SOLUTION: Suppose 3 of the 12 bottles in a case of wine are bad. If you randomly select 2 bottles, what is the probability that both are good? Both are bad? One is good and one is bad?

Question 1009677: Suppose 3 of the 12 bottles in a case of wine are bad. If you randomly select 2 bottles, what is the probability that both are good? Both are bad? One is good and one is bad?
Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!

Question:
Suppose 3 of the 12 bottles in a case of wine are bad. If you randomly select 2 bottles, what is the probability that both are good? Both are bad? One is good and one is bad?

Solution:
Can be solved using a tree diagram.
Let
G=event that one of the bottles is good
B=event that one of the bottles is bad

There are four possible outcomes.
Initially, there are 12 bottles, out of which 3 are bad.
P(G)=9/12=3/4
P(B)=3/12=1/4

After a good one is drawn, there are 8 good and 3 bad bottles, so
P(GG)=(3/4)(8/11)=24/44
P(GB)=(3/4)(3/11)=9/44

After a bad bottle is chosen, there are 9 good and 2 bad bottles, so
P(BG)=(1/4)(9/11)=9/44
P(BB)=(1/4)(2/11)=2/44

Total probability = (24+9+9+2)/44=44/44=1 ...ok

Answer:
A. P(GG)=24/44=6/11
B. P(BB)=2/44=1/22
C. P(BG or GB)=(9+9)/44=9/22

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