Exponential Distribution and radio active decay

[Bachelier]

Typical radio active decay question

Half time = 1 year
λ = ln 2 here

Q 1: if we have 1024 atoms at t=0, what is the time at which the expected number remaining is one.

Easy, I get 10 years

Q 2: The chance that in fact none of the 1024 atoms remains after the time calculated in c:

it should be P(T<10 years) = 1 - P(T>10 years) but I get different answers.

The answer given is e^-1

any hints please? I know it's easy, I'm just missing something

Thanks

 

[Bachelier]

I guess we can set up a new rv Y = number of surviving atoms.
p = 1/1024, all indep hence follows a Poisson dist. with parameter (1)

P(Y=0) = e^-1 ≈ 36%

 

[bpet]

Another way of looking at it: each atom survives one year with probability 1/2, so it survives for 10 years with probability 1/2^10 = 1/1024, and thus the probability that none of the 1024 survive for 1024 years is (1-1/1024)^1024 or approximately exp(-1).

 

[Bachelier]

Cool