exponential continuous distribution

Question

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

$λ=24$

$x$=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

Answer 1

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,

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

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