Decay probability

Question

If I have a nucleus that have a constant probability per unit time to decay $q$ (to simplify is the probability to decay per day),

Me and my friend decided to make a bet: the nucleus is alive at the end of the day?

I start to observe today, so I say:"I have a probability of $(1-q)$ that is alive!"

My friend, have been started to observe 4 days ago, say:"I have a probability of $(1-q)^5$ that the nucleus is alive at the end of the day!"

So who have the truth?

I think we want the exponential distribution here, but yes also some basic properties of conditional probability — innisfree, Nov 22 '21 at 13:37
As I mention here, the decay of the nucleus is like tossing a coin. — PM 2Ring, Nov 22 '21 at 13:44

score 2 · Answer 1 · edited Nov 22 '21 at 13:29

Provided that the probability of decay in 24h is $q$, the probability conditioned on the fact that it was alive 24h ago is $$P(\text{alive today}|\text{alive yesterday}) = 1 - q.$$

The probability that it is still alive after 5 days, provide dthat we didn't look at it in between is $$P(\text{alive today}|\text{alive yesterday}) = (1 - q)^5.$$

Whether you speak of a classical or quantum probability, if you have ascertained that it was alive yesterday, the history of previous observations does not influence the future - a good example of what is called Markov process (see also here).

Remark: Though the Halloween has passed, I also recommend my post about the death process

Should the condition in the second line be (…|alive 5 days ago)? — rob, Nov 22 '21 at 15:19

score 1 · Answer 2 · answered Nov 24 '21 at 20:45

If we are interested in the probability that the decay has taken place after a time $T$, we have three options.

Option 1: We could use the exponential distribution, because the exponential probability has the property that it is memoryless. This means the probability of an event does not change with the time we already waited. Using the exponential distribution the cumulated probability that the decay has taken place before time $T$ is given by $$ Pr(t\le T) = 1-exp(-\lambda T) $$ where $\lambda$ is the decay rate. This formula is valid for all times $T\ge 0$.

Option 2: If we know that

the decay probability in the time interval $[0, t^\prime]$ is $q$, and
we are only interested in the probability that the nucleus has decayed after some integer multiple of the time $t^\prime$,

we can $n$ independent Bernoulli trials to calculate the probability. The formula reads $$ Pr(t \le n\cdot t^\prime) = (1-q)^n ~~~ and ~~~ q = 1-Pr(t\le T) $$ and $n$ is an integer.

Option 3: Since the last formula is also the formula for the Binomial distr. with zero events and $n$ trials we could also use the Binomial distribution to calculate the probability.

In order to "convince" you that these formulas yield the same result let's plot the Binomial result and the result using the exponential distribution together

Decay probability

2 Answers2