Probability of measuring muons is a particular setup

[ChrisVer]

Homework Statement

You want to measure the momentum of muons with the help of a proportional chamber, that is a chamber
that applies an electric field inside its cavity and produces a signal current when a charged particle passes
through it. The probability of generation such a signal for a single chamber is 97%. For a momentum
measurement you need to measure the position of a muon in at least three points inside the detector, i.e.
three proportional chambers provide a signal. What is the minimum number of chambers required to provide
a momentum measurement for at least 99% of muons that pass through your detector?

Homework Equations

The Attempt at a Solution

The probability of a signal of a muon per chamber is: $p = 0.97 \times \frac{1}{3} = 0.323$.
My problem however is that I don't know the number of the muons... Any feedback of what method I could use? I thought about using a Gaussian and integrating it from $0.99N$ to $N$ ($N$ is the total number of muons passing through and an unknown parameter).

 

[mfb]

The probability of a signal of a muon per chamber is: $p = 0.97 \times \frac{1}{3} = 0.323$.

It is not. It is 97 % as given in the problem statement.

My problem however is that I don't know the number of the muons...

How would that matter? If you have a coin that shows head 50 % of the time, do you need the number of coin tosses to use that value of 50 %?

Any feedback of what method I could use? I thought about using a Gaussian and integrating it from $0.99N$ to $N$ ($N$ is the total number of muons passing through and an unknown parameter).

You are overthinking this.

What is the probability to detect a muon three times if you have three chambers (assuming muons pass through all of them)?

 

[ChrisVer]

It is not. It is 97 % as given in the problem statement.

Oops sorry, I was wrong.

What is the probability to detect a muon three times if you have three chambers (assuming muons pass through all of them)?

If I have 3 chambers the probability is $p= (0.97)^3$ and also that's the probability of a single signal. Now if I have to do it three times... I guess I can use the binomial $B= \begin{pmatrix} n \\ k \end{pmatrix} p^n (1-p)^{n-k}$ with $n=3$ but I don't understand what's the meaning of $k$ here.
In the coin toss example, $k$ stands for the number of successes in $n$ trials.

 

[mfb]

If I have 3 chambers the probability is $p= (0.97)^3$ and also that's the probability of a single signal.

Let's call it "probability to get a momentum measurement" as signal is used for single chambers in the problem statement.
Right. 0.97^3 < 0.99 so three chambers are not sufficient.

What is the probability to detect a muon three or four times if you have four chambers (assuming muons pass through all of them)?
The decimal number could be interesting here.

 

[ChrisVer]

It's that probability times the probability to get the right number of 3 signals out of 4 chambers?

 

[mfb]

No. You try 4 times, with 0.97 success probability each time. What is the probability to succeed 3 or 4 times?

 

[ChrisVer]

$0.97^3$ or $0.97^4$...

My conceptual problem comes when I add more chambers though... Because with four chambers I have the possible outcomes:
x : no signal ... o : signal in each chamber:

x x x x - fail
x x x o +3 permutations of o = fail
x x o o + 4 permutations of o = fail
x o o o + 3 permutations of x = success
o o o o = sucessThe appearence of each o is 97% probable... Looking at this scheme then the probability would be:
$p^4 + 4 p^3$ ? (this doesn't look right)

I think I'm trying to build up that:
$\begin{pmatrix} 4 \\ 4 \end{pmatrix} p^4 + \begin{pmatrix} 4 \\ 3 \end{pmatrix} p^3$
But I'm missing the $q=1-p$ probability in this.

 

[Ray Vickson]

$0.97^3$ or $0.97^4$...

My conceptual problem comes when I add more chambers though... Because with four chambers I have the possible outcomes:
x : no signal ... o : signal in each chamber:

x x x x - fail
x x x o +3 permutations of o = fail
x x o o + 4 permutations of o = fail
x o o o + 3 permutations of x = success
o o o o = sucessThe appearence of each o is 97% probable...

$0.97^3$ or $0.97^4$...

