Thomas bought a bag of jelly beans that contained 10 red jelly beans, 15 blue jelly beans, and 12 green jelly beans. What is the probability of Thomas reaching into the bag and pulling out a blue or green jelly bean and then reaching in again and pulling out a red jelly bean? Assume that the first jelly bean is not replaced.

A. 1/100

B. 3

C. 5/74

D. 0

Question

Sarah Browning

Mathematics

12 June, 20:21

Thomas bought a bag of jelly beans that contained 10 red jelly beans, 15 blue jelly beans, and 12 green jelly beans. What is the probability of Thomas reaching into the bag and pulling out a blue or green jelly bean and then reaching in again and pulling out a red jelly bean? Assume that the first jelly bean is not replaced.

A. 1/100

B. 3

C. 5/74

D. 0

+1

Answers (1)

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Harper Hutchinson · Answer 1

There are 37 jelly beans so p for the first one is 27/37. That leaves 36 in the bag. So p for the second jelly bean is 10/36=5/18.

The combine probability is 27/37*5/18=3/37*5/2=15/74.