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

*Hope the above answer has helped you in understanding the
problem. Please upvote the ans if it has really helped you. Good
Luck!!*

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