Probability

Question

A bag of jelly beans contains 10 red, 6 green, 7 yellow, and 5 orange jelly beans.  

 

a) What is the probability of selecting a red jelly bean, replacing it, and then selecting an orange jelly bean?

b) What is the probability of selecting a green, then yellow, then green jelly bean if you eat them instead of putting them back?

Answer 1

There are 10+6+7+5 = 28 jelly beans.  If you pick a jelly bean then replace it, you still have 28 jelly beans.  If you eat the jelly bean rather than replace it, you have only 27 jelly beans left.

 

a)  P(red) = 10/28 = 5/14.  P(orange) = 5/28.  P(red, orange) = P(red)*P(orange) = 5/14 * 5/28 = 25/392

 

b)  P(green) = 6/28

     P(yellow) = 7/27

     P(green) = 5/26  (remember you ate one green one already so there are only 5 greens left)

 

     P(green, orange, green) = (6/28)*(7/27)*(5/26) = _______

Answer 2

10+6+7+5=28

a)10*5/28^2 = 25/392 = .0638

b)6*7*(6-1)/28*27*26 = 5/468 = .0107