Question

Suppose that the time (in hours) that Adam spends on an untimed final exam follows an
exponential distribution with mean 1.75 hours, and the time that Ben spends on the same exam follows
an exponential distribution with mean 2.25 hours. Assume that their times are independent of each
other.
Using appropriate notation for random variables and events:
a) Determine the probability that Ben finishes in less than 2 hours. (Show your work; you may use either
the pdf or cdf.)
b) Determine the probability that both Adam and Ben finish the exam in less than 2 hours. State necessary
assumptions/rules along your steps.
c) Determine the probability that both Adam and Ben are still working on the exam after 2 hours. (Show
your work, and indicate which rules you use to combine probabilities.)
d) Determine the conditional probability that Ben is still working after 3 hours, given that he is still working
after 2 hours. (Explain/justify your answer.)
e) Determine the conditional probability that Ben is still working after 2 hours, given that Adam is still
working after 2 hours. (Explain/justify your answer.)

Answer 1

Solution :

a)

Let A and B be the time taken by Adam and Ben to complete the exm respectively.

A ~ Exp($\lambda$ = 1/1.75)

B ~ Exp($\lambda$ = 1/2.25)

P(B < 2) = 1 - exp(-2/2.25) = 0.5888877

b)

We assume that the time taken by Adam and Ben to complete the exm are independent.

Using multiplication rule of probability,

P(A < 2, B < 2) = P(A < 2) P(B < 2) = (1 - exp(-2/1.75)) * (1 - exp(-2/2.25))

= 0.6810934 * 0.5888877

= 0.4010875

c)

Using multiplication rule of probability,

P(A > 2, B > 2) = P(A > 2) P(B > 2) = exp(-2/1.75) * exp(-2/2.25)

= 0.3189066 * 0.4111123

= 0.1311064

d)

P(B > 3 | B > 2) = P(B > 3 and B > 2) / P(B > 2)

= P(B > 3) / P(B > 2)

= exp(-3/2.25) / exp(-2/2.25)

= exp(-1/2.25)

= 0.6411804

e)

Using Independence of time taken by Adam and Ben to complete the exm,

P(B > 2 | A > 2) = P(B > 2) = exp(-2/2.25)

= 0.4111123

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