"At least" probability

Txroadrunner

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Apr 27, 2013
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Hi, All,
I can't figure out how to work this problem... I've been messing with it for hours now... Here it is:
A customer from Cavallaro's Fruit Stand picks a sample of 4 oranges at random from a crate containing 70 oranges, of which 5 are rotten. What is the probability that the sample contains 1 or more rotten oranges? (Round your answer to three decimal places.)
S= 75C5 E=6C5 F= 69C5 Where S is sample space, E is picking rotten oranges, and F is picking good oranges.
From that, I did this: E*F/S. It didn't work... Am I setting it up wrong? Can you show me the right way to set it up?
 
I would set it up as follows:

Let X be the event that he picks at least 1 rotten orange, and Y be the event that he picks no rotten oranges. It is certain that one of the events will occur. We may then state:

\(\displaystyle P(X)+P(Y)=1\)

\(\displaystyle P(X)=1-P(Y)\)
 
And the angels sang!

Finally! Thanks so much... just seeing how to look at it differently made all the difference.:p
 
A customer from Cavallaro's Fruit Stand picks a sample of 4 oranges at random from a crate containing 70 oranges, of which 5 are rotten. What is the probability that the sample contains 1 or more rotten oranges? (Round your answer to three decimal places.)


\(\displaystyle \displaystyle{ {\frac{\binom{66}{4}}{{\binom{70}{4}}}}}\) is the probability that none of the four is rotten.

What does that have to do with this question?
 
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