##### About M&Ms and probability

A bowl contains a number of m&m’s, with equal numbers of each color. Adding 20 m&m’s of a new color to the bowl would not change the probability of drawing (without replacement) two m&m’s of the same color.

How many m&m’s are in the bowl before the extra m&m’s are added?

*………………Answer Preview…………..*

*This is under conditional probability*

*The formula is*

*P(B/A)= P(A and B)/P(A)*

*For the probability to remain the same after adding 20 m&m’s of a new color, then, the other colors must have been in groups of 20’s.*

*Thus, we can assume that there were 20 blue, 20 red and we added 20 green*

*The probability will thus be*

*P(B/A)= P(A and B)/P(A)*

*The total number of m&m’s in this case would be…*