(Hr/Ls)$ ❓

Please explain how here total probability is calculated 

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@Ray Tomlinson It is correct.
Here we considered all possibilities of the transmitted signals.
For $H_r$ there are $2$ cases signal transmitting from $H$ to $H$ and $L$ to $H$

So, total probability $=P(H_r)=P(H_r|H_s)P(H_s)+P(H_r|L_s)P(L_s)$
Carefully see the tree diagram we have to consider all cases for reciever side.

Answer 2

In question find Receiver H signal to send by H sender  P(Hs/Hr) probability of receiving signal by H to sender H

Probability sender to send signal Hs =0.1 and Ls = 0.9

Send signal by Hs to Hr = prob of send signal by H * probability of sender H to receive H signal

 = P(Hr/Hs) = 0.1*0.3

 receiver H to receive total signal = send signal by H + send signal by L 

Send signal by L = 0.9*0.8

Total signal received by H = 0.1*0.3+ 0.9*0.8

 

Probability of receiving signal by H is sending by H p(Hs/Hr) = P(Hs) *P(Hr/Hs) /P(Hs) *P(Hr/Hs) +P(Ls) *P(Hr/Ls)    Bayes theorem

P(Hs/Hr) = 0.1*0.3/0.1*0.3+0.9*0.8

P(Hs/Hr) = 0.04

Ans will be 0.04