I have 6 pieces of candy: two each of 3 different colors. They taste especially good if eaten two at a time, provided the colors are different.
These candies are in an opaque bag from which I pick two at a time. If they are different colors I eat them together (yum) but if they are the same I put them back in and draw again. I will repeat this process to eat two more.
What is the probability the last two candies will be of differing colors?
Repeat with two each of 4 colors.
Repeat with two each of 5 colors.
(In reply to
re: No Subject @ Charlie by Ady TZIDON)
I don't know what your "12 choices" are.
As armando pointed out, after the first choice of two unlike candies, there are four candies left. Call them, to be less confusing, colors a, b, c1 and c2, the last two matching.
Remember that if, on the second pair chosen, c1 and c2 are both found, they will be thrown back to start over, so there are 10 equally likely final choices of the second go-round:
ab, ac1, ac2, ba, bc1, bc2, c1a, c1b, c2a, c2b
leaving respectively:
c1c2, bc2, bc1, c1c2, ac2, ac1, bc2, ac2, bc1, ac1
In two of these 2 cases, what's left for the remaining two candies will be the match c1c2, and 2 our of 8 is 20%, or in other words 80% likelyhood that after the first two eating sessions the remaining last two candies will be non-matches.
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Posted by Charlie
on 2016-05-25 20:24:18 |
re(2): No Subject @ Charlie
