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.
Trying a manual but simplified way I arrive to a slightly different result, with P=0.882 (which I fear will be wrong as I see Charlie's numbers).
The first pair leaves always 2s 2d in the bag. There are 26 possible ways to pick a second pair to eat.
The second pair leaves 2d or 1d-2s or 4s.
2/26 ways leave 2d in the bag; 16/26 1d-2s; 8/26 4s.
When 2d are left in the bag there are 8 ways to eat another pair, none of them give same color candies in the bag.
When 1d-2s are left there are 10 ways to eat another pair but two of them leave same color candies in the bag.
When 4s are left there are 12 ways with no same color at least.
The total number of ways is 2*8+16*10+8*12=272. But in 32 of them the two last candies have same color.
So P=272-32/272=0.882. (aprox)
It could be that I'm wrong because I'm considering two different outcomes for ab candies and ba candies.
Edited on May 25, 2016, 5:43 pm
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Posted by armando
on 2016-05-25 17:34:34 |
Second part
