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This is a problem given in my Physics Textbook and I've been trying to solve it for the past hour. It's not something exceptionally challenging, but more conceptual in nature. Not much, connections have been made between Probability and Radioactivity by the author and so I've been reading up some articles on the web.

Coming back to this problem, I have mathematically guessed the answer but I don't understand why it really works. The Probability according to me is the following: $P={1\over2}+{1\over4}+{1\over8}=0.875$ And it turns out this is the correc answer, but what does this physically imply??

Student
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3 Answers3

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It means that after every half-life of time there is a 50% probability that any given nucleus will decay.

So after one half life, there is a 50% probability that a particular nucleus will have decayed. But after that time, if your particular nucleus has not decayed, then there is a further 50% probability that it will decay after another half life. Thus the total probability of decay is $0.5 + 0.5\times 0.5 =0.75$. The reason for the extra factor of 0.5 in the second term is that your nucleus must not have decayed during the first half-life of your trial in order to decay sometime between one and two half-lives.

Then, after a third half life, the probability of decay is $0.5 + 0.5\times 0.5 + 0.5\times0.5\times 0.5 = 0.875$. The reasoning behind the last term is that after two half-lives, the chance of having an undecayed nucleus is $1-0.75=0.25$ and that there is then a 50% chance of this decaying over the course of another half-life.

ProfRob
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    As an aside, lots of similar probability questions can be easily answered by looking at the probability that something hasn't happened, and the taking1 minus that answer to get the prob that the event has happened. – Dr Chuck Jan 15 '16 at 12:39
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    Probably worth expanding @DrChuck's comment to the general expression. If probability of decay in a given period is $p$, then probability of not decaying in that period is $(1-p)$. The probability of not decaying for $n$ periods is then $(1-p)^n$. The probability that something happens is 1 - probability that it did not happen, so the general answer is $1-(1-p)^n$ – Floris Jan 15 '16 at 14:29
  • @Rob Did I understand your reasoning correctly...$P(decay)=P(1st)+P(2nd)+P(3rd)$ – Student Jan 16 '16 at 09:54
  • @ShreyAryan That depends what you define P(2nd) and P(3rd) to be. P(1st) would be prob. of decay after a half-life, P(2nd) would be the combined probability that you have an undecayed nucleus after 1 half-life (0.5) that decays after another half-life. P(3rd) is the prob. that you have an undecayed nucleus after 2 half lives (0.25) that then decays after another half life. And so-on. – ProfRob Jan 16 '16 at 10:31
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About the physical implications, what this exercise shows you, is that the radiactive nucleus as a 50% chance to decay during its half-life period, but if it has not decayed, it remains unchanged: it has same number of protons and neutrons, same instability, and so during the next period it still have a 50% probability to decay, and so on.

ch7kor
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I'll try to give an answer that's as intuitive as possible. In probability you can often* replace any statement containing 'or' with addition. If you calculate the probability of throwing 1 or 2 with a dice you get $P(1\ \text{or}\ 2)=P(1\lor2)=P(1)+P(2)=1/6+1/6=1/3$. You can phrase the question 'what is the probability the atom has decayed after 3 half lives?' to 'what is the probability the atom has decayed after 1 half life or after 2 half lives or after three half lives'.

So we can write $$P(\text{decay in 3 half lives})=\\P(\text{decay during first half life or decay during second half life or decay during third half life})$$ To write this more neatly call $X$ the number of half lives after which the atom decays. Then we get \begin{align}P(X\leq3)&=P(X=1\ \lor X=2\ \lor X=3)\\ &=P(X=1)+P(X=2)+P(X=3) \end{align}

The probability of it decaying after one half life is obviously 1/2. To calculate the odds of it decaying during the third half life you have to calculate the probability of it surviving the first half live and surviving the second half life and then decaying during the third half life. I phrased this sentence awkardly on purpose so it is easier to translate into a probability. You can replace 'and' in probability often with multiplication.

\begin{align}P(X=3)=P(&\text{not decay during first half life and}\\ &\text{not decay during second half life}\\&\text{and decay during third half life})\\ =P(&\text{not decay during first half life})\times\\ P(&\text{not decay during second half life})\times\\ P(&\text{decay during third half life})\\ = 1/&2\cdot 1/2 \cdot 1/2=1/8\end{align} Similarly $P(X=2)=1/2\cdot1/2=1/4$. So finally $$P(X\leq3)=1/2+1/4+1/8$$ This is the most like your intuive answer, but there is a quicker way. Instead of calculating the probability that it decays in 3 half lives we can calculate the chance of it surviving longer than 3 half lives and invert that probability. Think of the dice again. The chance of not throwing a three is $P(\text{not 3})=1-P(3)$. Similarly $P(X\leq3)=1-P(X>3)=1-(1/2)^3=0.875$. Where $P(X>3)$ is the chance of surviving the first half life and the second and the third.

This answer has become quite long but I hope it's the most clear this way.

*There are cases where this isn't so simple, like in cases with A or B but not both.

AccidentalTaylorExpansion
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