Mia breaks glasses at the rate of 4 per week .Let t be the time in weeks between successive breakages of glasses .Then:

$f(t)= 4e^{-4t} \quad \text{when} \quad t\geq0 $

$ 0 \quad \quad \quad \quad \quad \quad \text{otherwise} $

What is the probability that a week goes by without Mia breaking any glasses ?

I tried to use integration with limits between Infinity and 1 ,since the minimum value of t seems to be 1 .However the series is divergent .How do I find the correct answer ?


  • $\begingroup$ The meaning of "a week goes by" is ambiguous: does it mean "the next week starting with any arbitrarily specified moment" or does it mean "at least a week's gap between broken glasses occurs within some specific time interval"? $\endgroup$ – whuber Oct 22 '19 at 15:58
  • $\begingroup$ I think the second assumption is correct $\endgroup$ – Sara Oct 23 '19 at 4:17
  • $\begingroup$ The answer that has been posted--and accepted by you--addresses the first interpretation, not the second. $\endgroup$ – whuber Oct 23 '19 at 13:59

You have the right idea, but the integral does in fact converge. We know that it has to since it is a probability density, which must integrate over the total domain to 1. The integral we want is (I will leave you to solve it yourself):

$$ \int_1^{\infty} 4e^{-4t} dt = e^{-4}$$.

So the probability is $e^{-4}$.

Another way of seeing this is noting that this is the exponential distribution which gives the probability of the interval between successive events generated by a Poisson point process. Hence we can use the Poisson distribution with the same rate as the exponential distribution ($\lambda=4$, number of glasses broken per week) and find the probability of no events occuring in the given interval (so no glasses breaking in a week) as

$$P(k=0) = \frac{\lambda^k e^{-\lambda}}{k!} = e^{-4}.$$


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