Blue eye X brown eye with unknown genotype?

[avasalie]

I am an instructor having a Problem trying to reconcile the sum rule with the answer to this Q: if a blue eyed (bb) woman marries a man with brown eyes (B?) where his genotype is unknown, what is the probability of having a blue-eyed child? I calculate it at 1/4 because he has 1/2 chance of being Bb and 1/2 chance of passing it on. But if you consider it as mutually exclusive he is either BB OR Bb, then if he is BB there is 0 chance of blue eyed child OR if he is Bb, there is 1/2 chance of blue eyed child, then 0 + 1/2 = 1/2. What is wrong here?

[imatfaal]

Surely you need to weight the averages?

 

There is a 50pct chance of him being Bb and if and only if this is the case then a 50pct chance that his child with bb mother will be blue eyed. That is two probabilities - both of which must come true; do you know how to calculate that single probability?

 

The other way is to draw a simple diagram listing all the equally likely possibilities (there will be eight - four that are the results of BB|bb and four that are the results of Bb|bb) You have eight possible outcomes - and you can count how many are blue-eyed

[Delta1212]

Yeah, you can't just add unweight odds like that. Let's say I have to flip a coin, and want to know what the probability that I will get heads is. First I have to pick which coin to flip from among four possible coins, one of which is a two-headed coin.

 

So coin A has a 1/2 chance of giving me heads, B has a 1/2, C has a 1/2 chance and D has a probability of 1.

 

If I add those the way you added 0 + 1/2, I would get :

 

1/2 + 1/2 + 1/2 + 1 = 2.5

 

And thus conclude that there is a 250% of my getting heads.

 

The proper way to do it is to see that there is a 1/4 chance of picking coin A, and then a 1/2 chance of getting heads. There is therefore a 1/8 chance of getting a coin A heads. Same with coins B and C. Coin D also has a 1/4 chance of being picked, but a 100% chance of getting heads if selected, so the odds of getting a coin D heads are 1/4.

 

That gives me 1/8 + 1/8 + 1/8 + 1/4 = 5/8 chance of getting heads, which is a little bit above 50%, or exactly what you'd expect when selecting from a group of coins that includes one double-head coin.

[Bruno527]

This seems like a bad question without enough information provided. You don't know that the probability of the man being Bb is 1/2 without knowing more about his parents or about the frequency of the b allele in the population. If you are to assume that his probability is 1/2, which is unfounded, then your reasoning is correct and the answer would be 1/4.