Question

The amount of time it takes Josslyn to wait for the train is continuous and uniformly distributed between 4 minutes and 11 minutes. What is the probability that it takes Josslyn between 5 and 6 minutes given that it takes less than 8 minutes for her to wait for the train?

Answer 1

X = uniform (4,11)

P(X < x) = (x-a)/(b-a)   when a < x<b

here a = 4 , b = 11

we need

P(5 <X< 6 | X < 8)

= P( 5 <X <6 and X < 8)/P(X < 8)

= P(5 <X< 6)/P(X < 8)

Now

P(X < 8) = (8 - 4)/(11-4) = 4/7

P(5 <X< 6) = (6-5)/(11-4) = 1/7

hence required probability = 1/4

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