SOLUTION: There are 12 items, 2 of which are special. An individual chooses 3 items.
a)Probability of getting exactly 1 special item?
My attempt: (2/12)+(10/11)+(9/10) is the probability o
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-> SOLUTION: There are 12 items, 2 of which are special. An individual chooses 3 items.
a)Probability of getting exactly 1 special item?
My attempt: (2/12)+(10/11)+(9/10) is the probability o
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Question 830625: There are 12 items, 2 of which are special. An individual chooses 3 items.
a)Probability of getting exactly 1 special item?
My attempt: (2/12)+(10/11)+(9/10) is the probability of getting exactly 1 special item.
My problem: This applies if the individual chooses 1 item, and then without replacement chooses another, and then another. However, if the individual were to choose 3 items all at once, which is what the question asks, I don't know what the answer would be.
b)What is the probability that the individual gets at least one special item?
My attempt: Probability of getting 1 special + P(2 special items)
My problem: I'm sure there's a faster method than my attempt. And part a's problem applies where I don't know how to calculate probability of getting exactly 1 or 2 special items. Answer by reviewermath(1029) (Show Source):
You can put this solution on YOUR website! Q:
There are 12 items, 2 of which are special. An individual chooses 3 items.
a)Probability of getting exactly 1 special item?
b)What is the probability that the individual gets at least one special item?
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A:
This is a hypergeometric distribution problem.
a. P(X = 1) = =
b. P(X ≥ 1) = =
