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An irs person randomly selects 3 tax returns without replacement from 47 returns of which 5 contain errors.
What is the probability that she selects none of those containing errors? Answer in decimal form.
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So, we have 47 items, in all.
Of them, there are 5 special items.
We select 3 arbitrary items from these 47 items.
The question is: what is the probability that we select none of those 5 special items.
Solution
In all, we can select = = 16215 different triples of 47 item, taking 3 at a time.
It is the entire space of events.
How many triples can we select avoiding these 5 special items ? - It means that we select triples
from 47-5 = 42 items of the complementary set.
The number of these "ideal" (or favorable) triples is therefore = = 11480.
Therefore, the probability under the question is = 0.708 (with 3 correct decimal places), or 70.8%.
Solved.
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On Combinations, see the lessons
- Introduction to Combinations
- PROOF of the formula on the number of Combinations
- Problems on Combinations
- OVERVIEW of lessons on Permutations and Combinations
in this site.
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- ALGEBRA-II - YOUR ONLINE TEXTBOOK.
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