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)?

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.

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.

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?

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 .

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 ...".

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

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?

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).