Genetics Question - Probability

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marcusnasland

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Hey guys i can't quite get my head around this probability question in genetics just wondering if i could get a little help... Here it is

- In the same family assuming the probability of blue eyes is 1/4 and the probability of green eyes is 3/4, what is the probability of having 2 blue eyed boys and 6 green eyed girls?

Thanks in advance!
 
haha k how did you determine that? Seems to me like it would be a lot less probablity to have exactly 2 boys and 6 girls and for them to have those specific eye colours
 
Hey guys i can't quite get my head around this probability question in genetics just wondering if i could get a little help... Here it is

- In the same family assuming the probability of blue eyes is 1/4 and the probability of green eyes is 3/4, what is the probability of having 2 blue eyed boys and 6 green eyed girls?

Thanks in advance!
You are only dealing with blue eyed boys and green eyed girls. The probability of a blue eyed boy is (1/4)(1/2) = 1/8. The 1/4 is for the blue eyes and the 1/2 for the boy. The probability of a green eyed girl is (3/4)(1/2) = 3/8.

So one possibility is (1/8)(1/8)(3/8)(3/8)(3/8)(3/8)(3/8)(3/8).

In how many ways can you order the births? (8 choose 2) = 28.

So the answer is 28 * (1/8)^2 * (3/8)^6 = 0.122%.
 
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Thank you very much! What is the formula you used there?

(nCr) * p^r * q^(n-r)

(nCr) is n! / [(n-r)!*r!]

n is the total # of events.
r is the # of events with probability p of occurring.
q is the probability of the other (n-r) events occurring.

Here n is 8. There are 8 events (births).
r is 2 and p is (1/8). There are 2 events (births) with probability (1/8) (boys + blue eyes).
(n-r) is 6 and q is (3/8). There are 6 events (births) with probability (3/8) (girls + green eyes).
 
Ok thanks a million i guess i was just a bit confused because i was thinking that we had to fit the probability of the boys having green eyes and the girls having blue eyes somewhere in there but this makes sense