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Official PYQ · 2021Set 1 · Q54Conditional Probability and Bayes TheoremNAT2 marks

GATE CS 2021 Set 1 Q54 — Conditional Probability and Bayes Theorem

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A sender (S)(\textsf{S}) transmits a signal, which can be one of the two kinds: HH and LL with probabilities 0.10.1 and 0.90.9 respectively, to a receiver (R)(\textsf{R}).

In the graph below, the weight of edge (u,v)(u,v) is the probability of receiving vv when uu is transmitted, where u,v∈{H,L}u,v \in \{H,L\}. For example, the probability that the received signal is LL given the transmitted signal was HH, is 0.70.7.

If the received signal is H,H, the probability that the transmitted signal was HH (rounded to 22 decimal places) is __________.

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Answer: 0.04 : 0.04

We are given the prior probabilities of the transmitted signals:

  • P(Htx)=0.1P(H_{\text{tx}}) = 0.1
  • P(Ltx)=0.9P(L_{\text{tx}}) = 0.9

From the transition graph, the conditional probabilities of receiving a signal given the transmitted signal are:

  • P(Lrx∣Htx)=0.7  ⟹  P(Hrx∣Htx)=1−0.7=0.3P(L_{\text{rx}} \mid H_{\text{tx}}) = 0.7 \implies P(H_{\text{rx}} \mid H_{\text{tx}}) = 1 - 0.7 = 0.3
  • P(Hrx∣Ltx)=0.8  ⟹  P(Lrx∣Ltx)=1−0.8=0.2P(H_{\text{rx}} \mid L_{\text{tx}}) = 0.8 \implies P(L_{\text{rx}} \mid L_{\text{tx}}) = 1 - 0.8 = 0.2

We need to find the posterior probability that the transmitted signal was HH given that the received signal is HH, i.e., P(Htx∣Hrx)P(H_{\text{tx}} \mid H_{\text{rx}}). Applying Bayes' theorem:

P(Htx∣Hrx)=P(Hrx∣Htx)⋅P(Htx)P(Hrx∣Htx)⋅P(Htx)+P(Hrx∣Ltx)⋅P(Ltx)P(H_{\text{tx}} \mid H_{\text{rx}}) = \frac{P(H_{\text{rx}} \mid H_{\text{tx}}) \cdot P(H_{\text{tx}})}{P(H_{\text{rx}} \mid H_{\text{tx}}) \cdot P(H_{\text{tx}}) + P(H_{\text{rx}} \mid L_{\text{tx}}) \cdot P(L_{\text{tx}})}

Substituting the values into the formula:

P(Htx∣Hrx)=0.3×0.1(0.3×0.1)+(0.8×0.9)=0.030.03+0.72=0.030.75=0.04P(H_{\text{tx}} \mid H_{\text{rx}}) = \frac{0.3 \times 0.1}{(0.3 \times 0.1) + (0.8 \times 0.9)} = \frac{0.03}{0.03 + 0.72} = \frac{0.03}{0.75} = 0.04

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