# Homework Help: Genetics-Two heterozygous brown-eyed (Bb)

1. Oct 26, 2005

### jena

Hi,

My Question:

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

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

Thank You

2. Oct 26, 2005

### jena

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

3. Oct 26, 2005

### Moonbear

Staff Emeritus
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.

4. Oct 27, 2005

### amcavoy

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?

5. Oct 27, 2005

### jena

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?

6. Oct 27, 2005

### gerben

No it is not 20%, why do you think that?

7. Oct 27, 2005

### amcavoy

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.

8. Oct 27, 2005

### jena

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

9. Oct 27, 2005

### gerben

Yes, it is right!