# Probability Equation That I am missing here

1) What is the probability that out of giving birth to 6 kids, 3 of them will be boys? The answer is 50% chance? Don't ask me why but I just logically see that it is 50% so why?

2) But what about birthing 4 boys out of 6 kids? Would it be that the probability would be 2/6 which equals a 1/3 chance? Once again why?

3) What if you had 4 different sexes to choose from- a boy a girl and 2 other sexes, and you gave birth to 10. How would you find the probability of birthing 3 boys?

Basically what is the overall equation that I am missing here? Thank You!

Does it have to do with Multichoosing? Probably not but you can repeat the boy so repetition is allowed and order does not matter?

• Under standard assumptions, the answer to (1) is $\frac{5}{16}$. I suggest you think a bit more about this before attempting the others. Apr 14 '14 at 0:20
• I have one word to offer - Pascal's Triangle Apr 14 '14 at 0:22
• @JoeTaxpayer, you can offer two words if you like ;-) Apr 14 '14 at 0:23
• "Don't ask me why but I just logically see that it is 50% so why?" That's a dangerous way to think about math.
– crf
Apr 14 '14 at 0:45
• @Adam - There are times that an 'extreme' example helps clarify a situation. Say you have 1000 coin flips. The results form a nice bell curve of all the possible HT combinations. Chances of 1000H0T is near zero, but appears as a result, as does 499H501T, etc. The chance of 500H500T actually feels pretty slim to me. In fact, it's about 1 in 40. calculator.tutorvista.com/coin-toss-probability-calculator.html is a calculator to play with the numbers. Apr 15 '14 at 15:33 This is Pascal's Triangle. I was given a book about it when I was 6, and I fell in love with math. It's as simple as looking at the results of n coin flips, or boy/girl births, or as complex as you wish. Either way, it answers your question with no factorials or any tougher math.

Line 1 - Ignore

Line 2 - One child, 1/2 boy 1/2 girl

Line 3 - Two Children - 1/4 boy 2/4 BG 1/4 girl

continue

The 7th level is 6 kids. 20/64 (5/16) chance of 3B/3G

4B is 15/64

• So you're saying 6 choose 3 divided by 2^6 is my answer for 1? How?
Apr 14 '14 at 0:30
• See my edit. Does that help? Read up on Pascal's Triangle. Apr 14 '14 at 0:33
• @Adam, as a way to see why this is your answer: You have a sequence of six letters, each B or G corresponding to which gender the kid in the sequence is. If you want to count how many of those sequences have exactly 3 boys, you need to choose 3 out of 6 slots to put the B's, and put G in the others. In total there are 2^6 sequences since each of the six slots has 2 choices. This corresponds exactly to choosing the correct term in Pascal's triangle, and dividing by the sum of that term's row, which is always 2^n. Apr 14 '14 at 0:41
• Yes Joe it helps but I'm just trying to see how. I'm thinking through it at the moment
Apr 14 '14 at 0:53
• rVitale thank you for breaking it down further. Could any of you tell me how to apply this so that I can make my own equation just like river_o6 did? In particular I mean for answering number 3.
The equation you are looking for is $$\frac{\begin{pmatrix}N \\ X \end{pmatrix}(S-1)^{N-X}}{S^N}$$ where $S$ is the number of sexes, $X$ is the number of boys, and $N$ is the number of total births. The denominator is the number of different sexes for the $N$ birth, and the numerator is the number of different sexes given $X$ of them are boys and the rest are not.
• You can analyze this problem by thinking $P(sample) = \frac{\text{size of sample}}{\text{size of sample space}}$. For each birth, there are $S$ possible outcomes, hence the total number is $S^N$. For the numerator, you can first decide which $X$ births to be boy. There are $N$ choose $X$ choices. The rest births can choose from any other sexes but boy, therefore there are $(S-1)^{N-X}$ choices for the rest births, which results in the total number of different choices as $\begin{pmatrix} N \\ X \end{pmatrix} \times (S-1)^{N-X}$, given $X$ of the births are boys. Apr 14 '14 at 1:01
• Hi @Adam. I think the answer is yes. For your new question, $S=6$ since each dice has six faces; $N=5$ since there are five rolls; and $X=2$ since you are getting 6 two times. Therefore the probability is $$P=\frac{\begin{pmatrix} 5 \\ 2 \end{pmatrix} (6-1)^{(5-2)}}{6^5}$$. Apr 28 '14 at 16:05