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Genetics-Two heterozygous brown-eyed (Bb)

  1. Oct 26, 2005 #1
    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:
     
  2. jcsd
  3. Oct 26, 2005 #2
    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
     
  4. Oct 26, 2005 #3

    Moonbear

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    Gold Member

    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.
     
  5. Oct 27, 2005 #4
    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?
     
  6. Oct 27, 2005 #5
    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?
     
  7. Oct 27, 2005 #6
    No it is not 20%, why do you think that?
     
  8. Oct 27, 2005 #7
    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.
     
  9. Oct 27, 2005 #8
    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:
     
  10. Oct 27, 2005 #9
    Yes, it is right!
     
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