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 :)