Exponential distribution of decay A container contains 13 particles, at time $t = 0$. The particles decay independently of each other and the time (unit: minutes) for a given particle's decay is a exponentially distributed random variable with expectation value $36.4$. Let $T$ denote the time that has passed when the number of particles has been reduced to 12. Calculate the probability $P(T > 1.035714)$
My solution is simple, but wrong. The pdf for the exponential distribution is given by
$$
f(x) = \frac{1}{\beta}\exp(-x/\beta), \beta > 0, 0 < x < \infty
$$
Numerically, we have
$$
f(x) = \frac{1}{36.4}\exp(-x/36.4)
$$ 
We want the probability 
$$
P(T > 1.035714) = 1 - F(1.035714) = 1 - \int_{0}^{1.035714}\frac{1}{36.4}\exp(-x/36.4)dx = \exp{-1.035714/36.4} = 0.97194731
$$
I know this is the wrong answer, but I can't see where I go wrong.

Hint: 


*

*That is the probability that a particular particle does not decay in that time  

*But there are $13$ particles, (a clue is that $13$ does not appear in your calculation) and you want the probability that none of the $13$ decay in that time

*You know that their decays are independent of each other 

Added after comments
There are at least two approaches giving the same answer: 


*

*If the probability a particular particle does not decay in that time is about $0.97194731$, the the probability that none of the $13$ independently decay is about $0.97194731^{13} \approx 0.6908$

*The rate of decay for one particle is $\frac{1}{36.4}$ so the rate for the first decay of $13$ independent particles is $\frac{13}{36.4}$, making the probability that this does not happen by time $1.035714$ be $\exp\left(−1.035714 \times \frac{13}{36.4}\right) \approx 0.6908 $
