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(3): Second part by Charlie)
Yes I mean cases or choices of the four candies.
If I have 2d (P=1/13) only in 4 of the 6 cases I'm going to eat the candies, in the other 2 cases I'm going to put them back in the bag (these 2 cases do not constitute a effective third event).
But if I have 1d 2s (P=8/13) I'm going to eat in 5/6 cases. The other case go back to the bag.
If I have 4s (P=4/13) I'll eat in 6/6 (all the cases).
I'd like to count the total number of possible sequences of three pairs of picks (a° c, b c°, b° d°) in order to know the total number of times that remain different color candies in the bag (a d).
I suppose that I have to consider the probability of each pattern 2d, (1d 2s), 4s (these are the possibles "outcomes" after the second pair have been picked), and the number (4, 5, 6) of effective picks for each of these patterns.
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Posted by armando
on 2016-05-26 16:51:32 |
re(4): Second part
