# Thread: Probability Distributions and Predicted Values

1. ## Probability Distributions and Predicted Values

WILL +REP AND THANK ALL WHO HELP!

There is an example for this problem, but like, it does not show an example of how to spit out all of these combinations.. it suggests using the tree diagram, but I don't really think the tree diagram would help...
So the example with the full solution is depicted in multiple screenshots below, I would just like someone to find a way that won't take me 25 minutes to do this one problem. It's not the probability that is getting me...it's the 32 equally likely outcomes....holy....

2. ## Re: Probability Distributions and Predicted Values

The purpose of this kind of problem is to make you understand the formula for and the assumptions behind the binomial distribution.

The assumptions are repeated trials, only two possible outcomes of a single trial, and independence of trials.

If n is the number of trials and each trial has exactly two possible outcomes, what is the number of possible outcomes? Obviously $2^n.$

It takes nowhere near 25 minutes to build that outcome chart in your problem. You will have $2^5 = 32$ outcomes. So using g and b to indicate the different outcomes, you start as follows

g g g g g g g g g g g g g g g g b b b b b b b b b b b b b b b b$\ \ \$ I wrote down $\dfrac{32}{2} = 16\ g's,\ then\ 16\ b's\ ONCE.$

g g g g g g g g b b b b b b b b g g g g g g g g b b b b b b b b$\ \ \$ Below that, I wrote down $\dfrac{16}{2} = 8\ g's, then\ 8 b's\ TWICE.$

g g g g b b b b g g g g b b b b g g g g b b b b g g g g b b b b$\ \ \$ Below that, I wrote down $\dfrac{8}{2} = 4\ g's, then\ 4 b's\ THRICE.$

g g b b g g b b g g b b g g b b g g b b g g b b g g b b g g b b$\ \ \$ Below that, I wrote down $\dfrac{4}{2} = 2\ g's, then\ 2 b's\ FOUR\ times.$

g b g b g b g b g b g b g b g b g b g b g b g b g b g b g b g b$\ \ \$ Below that, I wrote down $\dfrac{2}{2} = 1\ g, then\ 1 b\ EIGHT\ times.$

That is totally mechanical, no thinking required. Now reading down instead of across, I get a description of each possible outcome. The sixth column over reads downward ggbgb, meaning the first child was a girl, the second child was a girl, the third child was a boy, the fourth child was a girl, and the fifth child was a boy.

Now I can count how many g's and b's there are in each column and summarize how many columns have 5 g's, how many have 4 g's, how many have 3 g's, how many have 2 g's, how many have 1 g, and how many have no g's.

I get

1 case of 5 g's, which means 1 case of no b's.
5 cases of 4 g's, which means 5 cases of 1 b.
10 cases of 3g's, which means 10 cases of 2 b's.
10 cases of 2g's, which means 10 cases of 3 b's.
5 cases of 1 g, which means 5 cases of 4 b's.
1 case of no g's, which means 1 case of 5 b's.

If you are familiar with Pascal's triangle, you may see a pattern

Now there IS a formula for the related probabilities.

$\dbinom{n}{m} * p^m * q^{n - m}.$

So given that boys and girls have an equal probability of being born (which actually is not quite true), the probability of exactly three girls out of five births is

$\dbinom{5}{3} * \left(\dfrac{1}{2}\right)^3 * \left(\dfrac{1}{2}\right)^{(5 - 3)} = 10 * \dfrac{1}{8} * \dfrac{1}{4} = \dfrac{10}{32} \approx 31.25\%.$

3. ## Re: Probability Distributions and Predicted Values

Thank you, that was very informative and thorough. I really appreciate it. I did not realize I could do 2^n. Your way is so much more simple!! =]