Math question.

shilala

Lifer
Oct 5, 2004
11,437
1
76
Hi guys, long time lurker first time poster.
Love the show.

I know Dullard is the only guy smart enough to answer this, but I'm going to throw it out there anyway.
Here's the question...
Imagine that two brown eyed parents will produce four offspring.
The odds of the first child being born with blue eyes is 25% (4:1).
What are the odds of the second child being born with blue eyes?
What are the odds the third child will be born with blue eyes?
What are the odds that the fourth child will be born with blue eyes?
What are the odds that two children will be born with blue eyes?
What are the odds that three children will be born with blue eyes?
What are the odds that all four children will be born with blue eyes?
 

Heisenberg

Lifer
Dec 21, 2001
10,621
1
0
25%
25%
25%
25%^2
25%^3
25%^4

Edit: I'm assuming for the last 3 questions that there are only that number of children, not all 4 at once.
 

giantpinkbunnyhead

Diamond Member
Dec 7, 2005
3,251
1
0
too ambiguous! oh no! First we need to know if you need to account for dominant and recessive genes.

Then we need to know if, for example, the odds of the 2nd child being born with blue eyes: Is that IN ADDITION TO the first child having blue eyes, or independent of that?

 

Soccer55

Golden Member
Jul 9, 2000
1,660
4
81
Are we to assume that the probability of getting blue eyes is independent? If so, then each of the first 3 answers are 1/4. the last 3 use the binomial formula with n=4, p=1/4, q=3/4 with r=2,3,4.

-Tom
 

shilala

Lifer
Oct 5, 2004
11,437
1
76
Originally posted by: giantpinkbunnyhead
too ambiguous! oh no! First we need to know if you need to account for dominant and recessive genes.

Then we need to know if, for example, the odds of the 2nd child being born with blue eyes: Is that IN ADDITION TO the first child having blue eyes, or independent of that?

Part one of your question has already been covered by the stated 25%
Part two has been answered by my questions.
 

Heisenberg

Lifer
Dec 21, 2001
10,621
1
0
Originally posted by: kcthomas
Originally posted by: Heisenberg
25%
25%
25%
25%^2 * 75%^2
25%^3 * 75%^1
25%^4
See my edit. If we're talking about all 4 at once, then what you have is right, assuming you're talking about a specific set of children.
 

giantpinkbunnyhead

Diamond Member
Dec 7, 2005
3,251
1
0
OK then... we need to compare the qualifying outcomes to the total possible outcomes.

For two children to have blue eyes, there are six possibilities:
1,2
1,3
1,4
2,3
2,4
3,4

For 4 children, there are 16 POSSIBLE outcomes:
no one has blue eyes;
only one has blue eyes (1 possibility for each birth order, for 4 possibilities)
two have blue eyes: (6 possibilities, listed above)
three have blue eyes (1,2,3; 1,2,4; 1,3,4; 2,3,4)
all four have blue eyes.

So we have 16 outcomes, 6 of which result in 2 blue-eyed babies. 37.5%.

I forget the formulas behind it but I"m sure someone previously nailed it.


EDIT! Somehow I overlooked this tiny thing called probability. This post shall now serve as an example of how NOT to calculate statistics!
 

QED

Diamond Member
Dec 16, 2005
3,428
3
0
If I understand the question properly, then the answer is:

The odds that the second child will have blue eyes are the same that the first child will have blue eyes, 25%. Same thing with the third and fourth child-- they each have a 25% chance of being born with blue eyes.

The odds that exactly two children will have blue eyes is
(.25*.25*.75*.75) + (.25*.75*.75*.25) + (.25*.75*.25*.75) + (.75 * .25 * .25 * .75) + (.75 * .25 * .75 * .25) + (.75 * .75 * .25 * .25) = .2109375, or about 21%

The odds that exactly three children will have blue eyes is:

(.25 *.25 * .25 *.75) * 4 = .046875, or about 4.7%.

The odds that exactly four childred out of four will have blue eyes is:

.25 * .25 * .25 * .25 = 1/256, or about 0.39%