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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?

 May 5, 2015

Best Answer 

 #1
avatar+128475 
+16

 

 

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.....

 

  

 May 6, 2015
 #1
avatar+128475 
+16
Best Answer

 

 

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.....

 

  

CPhill May 6, 2015
 #2
avatar+906 
+3

?

probability?

 May 6, 2015
 #3
avatar+1833 
+1

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)

 May 6, 2015
 #4
avatar+1310 
+3

dunno, 50? or 70

 May 6, 2015
 #5
avatar+118609 
+11

 

$$\\$prob that they alternate$ \\\\
= 2\left(\frac{4}{8}\times\frac{4}{7}\times\frac{3}{6}\times\frac{3}{5}\times\frac{2}{4}\times\frac{2}{3}\times\frac{1}{2}\times\frac{1}{1}\right)\\\\
=2\times\frac{(4!)^2}{8!}$$

 

 $${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{4}}{!}}^{{\mathtt{2}}}}{{\mathtt{8}}{!}}} = {\frac{{\mathtt{1}}}{{\mathtt{35}}}} = {\mathtt{0.028\: \!571\: \!428\: \!571\: \!428\: \!6}}$$

 

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 :)

 May 6, 2015
 #6
avatar+128475 
+11

Hey......Melody and I got the same answer to a probability problem......!!!!!

 

 

  

 May 6, 2015
 #7
avatar+118609 
+6

YES it is proof - miracles really can happen !!   LOL

 May 7, 2015

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