Biology Genetics-Two heterozygous brown-eyed (Bb)

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The probability of three out of five children having blue eyes from two heterozygous brown-eyed parents (Bb) is calculated using the binomial expansion. The probability of a child having blue eyes is 1/4, while the probability of having brown eyes is 3/4. Using the binomial formula, the correct calculation yields a probability of approximately 8.78% for three children having blue eyes. The initial confusion arose from miscalculating the probability of blue eyes as 20% instead of the correct 25%. The final answer confirms the correct application of the binomial equation.
jena
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Hi,

My Question:

Two heterozygous brown-eyed (Bb) individual have five children.What ist he probability that three will have blue eyes?

Answer:

Would the possibility be zero for all three since blue eyes is recessive

Thank You:smile:
 
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Hi,

I looked over the question again and I know for sure this isn't the right answer. I do know that I should use a binomial expansion, so if the

Probability of blue eyes is 1/4 then I would square this and get

P(blue)=(1/4)^2=.0625*100=6.25%

Is this correct?

Thank You
 
Why are you squaring the probability?

This one is also a bit trickier because they aren't just asking the probability of 3 having blue eyes, but 3 out of 5. That will change how you do your calculations somewhat.
 
You might try to find the probability of 1/5 of the children having blue eyes. Having 5 kids would do what to that probability?
 
The probability of having 1 child having blue eyes is 20% right, so to figure the probability of three children having blue eyes would be

P(blue)=(1/5)^3 or .80 %

Is this correct?
 
jena said:
The probability of having 1 child having blue eyes is 20% right

No it is not 20%, why do you think that?
 
You had it right the first time when you said the probability of one child having blue eyes is 1/4, or 25%. Now try it.
 
I thought it was 20% because 1/5 is .20 multiply that by 100 and you get 20%, but I know what I did wrong this time all I have to do is use the binomial equation.

P=((n!)/(x!(n-x)!))*((p^x)(q^(n-x)))

I used the following steps to come up with an answer

Step 1: Calculate the individual probabilities
• P(blue eyes)= p=1/4
• P(brown eyes)=q=3/4

Step 2: Determine the number of events
• n=total number of children=5
• x= number of brown eyed children=3

Step 3: Substitute the values for p, q, x, and n in the binomial expansion equation(like above)

Finally

P=((5!)/(3!(5-3)!))*(((1/4)^3)((3/4)^(5-3)))
P=8.78%

I hope this time around that's right:smile:
 
Yes, it is right!
 
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