# Answer on Statistics and Probability Question for Revika

Question #5020

The probability of a cricket match to be played between India and Sri Lanka at one of the three selected venues X,Y and Z are 0.25, 0.35 and 0.40 respectively. The probability that the match being disturbed at these three venues are 0.05, 0.04 and 0.02 respectively. A match is played and disturbed, what is the probability that the match was played at X?

Expert's answer

Let's formalize the problem statements:

P(X) = 0.25

P(Y) = 0.35

P(Z) = 0.40

P(D|X) = 0.05

P(D|Y) = 0.04

P(D|Z) = 0.02

Now let's use the Bayes' theorem to determine the probability that the match was played at X:

P(X|D) = P(D|X)P(X) / [P(D|X)P(X)+P(D|Y)P(Y)+P(D|Z)P(Z)] = 0.05*0.25 / (0.05*0.25 + 0.04*0.35 + 0.02*0.40) = 0.0125/0.0345 ≈ 0.362.

P(X) = 0.25

P(Y) = 0.35

P(Z) = 0.40

P(D|X) = 0.05

P(D|Y) = 0.04

P(D|Z) = 0.02

Now let's use the Bayes' theorem to determine the probability that the match was played at X:

P(X|D) = P(D|X)P(X) / [P(D|X)P(X)+P(D|Y)P(Y)+P(D|Z)P(Z)] = 0.05*0.25 / (0.05*0.25 + 0.04*0.35 + 0.02*0.40) = 0.0125/0.0345 ≈ 0.362.

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