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) denote 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 P(temperature of Delhi | temperature of Guwahati):
H_D M_D L_D
H_G: 0.40 0.48 0.12
M_G: 0.10 0.65 0.25
L_G: 0.01 0.50 0.49
Consider the first entry: P(H_D | H_G) = 0.40. It means if Guwahati has high temperature then the probability that Delhi also has high temperature is 0.40.
If it is known that P(H_G) = 0.2, P(M_G) = 0.5, and P(L_G) = 0.3, the probability (correct to two decimal places) that Delhi has high temperature given that Guwahati has high temperature is __________.
A. 0.15
B. 0.25
C. 0.35
D. 0.45
GATE 2018 · Engineering Mathematics · Conditional Probability · medium
Answer: 0.133 (approx 0.15, answer A)
Apply Law of Total Probability: P(H_D) = 0.40*0.2 + 0.10*0.5 + 0.01*0.3 = 0.080 + 0.050 + 0.003 = 0.133.
Select the closest answer: Among A=0.15, B=0.25, C=0.35, D=0.45, the value 0.133 is closest to A=0.15. Official answer is A.