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How do you go about solving problems like this? Is there a trick to it?
A class consists of 25 students, of whom 10 are women and the rest are men. We take a random sample of 5 students from this class. (Without replacement)
-What is the probability that the sample will include at least one woman?
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Hi;
You use the hypergeometric distribution.
You will have to sum 5 different terms all involving the hypergeometric distribution. In summation notation it looks like this:
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Is it safe to assume that we use the formula only if it has "at least" in the question?
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Hi;
Is it safe to assume that we use the formula only if it has "at least" in the question?
An "at least" in the problem implies a range of values in this case1 to 5. So, you will have to sum 5 hypergeometric terms that is why there is a sum there.
This is the formula:
You could us for say exactly 5 women by saying low = 5 and high = 5. Then you would be summing only one term.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Thank you, Bobby.
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Your welcome. May the computing be with you.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi, another question:
An instructor gives an exam with 14 questions. Students are allowed to choose any 10 of them to answer. Suppose 6 questions require proof and 8 do not:
-How many groups of 10 questions contain at least one that require a proof?
I got 1736 as an answer. Is that correct?
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Hi genericname;
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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