# Probability to get an orphan disease

• December 17th 2012, 03:38 PM
fixisnotusedyet
Probability to get an orphan disease
Hello all,

p : Probability of population with a specific orphan disease (p=10^-7)
q : Probability to get one allele called L1 on his X chromosome (q=0.01)

The only 3 known cases of this orphan desease occured with people having the L1 allele on their X chromosome.

What is the probability for someone to have this orphan disease knowing he has the L1 allele (with an associated confidence interval)?
• December 17th 2012, 04:23 PM
chiro
Re: Probability to get an orphan disease
Hey fixisnotusedyet.

Hint: to start you off can you express the probability in terms of a conditional statement (in other words can you define the probability as P(A|B) where you give the definitions of A and B)?

This is half the battle in mathematics and if you do this you'll probably the solve the problem all by yourself.
• December 17th 2012, 04:44 PM
fixisnotusedyet
Re: Probability to get an orphan disease
Thank you chiro.
A=i am sick
B=i have allele L1
P(A)=10^-7
P(B)=10^-2

We are trying to calculate P(A/B)

P(A/B)=P(A and B)/P(B)=10^2.P(A and B)

But I have no idea how to calculate P(A and B) based on this context.
Based on my sample of only 3 cases, observed P(A and B)=10^-7

I am 100% sure that P(A/B) is somewhere between 0 and 10^-5.
But giving you an answer with an 80% confidence interval... no idea if this is possible

I know the sample is very poor (n=3) but probabilities p and q are so small that I am convinced we can still get something out of this.
• December 17th 2012, 04:54 PM
chiro
Re: Probability to get an orphan disease
Well since P(A|B) = P(A and B)/P(B) it would make sense that P(A|B) = 10^2 * 10^(-7) = 10^(-5).

If you want to do inference, can you specify the attribute you wish to do inference on? Is it sickness or having a particular allele?
• December 17th 2012, 04:56 PM
Plato
Re: Probability to get an orphan disease
Quote:

Originally Posted by fixisnotusedyet
Thank you chiro.
A=i am sick
B=i have allele L1
P(A)=10^-7
P(B)=10^-2
We are trying to calculate P(A/B)
P(A/B)=P(A and B)/P(B)=10^2.P(A and B)
But I have no idea how to calculate P(A and B) based on this context.
Based on my sample of only 3 cases, observed P(A and B)=10^-7
I am 100% sure that P(A/B) is somewhere between 0 and 10^-5.
But giving you an answer with an 80% confidence interval... no idea if this is possible
I know the sample is very poor (n=3) but probabilities p and q are so small that I am convinced we can still get something out of this.

"I know the sample is very poor (n=3)" but surely that is the whole point.
It a heck of a difference if the population is $10^3\text{ or }10^{6}$.
• December 17th 2012, 05:01 PM
fixisnotusedyet
Re: Probability to get an orphan disease
As Plato hinted, P(A and B) =10^-7 is not true. This is only the assumption based on the 3 known cases...
This is the problem with orphan disease. We do not have a population of 10^6 to confirm the assumption !

Any idea how we can transform this false statement P(A and B) =10^-7 into a "With a probability of 80%, P(A and B) is between x and y" ? :)

To clarify, the objective of my exercice is to complete the following sentence "With a probability of 80%, the probability to be sick when we have the allele L1 is between ... and ...".
• December 17th 2012, 06:06 PM
fixisnotusedyet
Re: Probability to get an orphan disease
Hi i may have found the solution.

T=proba of being sick having L1 allele
t=proba of being sick.
I calculated probabilities of having 3 cases with L1 allele for every value of T between 0 and 10^-5.
Then I calculated the density of probability:

With a confidence of 50% : T > 80t
With a confidence of 80%: T>59t
With a confidence of 95%: T>39t
With a confidence of 99%: T>25t
With a confidence of 99,9%: T>6t
• December 18th 2012, 06:08 AM
fixisnotusedyet
Re: Probability to get an orphan disease
I reviewed what i said, it is probably wrong.

t Probability to be sick
p Probability to have a l1 allele
T P(t/p)=Probability to be sick knowing we have a l1 allele
f(T) Probability to have 3 people with a l1 allele

For every value of T, i have calculated f(T)=(T/t*p)^3

Si T=0.00001, f(T)=100%
Si T=0.0000099, f(T)=97.03%
0.0000098 94.12%
0.0000097 91.27%
0.0000096 88.47%
0.0000095 85.74%
0.0000094 83.06%
0.0000093 80.44%
0.0000092 77.87%
0.0000091 75.36%
...
0.0000006 0.02%
0.0000005 0.01%
0.0000004 0.01%
0.0000003 0.00%
0.0000002 0.00%
0.0000001 0.00%
0 0.00%

I am able to affirm stuff like:
If T=100t, probability to have 3/3 cases with an l1 allele is 100%.
If T=21t, probability to have 3/3 cases with an l1 allele is 0.93%.

(As a reminder T=100t means it is 100 times more like likely to be sick when you have a l1 allele)

Any idea how to pursue the analysis?
• December 18th 2012, 06:10 PM
chiro
Re: Probability to get an orphan disease
What exactly are you making an inference on? (I apologize if you already stated this, but I want to just double check).