Basic probability question

tres · 22 May 2012 at 12:02

My friend and I are having a discussion re. Mendelian probability

We want to find out the probability of the ? child being a sufferer of an autosomal recessive condition.

We do not know the full genotype of the parents - only the first letter, which is the clear dominant.

Here is where we are disagreeing:

Before the parent's birth, we both agree (simple Mendelian) that the odds of a parent being Clear [RR] 25%, Carrier [Rr] 50%, Affected [rr] 25%

* I think that the odds of a parent (both shown as P/-) of ? carrying the recessive gene is 66% (and being clear [RR] as 33%).
This is because we know, for certain, that they are not sufferers [rr]. There is a 0% chance of them being [rr] Therefore the odds change.

* My friend thinks that their initial odds before birth affect these odds now - similar to the Monty Hall gameshow problem.
I think he's wrong, and they're two different probabilities.

Someone help please! We're both relatively new at this!

I've attached my workings here

[FnG]magnolia · 22 May 2012 at 12:05

Are the two P/- symbols above the ? symbol supposed to form a face? :eek:

I can't help you bro. I've forgotten almost everything I ever knew about stats.

Josh P · 22 May 2012 at 12:25

I think you're right, and then the probability of the child being a sufferer would be 1/4 * (2/3)^2 = 1/9. I think.

yer_averagejoe · 22 May 2012 at 12:37

Yes, after making the observation that the parents are not affected, the odds shift as you have calculated. Josh P's calculation is correct about the child being a sufferer.

estebanrey · 22 May 2012 at 12:40

tres said:

From your image, I can deduce that there is a 10% the child will be a homosexual.

Morbius · 22 May 2012 at 14:07

tres said:
* My friend thinks that their initial odds before birth affect these odds now - similar to the Monty Hall gameshow problem.
I think he's wrong, and they're two different probabilities.

You want to calculate the conditional probability, P(A|B) i.e. the probability of A, given B.

P(A|B) = P(A n B)/P(B)

So P(Carries recessive gene given not sufferer) = P(Carries recessive gene and not sufferer)/P(Not sufferer) = (1/2)/(3/4) = 4/6 = 2/3.

The Monty Hall problem is fairly unique in that it requires the contestant to make a pick before we gain new information. The probability of the contestant picking the winning door does not change retroactively.