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Suppose that the time between calls from your best friend has an exponential distribution with a mean time of $3$ days.

(a) If you just received a call from her, what is the probability that you will receive the next call within the next $2$ days?

(b) You realize that you have not received a call in at least $2$ days. What is the probability that she will not call you within the next $2$ days?

My attempt:

if $\mu=3$ then $m=1/\mu = {1/3}$ hence $X \sim \exp(1/3)$

(a) $P(X\leq2) = 1-e^{-1/3*2} = 0.4866$

(b) $P(X > 2) = 1-P(X\leq2) = 1-(1-e^{-1/3*2}) = e^{-2/3} = 0.5134$

I think my answers are incorrect. I also tried $X \sim \exp(3)$ but those answer make even less sense than the ones above.

Will anyone please point me in the correct direction if my approach is incorrect please?

Stanley
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Jose soriano
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1 Answers1

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Your answers are correct. Perhaps tell us why you thought they were incorrect, because it's important to improve on the intuition that led you to think this.

Stanley
  • 3,094
  • I feel like I'm ignoring the first part of each question, like if i receive a call or not receive a call then calculate each probability, is like something is missing. Does that make sense? – Jose soriano Jul 07 '15 at 12:19
  • When waiting times are exponentially distributed, anything in the past is irrelevant. This is called the 'memorylessness' property and the exponential distribution is the only continuous distribution that has it. So, they put information about the past, probably to play games with your intuition. :) If you know a bit about conditional probabilities, e.g. that $f_{X|Y}(x|y) = f_{X,Y}(x,y)/f_{Y}(y)$ then you can prove that conditioning on the past doesn't change the result i.e. in (b) that $Pr[X>4|X>2] = Pr[X>2]$ where $X \sim \exp(1/3)$. – Stanley Jul 07 '15 at 12:34