related to probability

Question

A public-interest group was planning to file a case in consumer court to challenge the of charging "surge pricing" by urban ride hailing companies in one of three cities. New Delhi, Mumbai, or Kolkata. The probability that it would choose New Delhi was 0.40; Mumbai, 0.35; and Kolkata, 0.25. The group also knew that it had a 60 percent chance of a favourable ruling if it chose Mumbai, 45 percent if it chose New Delhi, and 35 percent if i chose Kolkata. If the group did receive a favourable ruling, which city did it most likely choose?

Answer 1

Let N = event that New Delhi is chosen, M = event that Mumbai is chosen, K = event that Kolkata is chosen, and F = event that a favorable ruling occurs. Then it is given that

P(N) = 0.40

P(M) = 0.35

P(K) = 0.25

P(F | N) = 0.45

P(F | M) = 0.60

P(F | K) = 0.35

We require P(N | F), P(M | F), and P(K | F). We compute each of these using Bayes' theorem in conjunction with the definition of conditional probability.

P(N | F) = P(N ∩ F) / P(F)

= ( P(N) x P(F | N) ) / ( P(N) x P(F | N) + P(M) x P(F | M) + P(K) x P(F | K) )

= (0.40 x 0.45) / (0.40 x 0.45 + 0.35 x 0.60 + 0.25 x 0.35 )

= 0.3770

P(M | F) = P(M ∩ F) / P(F)

= ( P(M) x P(F | M) ) / ( P(N) x P(F | N) + P(M) x P(F | M) + P(K) x P(F | K) )

= (0.35 x 0.60) / (0.40 x 0.45 + 0.35 x 0.60 + 0.25 x 0.35 )

= 0.4398

P(K | F) = P(K ∩ F) / P(F)

= ( P(K) x P(F | K) ) / ( P(N) x P(F | N) + P(M) x P(F | M) + P(K) x P(F | K) )

= (0.25 x 0.35) / (0.40 x 0.45 + 0.35 x 0.60 + 0.25 x 0.35 )

= 0.1832

The largest of these three probabilities is P(M | F). Therefore, it is most likely that Mumbai was chosen given that a favorable ruling occurred.

Hope this helps!