Consider Guwahati (G) and Delhi (D) whose temperatures can be classified asβ¦
2018
Consider Guwahati (G) and Delhi (D) whose temperatures can be classified as high (π»), medium (π) and low (πΏ). Let π(π»πΊ ) denote the probability that Guwahati has high temperature. Similarly, π(ππΊ ) and π(πΏπΊ ) denotes the probability of Guwahati having medium and low temperatures respectively. Similarly, we use π(π»π·), π(ππ·) and π(πΏπ·) for Delhi.
The following table gives the conditional probabilities for Delhiβs temperature given Guwahatiβs temperature.
Consider the first row in the table above. The first entry denotes that if Guwahati has high temperature (π»πΊ) then the probability of Delhi also having a high temperature (π»π·) is 0.40; i.e., π(π»π·|π»πΊ ) = 0.40. Similarly, the next two entries are π(ππ·|π»πΊ ) = 0.48 and π(πΏπ·|π»πΊ ) = 0.12. Similarly for the other rows.
If it is known that π(π»πΊ ) = 0.2, π(ππΊ ) = 0.5, and π(πΏπΊ ) = 0.3, then the probability (correct to two decimal places) that Guwahati has high temperature given that Delhi has high temperature is _______.
Attempted by 43 students.
Show answer & explanation
Correct answer: 0.6 to 0.62
Goal: find the probability that Guwahati has high temperature given that Delhi has high temperature.
Use Bayes' theorem: P(Guwahati is high | Delhi is high) = P(Delhi is high | Guwahati is high) Γ P(Guwahati is high) / P(Delhi is high).
Compute P(Delhi is high) by total probability over Guwahati's states:
P(Delhi is high) = P(Delhi high | Guwahati high)P(Guwahati high) + P(Delhi high | Guwahati medium)P(Guwahati medium) + P(Delhi high | Guwahati low)P(Guwahati low).
Substitute the numbers: 0.40Γ0.20 + 0.10Γ0.50 + 0.01Γ0.30 = 0.08 + 0.05 + 0.003 = 0.133.
Compute the numerator: P(Delhi high | Guwahati high)P(Guwahati high) = 0.40 Γ 0.20 = 0.08.
Therefore the conditional probability = 0.08 / 0.133 β 0.6015β¦
Rounded to two decimal places this is 0.60.
Final answer: 0.60