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(2): Second part by armando)
I don't understand
"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."
2d has only one outcome: 2s
1d 2s has two possible outcomes: 1d or 2s
4s has only one outcome: 2s
Or, if you're talking about cases of choices of the four candies, they are all 6, if you identify the candies in 2d, for example, as a1, a2, b1, b2, there are 6 combinations of these 4 taken 2 at a time.
Edited on May 26, 2016, 3:14 pm
|
Posted by Charlie
on 2016-05-26 15:13:50 |