A box contains 4 white b***s and 4 black b***s. I draw them out of the box, one at a time. What is the probability that all of my draws alternate colors?
Mellie.....others on this site (Melody, for instance) are better at these than I am.....but I'll venture an answer
I see all the white b***s as being indistiguishable from each other....and the same with the black b***s
So.....this amounts to the same thing as the number of identifiable "words" we could form with, say, 4 e's and 4 h's'
And this is given by 8! / (4! * 4!) = 70
There are only 2 ways to draw alternate colors .... BWBWBW.......etc. or WBWBWB......etc.
So.....the probability of alternating colors on the draws is just 2 / 70 or 1 / 35
If this isn't correct......maybe wait until Melody comes on later.....
Mellie.....others on this site (Melody, for instance) are better at these than I am.....but I'll venture an answer
I see all the white b***s as being indistiguishable from each other....and the same with the black b***s
So.....this amounts to the same thing as the number of identifiable "words" we could form with, say, 4 e's and 4 h's'
And this is given by 8! / (4! * 4!) = 70
There are only 2 ways to draw alternate colors .... BWBWBW.......etc. or WBWBWB......etc.
So.....the probability of alternating colors on the draws is just 2 / 70 or 1 / 35
If this isn't correct......maybe wait until Melody comes on later.....
Even though you might think others are better than you, you are still very intelligent. (Not putting the others down, just saying that you are very smart)
$$\\$prob that they alternate$ \\\\
= 2\left(\frac{4}{8}\times\frac{4}{7}\times\frac{3}{6}\times\frac{3}{5}\times\frac{2}{4}\times\frac{2}{3}\times\frac{1}{2}\times\frac{1}{1}\right)\\\\
=2\times\frac{(4!)^2}{8!}$$
$${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{4}}{!}}^{{\mathtt{2}}}}{{\mathtt{8}}{!}}} = {\frac{{\mathtt{1}}}{{\mathtt{35}}}} = {\mathtt{0.028\: \!571\: \!428\: \!571\: \!428\: \!6}}$$
I times it by 2 because either the black can come first OR the white can come first
CPhill and I got the same answer - how about that !
CPhill's logic is different from mine but both make perfect sense to me :)