finding the # of possibilities for a probability distribution

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lldwarrior

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We are to find the probability that a family having 4 children will have all girls. He said to list all the possibilities, i.e. BBBG, BBGB, BGBB, BBBB...etc. Is there an easier way to figure out the # of possibilities without mapping it out like this? I am sure to miss a combination if I have to do it that way!!:-x
 
thank you!

Thank you very much for your help!
JeffM said:
The number of possibilities is 24 = 16 because there are two possibilities for each child and 4 children.

You will not miss a possibility if you use an orderly process for writing them down.

Start with BBBB (you could equally well start with GGGG).
On the next one, write down BBBG, which is the only other possibility if the first three children are boys.

On the third one write down BBGB

On the fourth BBGG. The third and fourth entries are all that is possible if the first two children are boys and the third is a girl

On fifth BGBB

You take it from here. Once you know the total number and set up this process you can hardly miss a possibility.
Click to expand...
 
Hello, lldwarrior!

We are to find the probability that a family having 4 children will have all girls.
He said to list all the possibilities, i.e: BBBG, BBGB, BGBB, BBBB, ... etc.
Is there an easier way to figure out the # of possibilities without mapping it out like this?
I am sure to miss a combination if I have to do it that way!
Click to expand...

There is a mechanical routine guarenteed to give you all the possibilities.

First, note that there are: \(\displaystyle 2^4 = 16\) possible outcomes.


In the first column, write 8 B's, then 8 G's.
. . That is, "change every eight terms."

. . \(\displaystyle \begin{array}{c}1 \\ \hline B \\ B \\ B \\ B \\ B\\B\\B\\B\\G \\ G \\ G \\ G\\G\\G\\G\\G \end{array}\)


In the second column, write 4 B's, 4 G's, 4 B's, 4 G's.
. . That is, "change every four terms."

. . \(\displaystyle \begin{array}{c}2 \\ \hline B \\ B \\ B \\ B \\ G \\ G \\ G \\ G \\ B \\ B \\ B \\ B \\ G \\ G \\ G \\ G \end{aray}\)


In the third column, write 2 B's, 2 G's, 2 B's, 2G's, 2B's, 2 G's, 2B's, 2G's.
. . That is, "change every two terms."

. . \(\displaystyle \begin{array}{c}3 \\ \hline B \\ B \\ G \\ G \\ B \\ B \\ G \\ G \\ B\\B\\G\\G\\B\\B\\G\\G \end{array}\)


In the fourth column, alternate the B's and g's.
. . That is, "change every one term."

. . \(\displaystyle \begin{array}{c}4 \\ \hline B\\G\\B\G\\B\\G\\B\\G\\B\\G\\B\\G\\B\\G\\B\\G \end{array}\)


Combine them in one chart:

. . \(\displaystyle \begin{array}{cccc}1 & 2 & 3 & 4 \\ \hline B&B&B&B \\ B&B&B&G \\ B&B&G&B \\ B&B&G&G \\ B&G&B&B \\ B&G&B&G \\ B&G&G&B \\ B&G&G&G \\ G&B&B&B \\ G&B&B&G \\ G&B&G&B \\ G&B&G&G \\ G&G&B&B \\ G&G&B&G \\ G&G&G&B \\ G&G&G&G \end{array}\)


Each row is one of the 16 possible orders for the genders.
 
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