Genetics-Two heterozygous brown-eyed (Bb)

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Discussion Overview

The discussion revolves around calculating the probability of three out of five children having blue eyes, given that both parents are heterozygous brown-eyed (Bb). The conversation includes various approaches to the problem, including the use of binomial expansion and probability calculations.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant initially suggests that the probability of three children having blue eyes is zero, as blue eyes are recessive.
  • Another participant proposes using binomial expansion, calculating the probability of blue eyes as 1/4, and attempts to derive the probability for three children having blue eyes.
  • A participant questions the squaring of the probability and notes that the calculation should consider the specific scenario of three out of five children.
  • There is a suggestion to find the probability of one child having blue eyes and how that relates to having five children.
  • One participant incorrectly states that the probability of one child having blue eyes is 20%, leading to confusion in subsequent calculations.
  • Another participant corrects this by reaffirming that the probability is actually 1/4 or 25% and encourages the use of the binomial equation for the calculations.
  • A later reply details the steps of using the binomial expansion to arrive at a final probability of approximately 8.78% for three children having blue eyes.
  • One participant confirms that the final calculation is correct.

Areas of Agreement / Disagreement

Participants express disagreement regarding the initial probability calculations, particularly the interpretation of the probability of blue eyes. There is no consensus on the initial approach, but later calculations using the binomial expansion receive agreement.

Contextual Notes

Some participants express confusion over the correct interpretation of probabilities and the application of the binomial expansion, indicating a need for clarity in the steps taken. The discussion reflects varying levels of understanding of probability concepts.

Who May Find This Useful

Individuals interested in genetics, probability theory, or those seeking assistance with similar homework problems may find this discussion beneficial.

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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