Question

4.50. Suppose an alarm clock has been set to ring after T hours, where T ∼ Exp(1/3). Suppose further that your friend has been staring at the clock for exactly 7 hours and can confirm that it has not yet rung. At this point, your friend wants to know when the clock will finally ring. Calculate her conditional probability that she needs to wait at least 3 more hours, given that she has already waited 7 hours. More generally, calculate the conditional probability that she needs to wait at least x more hours, given that she has already waited 7 hours.

Answer 1

Answer:

$P[T \geq x]=e^{-\frac{1}{3}x}$

For x = 3 then

$P[T \geq 3]=0.368$

Step-by-step explanation:

First we must consider the “memoryless” or “forgetfulness” property of the exponential distribution, which means that the likelihood of something happening in the future has no relation with past events.  

Taking the above statement into account we have that the probability that T lasts more than a+b hours given that an event has lasted at least “a” hours equals the probability that T lasts more than “b” hours:

$P[T \geq a+b | T \geq a]=P[T \geq b]$

For our case we have:

$P[T \geq x+7 | T \geq 7]=P[T \geq x]$

To make calculations easier we resolve using the Exponential cumulative distribution, therefore:

$P[T \geq x]=1-P[T < x]$

$=1 - (1 - e^{-\frac{1}{3}x})$

$=1 - 1 + e^{-\frac{1}{3}x}}$

$=e^{-\frac{1}{3}x}$

Now, if x = 3 then,

$P[T > 3+7 | T> 7]=P[T \geq 3]=e^{-\frac{1}{3}3 }=e^{-1}=0.368$

So, we can say that the probability that our friend waits at least 3 hours for the alarm to ring is 36.8%