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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 __________.

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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’,

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

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

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

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nice explanation
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5 votes

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

Answer:

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