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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.

Kristoffer Jerzy Linder
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1 Answers1

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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 $

Henry
  • 157,058
  • Yeah. But where should I put $13$, and where should I put $12$. From the Physics I always learnt that the number of particles should be put as a coefficient before the exponential. Like

    $$ 12\exp(-x/36.4), or \ \ 12\exp(-x/36.4) $$

    But according to the definition of the exponential distribution the exponent, and the coefficient should be the same. I'm confused.

     – Kristoffer Jerzy Linder Nov 26 '18 at 20:34
  • @KristofferJerzyLinder - multiplying your answer by $12$ will give a value over $1$ so it cannot be a probability. I have added a couple of approaches to my answer – Henry Nov 26 '18 at 21:13