A box of candy contains 24 bars. The time between demands for these candy bars is exponentially distributed with a mean of 10 minutes. What is the probability that a box of candy bars opened at 8:00 AM will be empty by noon?

My attempt

1)Transform the exponential distribution to Poisson. λ from 8 am till noon is 24


$x$=number of chocolates demanded=24

$P(X=x)= \dfrac{λ^x * e^{-λ}}{x!}$

$P(X=24)= \dfrac{24^{24} * e^{-24}}{24!}=0.081$

However the correct answer is (0.632)

I hope someone can help me.Thnks a lot

edit: I also took the take case where $P(x\ge 24)=1-P(x\le 23)$ which gave me 0.53 which is also wrong

  • $\begingroup$ Shouldn't you also count the events where demand exceeds 24? That is, you actually want to calculate $\Pr (X \geq 24)$. $\endgroup$ – Theoretical Economist Dec 11 '16 at 2:18
  • $\begingroup$ I thought about it but it is very consuming to solve such question during exam so I assumed that the demand stops once chocolates are consumed. Should I calculate [ 1- Σ (λ^(x) * e^(-λ))/x!)] where x is from 0 to 23 and lambda is 24? $\endgroup$ – Basil Bassam Dec 11 '16 at 2:22
  • $\begingroup$ What's your source material? I agree with @TheoreticalEconomist that it should be $\Pr (X \geq 24) = 0.52715\ldots \approx 0.53$ as you calculated. $\endgroup$ – Lee David Chung Lin Dec 11 '16 at 4:25
  • $\begingroup$ Yes, I made that calculation as well. Where are you getting these exercises and solutions? It seems to have gotten the last exercise you posted about wrong as well. $\endgroup$ – Theoretical Economist Dec 11 '16 at 4:28
  • $\begingroup$ Hmmm Thank you very much for your time. These are some questions sent to me by my Professor.I'll let her know about the mistakes. $\endgroup$ – Basil Bassam Dec 11 '16 at 13:40

I agree with the above comments about the probability being $.53$

Since the only way the box can be empty is if $24$ are eaten,

$$P (x\geq 24)$$ = $$1 - P(x <24)$$ = $$1 - P(x\leq23)$$

enter image description here

This is the density plot (i.e. probability at a specific value)

We use the CDF to compute area to the right of $24$ (including $24$) which is just

$$1 - P(x\leq23)$$

 1-ppois(23,24) = .52715

Please give an update when you figure out the answer

  • $\begingroup$ I sent my Professor message she sent me a solution I didn't understand P(x>=1)=1-P(X=0) = 0.632 Any ideas ?? $\endgroup$ – Basil Bassam Dec 11 '16 at 18:01
  • $\begingroup$ I think you mean P(x=>24) = 1 - P(x <=23). I'm willing to bet it's an error. The .527 is correct $\endgroup$ – Brandon Dec 11 '16 at 18:44
  • $\begingroup$ She is like think of the 24 candies as one bundle. I am totally going to ignore what she says. Thank you for your help $\endgroup$ – Basil Bassam Dec 11 '16 at 19:08
  • $\begingroup$ Hmmm that is interesting.. I understand what she is saying, but I don't know why she is doing it that way. 1-ppois(0,1) is in fact .632, If you don't have R (which is what I'm using), you can do it in a graphing calculator. "PoissonCDF"..Just use the 2 values (lambda and x). Youll see that using 0 and 1 does produce .632, whereas the 23 and 24 produces .527. I still prefer the way that it is currently done (the .527). Glad I could help! $\endgroup$ – Brandon Dec 11 '16 at 19:14

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