Exponential distribution of decay












1














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.






exponential-distribution







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    1














    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.






    exponential-distribution







    share|cite|improve this question

























      1












      1








      1







      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.






      exponential-distribution







      share|cite|improve this question













      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.






      exponential-distribution




      exponential-distribution






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      share|cite|improve this question











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      share|cite|improve this question










      asked Nov 26 '18 at 19:34









      Kristoffer Jerzy Linder

      316




      316






















          1 Answer
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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 $expleft(−1.035714 times frac{13}{36.4}right) approx 0.6908 $







          share|cite|improve this answer























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











          Your Answer





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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          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 $expleft(−1.035714 times frac{13}{36.4}right) approx 0.6908 $







          share|cite|improve this answer























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
















          1














          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 $expleft(−1.035714 times frac{13}{36.4}right) approx 0.6908 $







          share|cite|improve this answer























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














          1












          1








          1






          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 $expleft(−1.035714 times frac{13}{36.4}right) approx 0.6908 $







          share|cite|improve this answer














          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 $expleft(−1.035714 times frac{13}{36.4}right) approx 0.6908 $








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 26 '18 at 21:12

























          answered Nov 26 '18 at 19:43









          Henry

          98.3k475162




          98.3k475162












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


















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
















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






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




          @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


















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