My conceptual problem comes when I add more chambers though... Because with four chambers I have the possible outcomes:
x : no signal ... o : signal in each chamber:

x x x x - fail
x x x o +3 permutations of o = fail
x x o o + 4 permutations of o = fail
x o o o + 3 permutations of x = success
o o o o = sucessThe appearence of each o is 97% probable... Looking at this scheme then the probability would be:
$p^4 + 4 p^3$ ? (this doesn't look right)

I think I'm trying to build up that:
$\begin{pmatrix} 4 \\ 4 \end{pmatrix} p^4 + \begin{pmatrix} 4 \\ 3 \end{pmatrix} p^3$
But I'm missing the $q=1-p$ probability in this.

Right, because you copied down the $B$-formula incorrectly. Go back and read your post #3 again, putting in $n = 4$ and $k = 3$ plus $k = 4$.

 

[ChrisVer]

My problem then is why didn't 1-p appear? Where did I lose it in the logic?

 

[mfb]

You just forgot to add it? It was present in the original formula.

 

[ChrisVer]

Yes it was...
So for 4 chambers I obtain:
$0.97^4 + 4 \times 0.97^4 \times (0.03) =0.991..$
I think I got the idea of the problem: "need the k>=3 success in n trials probability to be > 99%"...

 

[Ray Vickson]

My problem then is why didn't 1-p appear? Where did I lose it in the logic?

As I said: go back and read your own post #3 again. Really do it, and carefully this time.

 

[ChrisVer]

As I said: go back and read your own post #3 again. Really do it, and carefully this time.

No it's OK... In each row I didn't count the probability (when it appeared) that the muon wouldn't interact, and that's why I was losing the (1-p) factors...

Eg for this: o o - o , I would have p*p*(1-p)*p ...

 

[ChrisVer]

Well the solution would be:

$0.99 \le P = \sum_{k=3}^n \begin{pmatrix} n \\ k \end{pmatrix} p^k (1-p)^{n-k} = 1 - \sum_{k=0}^2 \begin{pmatrix} n \\ k \end{pmatrix} p^k (1-p)^{n-k}$

Where I would have to solve for $x$ (and $p=0.97$):

$(1-p)^x + p (1-p)^{x-1} x +\frac{x(x-1)}{2} p^2 (1-p)^{x-2} \le 0.01$

Wolfram solves this with $x \in [0.00202302, 0.936647]$ (which actually doesn't make sense since the number of chambers must be an integer... and $x\ge 3.77 ...$ whose right next integer is 4 (verifying the result of 0.991... I got above)

 

[Ray Vickson]

Well the solution would be:

$0.99 \le P = \sum_{k=3}^n \begin{pmatrix} n \\ k \end{pmatrix} p^k (1-p)^{n-k} = 1 - \sum_{k=0}^2 \begin{pmatrix} n \\ k \end{pmatrix} p^k (1-p)^{n-k}$

Where I would have to solve for $x$ (and $p=0.97$):

$(1-p)^x + p (1-p)^{x-1} x +\frac{x(x-1)}{2} p^2 (1-p)^{x-2} \le 0.01$

Wolfram solves this with $x \in [0.00202302, 0.936647]$ (which actually doesn't make sense since the number of chambers must be an integer... and $x\ge 3.77 ...$ whose right next integer is 4 (verifying the result of 0.991... I got above)

For your given $p$ the equation $P_n \equiv (1-p)^n + p(1-p)^{n-1}\,n + \frac{1}{2} p^2 (1-p)^{n-1} \,n(n-1) = 0.01$ has three solutions for $n \geq 0$, namely, $n \doteq 0.002023$, $n \doteq.93616$ and $n \doteq 3.77439$. Wolfram has found the two smallest and has taken the interval between the two. Algebraically, that is correct, but probabilistically it is invalid. If there is a way in Wolfram to tell it to restrict n to $n \geq 1$ then it should find the other root, and all $n$ values greater than that root will satisfy the inequality (because of the shape of the graph of $P_n$ vs. $n$.

In Maple, the command 'fsolve(Pn = 0.01, n=1..10)' finds the largest root; here Pn is the formula above, with p = 0.97 and q = 0.03 in it.