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: Second part by Charlie)
Thanks for the test and for the text.
It is possible I'm misunderstanding below your solution post containing the formulas. If that would be the case I apologize.
Four colours candies: in the beginning there are 4d and 0s.
First valid choose: Applying your third formula the P of going from 4d to (2d 2s) is 1.
Second valid choose: From (2d 2s) applying your formulas the P of going to:
2d=1/13 (first formula)
1d 2s=8/13 (second formula)
4s=4/13 (third) Total P=1
Third valid choose:
From 2d to 2s: P=1 (third formula)
From (1d 2s) to 2s: P=0.8 (second formula)
From 4s to 2s: P=1 (first formula)
Then to finish with 2s in the bag P=1/13*1+8/13*0,8+4/13*1=0,87692 (1)
But I don't find this P when I try to do it manually (counting global outcomes and substracting those who finish with the same color in the bag). This is because 2d, (1d 2s) and 4s have each of them a different number of outcomes (4, 5 and 6 respectively), despite the fact that they are always four candies.
|
|
Posted by armando
on 2016-05-26 10:07:33 |
re(2): Second part
