Quick probability question

[Cheezeit]

Hey guys,

I have this simple probability question my bro is asking me, but i just can't think of how to do it right now. Here's the question:

Suppose that a husband and wife are both heterozygous for a recessive gene for albinism. If they have fraternal twins, what is the probability that both of the twins will have the same phenotype for pigmentation?

So basically the parents are Aa and Aa

What is the probability that the kids will have the same phenotype (Normal- AA or Aa) or (albino- aa)?

Any help would be appreciated

Thanks

 

[oiprocs]

100% probability that both of the twins will have chromosomes.

No need to thank me. You're welcome.

 

[MartyMcFly3]

Aa x Aa will produce possible combos for 2 children:

AA-AA
AA-Aa (Dad A)
AA-Aa (Mom A)
AA-aa
Aa-Aa (Dad A, Dad A)
Aa-Aa (Mom A, Mom A)
Aa-Aa (Dad A, Mom A)
Aa-Aa (Mom A, Dad A)
Aa-aa (Dad A)
Aa-aa (Mom A)
aa-aa

1/11 chance both would be albino. 7/11 chance they will be normal. 3/11 chance only one of them is albino. 6/11 that they will have the same phenotype.

I think.

Wait, does order matter??? If it does:

AA-AA
AA-Aa (Dad A)
AA-Aa (Mom A)
AA-aa
Aa-Aa (Dad A, Dad A)
Aa-Aa (Mom A, Mom A)
Aa-Aa (Dad A, Mom A)
Aa-Aa (Mom A, Dad A)
Aa-aa (Dad A)
Aa-aa (Mom A)
Aa-AA (Mom A)
Aa-AA (Dad A)
aa-aa
aa-Aa (Mom A)
aa-Aa (Dad A)
aa-AA

1/16 both are albino. 9/16 chance they will both be normal. 3/8 only one is albino. 3/8 they will have the same phenotype.

 

[Skeeedunt]

Wait that's completely wrong, nm

 

[Cheezeit]

I think i figured it out:

The chances that both kids are normal are 3/4 times 3/4 = 9/16
The chances that both kids are albino are 1/4 times 1/4 = 1/16

Since it can be either albino or normal just the same for both, you add them to 10/16.
Simplify to 5/8

Anyone else get this?

 

[FleshLight]

P(aa) = n/P = 1/4

P(aa) intersect P(aa) = 1/4 * 1/4 = 1/16

P(Aa U aA U AA) = P(Aa) + P(aA) + P(AA) - P(Aa)*P(aA) - P(aA)*P(AA) -P(Aa)*P(AA) + P(aA)*P(Aa)*P(AA) = figure it out

P(Aa U aA U AA) intersect P(Aa U aA U AA) = answer

 

[Cheezeit]

Originally posted by: MartyMcFly3
Aa x Aa will produce possible combos for 2 children:

AA-AA
AA-Aa (Dad A)
AA-Aa (Mom A)
AA-aa
Aa-Aa (Dad A, Dad A)
Aa-Aa (Mom A, Mom A)
Aa-Aa (Dad A, Mom A)
Aa-Aa (Mom A, Dad A)
Aa-aa (Dad A)
Aa-aa (Mom A)
aa-aa

1/11 chance both would be albino. 7/11 chance they will be normal. 3/11 chance only one of them is albino. 6/11 that they will have the same phenotype.

I think.

Wait, does order matter??? If it does:

AA-AA
AA-Aa (Dad A)
AA-Aa (Mom A)
AA-aa
Aa-Aa (Dad A, Dad A)
Aa-Aa (Mom A, Mom A)
Aa-Aa (Dad A, Mom A)
Aa-Aa (Mom A, Dad A)
Aa-aa (Dad A)
Aa-aa (Mom A)
Aa-AA (Mom A)
Aa-AA (Dad A)
aa-aa
aa-Aa (Mom A)
aa-Aa (Dad A)
aa-AA

1/16 both are albino. 9/16 chance they will both be normal. 3/8 only one is albino. 3/8 they will have the same phenotype.

The second half is what makes sense. Why 3/8 they will have the same phenotype though? I have 1/16 and 9/16 like you, but i added them to get 5/8.

 

[MartyMcFly3]

AA-AA
Aa-Aa
Aa-Aa
Aa-Aa
Aa-Aa
aa-aa

Those are the ones with the exact same phenotypes for pigmentation. 6/16 or 3/8...

Though yes if he meant what are the odds they are both normal or both albino, then you would be right.

 

[FleshLight]

Or simply put, P(aA U Aa U AA)^2 = 1 - P(aa)^2

 

[Cheezeit]

Yeah, the question was what was the probablity they would be both normal or both albino. So in your lists, AA -Aa etc would be both normal, which makes up for the other 4 combos to make i also 5/8.

Fleshlight- what the heezy?

Thanks for the help guys, 5/8 it is.

 

[FleshLight]

The answer is 1/16 for both being albino
P(aa intersect aa)

For both being normal (aA, Aa, or AA), the answer is 15/16
P(aA U Aa U AA) intersect P(aA U Aa U AA) = 1 - P(aa intersect aa)

This is assuming P(aa)=P(aA)=P(Aa)=P(AA) = 1/4

 

[MartyMcFly3]

Originally posted by: FleshLight
The answer is 1/16 for both being albino
P(aa intersect aa)

For both being normal (aA, Aa, or AA), the answer is 15/16
P(aA U Aa U AA) intersect P(aA U Aa U AA) = 1 - P(aa intersect aa)

This is assuming P(aa)=P(aA)=P(Aa)=P(AA) = 1/4

Uhh no.

 

[FleshLight]

Yep, I'm wrong.

I just did the math:

P(aA U Aa U AA) = 1/4 + 1/4 + 1/4 - 1/4*1/4 - 1/4*1/4 - 1/4*1/4 + (1/4*1/4*1/4) = 37/64

Intersect that with itself and u get 0.334

So probability of both being normal is 0.334