A bag holds 20 yellow mints and 80 other green or pink mints. You choose a mint at random, eat it, and choose another.
a. Find the number of pink mints if P(yellow then pink) = P(green then yellow).
b. What is the least number of pink mints if P(yellow then pink) > P(green then yellow)?
There are
20 yellow
x green
80-x pink
Total 100
P(YP)=(20/100)(80-x)/99
P(GY)=(x/100)(20/99)
If P(YP)=P(GY) then
20(80-x)=20x after cancelling identical denominators
Solve for x:
x=1600/40=40
So there are 40 green and 40 pink mints.
If P(YP)>P(GY) then
20(80-x)>20x
40x<1600
x<40, or pink=(80-x)>40
To solve this problem, we will use the concept of probability.
a. Find the number of pink mints if P(yellow then pink) = P(green then yellow):
Let's assume that the number of pink mints is P. The probability of selecting a yellow mint and then a pink mint is given by:
P(yellow then pink) = (number of yellow mints / total number of mints) * (number of pink mints / total number of mints)
= (20 / (20 + 80)) * (P / (20 + 80))
Similarly, the probability of selecting a green mint and then a yellow mint is given by:
P(green then yellow) = (number of green mints / total number of mints) * (number of yellow mints / total number of mints)
= (80 / (20 + 80)) * (20 / (20 + 80))
Given that P(yellow then pink) = P(green then yellow), we can set up the equation:
(20 / (20 + 80)) * (P / (20 + 80)) = (80 / (20 + 80)) * (20 / (20 + 80))
Simplifying this equation, we get:
20P = 80(20)
P = 80
Therefore, the number of pink mints in the bag is 80.
b. What is the least number of pink mints if P(yellow then pink) > P(green then yellow):
Using the same approach as before, we need to find the minimum number of pink mints, such that:
(20 / (20 + 80)) * (P / (20 + 80)) > (80 / (20 + 80)) * (20 / (20 + 80))
Simplifying this inequality, we get:
20P > 80(20)
P > 80
Therefore, the least number of pink mints must be greater than 80 to satisfy the condition where P(yellow then pink) > P(green then yellow).