A jar contains 60 candies, 10 reds, 20 blues and 30 yellows. If one takes out candies one by one, what is the probability that there is at least one yellow and one blue left after all the red candies have been taken out?
The solution is a combination of two mutually exclusive probabilities.
1) Probability that yellow is the 60th candy and blue is the last candy among the bunch of 10 reds and 20 blues.
OR
2) Probability that blue is the 60th candy and yellow is the last candy among the bunch of 10 reds and 30 yellows.
1a. The Probability that one of the 30 yellows is the last candy among 60 candies = 30/60
1b. The Probability that one of the 20 blues is the last candy among 30 candies = 20/30
2a. The Probability that one of the 20 blues is the last candy among 60 candies = 20/60
2b. The Probability that one of the 30 yellows is the last candy among 40 candies = 30/40
The first probability is a joint probability (‘AND’ rule) of 1a and 1b
The second probability is a joint probability of 2a and 2b
The final probability is the sum (‘OR’ rule) of the two.
(30/60) x (20/30) + (20/60) x (30/40) = 0.58
