Question 1086116:
An ice chest contains 8 cans of apple juice, 6 cans of grape juice, 7 cans of orange juice, and 8 cans of mango juice. Suppose that you reach into the container and randomly select three cans in succession. Find the probability of selecting three cans of orange juice.
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website! Question:
An ice chest contains 8 cans of apple juice, 6 cans of grape juice, 7 cans of orange juice, and 8 cans of mango juice. Suppose that you reach into the container and randomly select three cans in succession. Find the probability of selecting three cans of orange juice.
Solution:
This is NOT a case of binomial distribution because the probability of success in reaching for the first can of orange juice is different from that of the second, and so on.
This can be solved by logic, or by the hypergeometric distribution. We'll do both.
First by logic:
Total number of cans, N=8+6+7+8=29, out of which 7 are orange juice.
Probability of getting
1st can of orange juice = 7/29
2nd can of orange juice = 6/28
3rd can of orange juice = 5/27
Probability of success of all three steps can be found using the multiplication rule,
P(orange=3)=7/29*6/28*5/27=5/522 (=0.00958 approx.)
By Hypergeometric distribution:
N=29
O=7 (orange juice in ice chest)
M=29-7=22 (misc. in ice chest)
o=3 (orange juice picked)
m=0 (misc. juice picked)
P(o=3)=C(O,o)C(M,m)/C(O+M,o+m)
=C(7,3)C(22,0)/C(29,3)
=35*1/3654
=5/522 as before
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