Question

Consider Guwahati (G) and Delhi (D) whose temperatures can be classified as high (HH), medium (MM) and low (LL). Let P(HG)P(H_G) denote the probability that Guwahati has high temperature. Similarly, P(MG)P(M_G) and P(LG)P(L_G) denotes the probability of Guwahati having medium and low temperatures respectively. Similarly, we use P(HD)P(H_D), P(MD)P(M_D) and P(LD)P(L_D) for Delhi.

The following table gives the conditional probabilities for Delhi’s temperature given Guwahati’s temperature.

Guwahati \ DelhiHDH_DMDM_DLDL_D
HGH_G0.400.480.12
MGM_G0.100.650.25
LGL_G0.010.500.49

Consider the first row in the table above. The first entry denotes that if Guwahati has high temperature (HGH_G) then the probability of Delhi also having a high temperature (HDH_D) is 0.40; i.e., P(HD∣HG)=0.40P(H_D|H_G) = 0.40. Similarly, the next two entries are P(MD∣HG)=0.48P(M_D|H_G) = 0.48 and P(LD∣HG)=0.12P(L_D|H_G) = 0.12. Similarly for the other rows.

If it is known that P(HG)=0.2P(H_G) = 0.2, P(MG)=0.5P(M_G) = 0.5, and P(LG)=0.3P(L_G) = 0.3, then the probability (correct to two decimal places) that Guwahati has high temperature given that Delhi has high temperature is _______.

Official answer and solution
Accepted answers
0.6 to 0.62
From
Official GATE 2018 answer key

Solution

Apply the Law of Total Probability to find P(HD)P(H_D) and Bayes' Theorem to find P(HG∣HD)P(H_G | H_D).

Given priors:

P(HG)=0.2,P(MG)=0.5,P(LG)=0.3P(H_G) = 0.2, \quad P(M_G) = 0.5, \quad P(L_G) = 0.3

Conditional probabilities for Delhi high temperature from table:

P(HD∣HG)=0.40,P(HD∣MG)=0.10,P(HD∣LG)=0.01P(H_D \mid H_G) = 0.40, \quad P(H_D \mid M_G) = 0.10, \quad P(H_D \mid L_G) = 0.01

Calculate the total probability of Delhi having high temperature:

P(HD)=P(HD∣HG)P(HG)+P(HD∣MG)P(MG)+P(HD∣LG)P(LG)P(H_D) = P(H_D \mid H_G)P(H_G) + P(H_D \mid M_G)P(M_G) + P(H_D \mid L_G)P(L_G)
P(HD)=(0.40)(0.2)+(0.10)(0.5)+(0.01)(0.3)=0.08+0.05+0.003=0.133P(H_D) = (0.40)(0.2) + (0.10)(0.5) + (0.01)(0.3) = 0.08 + 0.05 + 0.003 = 0.133

By Bayes' Theorem, the posterior probability is:

P(HG∣HD)=P(HD∣HG)P(HG)P(HD)=0.080.133=80133≈0.6015P(H_G \mid H_D) = \frac{P(H_D \mid H_G)P(H_G)}{P(H_D)} = \frac{0.08}{0.133} = \frac{80}{133} \approx 0.6015

Rounded to two decimal places, the probability is 0.60.

Final answer

0.60

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

Source

Paper
GATE 2018 Computer Science and Information Technology (CS)
Question
54 of 65, printed as CS Q.44
Section
Computer Science and Information Technology

Report an error in this question

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.