Scholarship Probability

[Suspence]

I was reading "Ask Marilyn" recently, and there was a question about probability. I'm interested in probability, but not very good at it, and I didn't really understand Marilyn's answer, though I trust she is correct.

I'm sure someone here could answer it in a way where I understood it much better. I'll restate the question here without her answer, so that it also poses as a puzzle

A student is one of five finalists for a grant that will award a scholarship to two of the finalists. That same student is also one of five finalists for a separate grant that awards only one scholarship.

What is the probability that she will win at least one of the scholarships? Assume the two grants are unrelated, i.e. a student could potentially win a scholarship from both grants.
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[Trojan Horse]

I'm not going to put this one in spoiler tags.

Here's now I would explain it to my math students. A better question than "what is the probability she wins at least one scholarship" is "what is the probability she wins NO scholarships?" She has a 3/5 chance of losing out on the first scholarship, and a 4/5 chance of losing out on the second scholarship. So she has a (3/5)(4/5)=12/25 chance of winning no scholarships.

That leaves a 1-(12/25)=13/25 chance that she wins at least one scholarship.

How did Marilyn do it?

[Suspence]

That makes better sense to me. Marilyn also went backward, but her description seemed more convoluted.

For the record, Marilyn's answer was:
An easier way to figure out the answer is to go backward. The probability your daughter will not win one of the Grant A awards is 80 percent (4 in 5). Say that occurs. As the probability she will not win the other one is 75 percent (3 in 4), the likelihood she won’t win either one is 60 percent (.80 x .75 = .60). Say that occurs, too. The probability she won’t win Grant B is 80 percent (4 in 5), so the likelihood she won’t win any grant at all is 48 percent (.60 x .80 = .48). This means she must have a 52 percent chance of winning at least one.

What would be the process for answering the question forward based on the way it was phrased, rather than the finding the chances of winning nothing and working backwards?

As an aside, I'm currently re-reading The Drunkard's Walk and I'm in the section on probability. I'm really enjoying it.
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I hate people who try to write interesting things in their signature.

[novice*]

Forward answer: 1/5 chance of scolarship A, plus 2/5 of scolarship B = 3/5 total. But now we've counted the case of winning borh A and B twice, so we subtract it once. Chance of that is 1/5 * 2/5 = 2/25, so final answer is 3/5 - 2/25 = 13/25.

[Zag]

I posted this probability problem on Braingle a while back. The solution to #2 uses the same approach -- that is, find the chance that all five games on the ticket are different, and then subtract from 1. Note that novice's approach becomes quite a bit trickier where there are more than two opportunities.

[referee]

Yeah. Zag's #1 is really asking how many combinations there are. That is, "the number of combinations of six, taken out of forty-two". Since that is a mouthful, we write 42C6, and that evaluates to (42*41*40*39*38*37)/(6*5*4*3*2*1) (note that there are six numbers on the numerator (Namely, the first pick may come from 42 different numbers, second one from 41, because no repeats, and so on. We divide by 6! because the order in which the numbers were picked won't matter). That yields 5,245,786. Since that's another mouthful, I'll call that number 'n'.

So, the answer to the first question is 1/n. The answer to the second question is easier with the working backwards. the probability of two games being different is p = 1 - 1/n. And the probability of the third being different to the other two is q = 1 - 2/n, and so on. The final answer evaluates to p* = (n-1)(n-2)(n-3)(n-4) / n^4 For all five games being different (only four, because the first one is guaranteed to be different to all the previous ones), or 1 - p* for the answer you're being asked to find.

Number three involvings finding the moment t in which (n-1)(n-2)...(n-t) / n^t falls below 1/2, and then the answer is t+1.
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