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

 

[blue_raver22]

for your question. The exponential distribution is commonly used to describe the decay of radioactive substances over time. In this case, the half-life of the substance is 1 year, which means that after 1 year, half of the original atoms will have decayed. The decay rate, or λ, is equal to ln 2 here.

For Q1, you are correct in saying that it would take 10 years for the expected number of remaining atoms to be one. This is because after each half-life, the number of remaining atoms is halved. So after 1 year, there would be 512 atoms remaining, after 2 years there would be 256, and so on until we reach 1 atom after 10 years.

For Q2, you are correct in using the formula P(T<10 years) = 1 - P(T>10 years). However, the answer given of e-1 is not accurate. The correct answer would be (1/2)^10 = 1/1024, which is the probability that none of the original 1024 atoms would remain after 10 years. This can also be seen by plugging in 10 for t in the exponential distribution formula, P(t) = e^(-λt), which would give us e^-10ln2 = (1/2)^10.

I hope this helps clarify any confusion. Remember that the exponential distribution is a continuous probability distribution, so it's important to use the correct formula and units when solving these types of problems.