Official PYQ · 2018Q44Conditional Probability and Bayes TheoremNAT2 marks

GATE CS 2018 Q44 — Conditional Probability and Bayes Theorem

Consider Guwahati,

(G)

and Delhi

(D)

whose temperatures can be classified as high

(H)

, medium

(M)

and low

(L)

. Let

P(H_G)

denote the probability that Guwahati has high temperature. Similarly,

P(M_G)

and

P(L_G)

denotes the probability of Guwahati having medium and low temperatures respectively. Similarly, we use

P(H_D),\: P(M_D)

and

P(L_D)

for Delhi.

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

\begin{array}{|c|c|c|c|} \hline \text{} & H_D & M_D & L_D \\ \hline H_G & \text{0.40}& \text{0.48} & \text{0.12} \\ \hline M_G & \text{0.10}& \text{0.65} & \text{0.25} \\ \hline L_G & \text{0.01}& \text{0.50} & \text{0.49} \\ \hline \end{array}

Consider the first row in the table above. The first entry denotes that if Guwahati has high temperature

(H_G)

then the probability of Delhi also having a high temperature

(H_D)

0.40;

i.e.,

P(H_D \mid H_G) = 0.40

. Similarly, the next two entries are

P(M_D \mid H_G) = 0.48

and

P(L_D \mid H_G) = 0.12

. Similarly for the other rows.

If it is known that

P(H_G) = 0.2, \: P(M_G) = 0.5

, and

P(L_G)=0.3

, then the probability (correct to two decimal places) that Guwahati has high temperature given that Delhi has high temperature is ________.

View worked solution

Answer: 0.60 : 0.61

We need to find the conditional probability that Guwahati has a high temperature given that Delhi has a high temperature, denoted as $P(H_G \mid H_D)$ .

By Bayes' Theorem, we have:

P(H_G \mid H_D) = \frac{P(H_D \mid H_G) \cdot P(H_G)}{P(H_D)}

Step 1: Identify the given probabilities

Prior probabilities for Guwahati:
- $P(H_G) = 0.2$
- $P(M_G) = 0.5$
- $P(L_G) = 0.3$
Conditional probabilities for Delhi having a high temperature ( $H_D$ $H_{D}$ ):
- $P(H_D \mid H_G) = 0.40$
- $P(H_D \mid M_G) = 0.10$
- $P(H_D \mid L_G) = 0.01$

Step 2: Calculate the total probability $P(H_D)$

Using the law of total probability:

\begin{aligned} 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) \\ &= (0.40 \times 0.2) + (0.10 \times 0.5) + (0.01 \times 0.3) \\ &= 0.08 + 0.05 + 0.003 \\ &= 0.133 \end{aligned}

Step 3: Calculate $P(H_G \mid H_D)$

\begin{aligned} P(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 \end{aligned}

Rounding to two decimal places, the probability is 0.60 (or 0.61 depending on intermediate rounding).

Free practice · Your answers on this page are not saved.

Published 10 Oct 2026

Report a correction

Browse this topic

More Conditional Probability and Bayes Theorem questions

GATE CS 2018 Q44 — Conditional Probability and Bayes Theorem

Step 1: Identify the given probabilities

Step 2: Calculate the total probability P(HD)P(H_D)P(HD​)

Step 3: Calculate P(HG∣HD)P(H_G \mid H_D)P(HG​∣HD​)

Related questions

Step 2: Calculate the total probability $P(H_D)$

Step 3: Calculate $P(H_G \mid H_D)$