Expert: Josh - 12/10/2009

Question
A package of M&M contains 35 pieces of candy : 5red, 12 yellow, 10 brown,& 8 orange.
Find the probability:
If two pieces of candy are taken from the bag-
A)both orange
B) one brown and one yellow
C) 1st a red and 2nd a brown

Answer
Hi Lisa,

This question is about selection without replacement. The best way to understand how the probabilities are calculated is by drawing pictures.

A) Initially, 8 out of 35 pieces are orange. Thus, the first pick has a probability of 8/35 of being orange. Having removed one "orange" candy, there are 7 orange pieces remaining out of a total of 34. So, for the second pick, the probability of being orange is 7/34. To get "orange" in the first pick AND another "orange" in the second pick, we multiply (8/35) by (7/34) to get the joint probability. Final answer is (8*7)/(35*34). Please use a calculator to get an approximate numerical answer if necessary.

B) is done using similar ideas. To end up with one "brown" and one "yellow", there are two possibilities to consider.
- 1st possibility: First pick is "brown" and second pick is "yellow".
Probability of "brown" for first pick is 10/35. Probability of "yellow" for second pick is 12/34. Thus, overall probability of this event is (10/35)*(12/34).

- 2nd possibility: First pick is "yellow" and second pick is "brown".
Probability of "brown" for first pick is 12/35. Probability of "yellow" for second pick is 10/34. Thus, overall probability of this event is (12/35)*(10/34).

- The overall probability of seeing 1 "brown" and 1 "yellow" is given by (10/35)*(12/34)+(12/35)*(10/34).

C) This time, the order of selection MATTERS. Initially, we have 5 "red" out of 35 pieces. Thus, the first pick has a probability of (5/35) of being "red". Subsequently, there are 10 "brown"'s remaining, so the second pick has a probability of (10/34) of being "brown". All in all, the probability of a "red" followed by a "brown" is (5/35)*(10/34).