# Thread: [SOLVED] Gender distribution. Wrong (official) solution?

1. ## [SOLVED] Gender distribution. Wrong (official) solution?

Couple are planning four kids. What is more likely: both genders equal (2 boys, 2 girls) or 3 of one gender and 1 of the other gender?
$P(\text{2 boys, 2 girls})=\frac{6}{16}$ I agree with that.

But is $P(\text{3 boys, 1 girl or vice-versa})$ really $\frac{8}{16}$? I'd say it is rather $\frac{{4\choose 1}}{16}=\frac{4}{16}$.

How could they get $8$ in the numerator?!

Is there already any [tex] keyboard shortcut? I vote for FN/CTRL+M.

2. Isn't the combination of 3 one gender and 1 the other gender 4C1*3C3? However that only takes into account ONE gender combo.

3. In this one, there ARE only 16. Enumerate them.

4 vs 0
BBBB
GGGG

3 vs 1
BBBG
BBGB
BGBB
GGBG
GGGB
BGGG
GBBB
GBGG

2 vs 2
GBBG
GBGB
GGBB
BGBG
BGGB
BBGG

Of course, had you pondered one important thing, you would have kept trying. After counting 2 vs 2, you could have counted 4 vs 0 and subtracted, since there are only three types. This would have discouraged your result.

Further, a little binomial expansion would reveal it.

$(B+G)^{4}\;=\;B^{4}+4B^{3}G+6B^{2}G^{2}+4BG^{3}+G^ {4}$

It's pretty obvious from that.

The REAL lesson is that there are many ways to proceed. If you know only one way, you will not have a backup plan or an alternative approach.

4. Originally Posted by courteous
$P(\text{2 boys, 2 girls})=\frac{6}{16}$ I agree with that.

But is $P(\text{3 boys, 1 girl or vice-versa})$ really $\frac{8}{16}$? I'd say it is rather $\frac{{4\choose 1}}{16}=\frac{4}{16}$.

How could they get $8$ in the numerator?!

Is there already any [tex] keyboard shortcut? I vote for FN/CTRL+M.
The number of girls has a binomial distribution B(4,0.5)

So probability of exactly two girls is:

$p_1=b(2;4;0.5)=\frac{4!}{2!2!}0.5^4=\frac{3}{8}$

Probability of three of one gender and the one of the other gender is equal to the sum of the probability of exactly one girl and of exactly three girls:

$p_2=b(1;4,0.5)+b(3;4,0.5)=1/2$

CB

5. Hello, courteous!

A couple is planning on four chldren.
What is more likely: 2 boys and 2 girls, or 3 of one gender, 1 of the other?
If you're not convinced, we can LIST the possible outcomes.

. . $\begin{array}{c} \text{Two of each} \\ \hline BBGG \\BGBG \\ BGGB \\ GBBG \\ GGBB \\ GBGB \end{array} \qquad \text{ 6 ways}$

. . $\begin{array}{c}\text{Three and one} \\ \hline
BBBG \\ BBGB \\ BGBB \\ GBBB \\ BGGG \\ GBGG \\ GGBG \\ GGGB \end{array} \qquad \text{ 8 ways}$

6. Originally Posted by courteous
$P(\text{2 boys, 2 girls})=\frac{6}{16}$ I agree with that.

But is $P(\text{3 boys, 1 girl or vice-versa})$ really $\frac{8}{16}$? I'd say it is rather $\frac{{4\choose 1}}{16}=\frac{4}{16}$.

How could they get $8$ in the numerator?!

Is there already any [tex] keyboard shortcut? I vote for FN/CTRL+M.
Hi courteous,

2 boys, 2 girls

arrange 4, with 2 sets of 2 the same $\Rightarrow\ P(2 girls)=\frac{\left(\frac{4!}{2!2!}\right)}{4!}$

3 boys, 1 girl

arrange 4, with a set of 3 the same $\Rightarrow\ P(3 boys)=\frac{\left(\frac{4!}{3!}\right)}{4!}$

3 girls, 1 boy.... same calculation.

You neglected to count both cases.

7. I should have counted the other gender "combination" (instead of trying to dispute their solution). Thank you all for your help, I was surprised to see 5 replies. I also like TKHunny's quick binomial way of showing it.