Question

Along the road near Bundamba State High School, there are traffic lights at two different locations. At the first location, the light is green 80.9% of the time. At the second location, the light is green 79.9% of the time. The probability a vehicle gets a green light at the first location and the second location is 71.4%. What is the probability that the light will be green at the second location if the light was green at first location? (3 decimal places)

Answer 1

P(G 1st) = 0.809

P(G 2nd) = 0.799

P(G 1st G 2nd)=0.714

P(G 2nd | G 1st) =? (We need to find Prob of green light at 2nd location given that light was green at first location

As per bay's theorm

P(A|B) = P(AB)/P(B)

Here

P(G 2nd | G 1st) = P(G 2nd G 1st) / P(G 1st)

=0.714/0.809

=0.8825711

=0.883 (3 decimal places)

Hence we can say that 0.883 is the probability that the light will be green at the second location if the light was green at first location

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