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Exponential Distribution and radio active decay

  1. Nov 12, 2011 #1
    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

  2. jcsd
  3. Nov 13, 2011 #2
    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%
  4. Nov 14, 2011 #3
    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).
  5. Nov 14, 2011 #4
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