Student and Donald Duck

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Hi to all!
I don`t know to do this tasks, so I would much apriciate if someone can help

1) Student needs to answer on n independent questions (n is even). Probability that he`ll give the correct answer is P1 for first n/2 questions, and P2 is probability that he`ll give the correct answer for the following questions (n/2). Z is total number of correct answers. Determine the distribution of random variable Z.

2) Donald Duck had hidden $100 bill in one of three books that he currently reads, but he forgot in which one. Pi is the probability that he will find the bill by quickly browsing the i-th book (i = 1,2,3) if the bill really is in that book (it is possible Pi <1).Let`s suppose that Donald Duck quickly browsed through the book and didn`t found the bill. What is the probability that bill really is in that book?

Desperately need help Sad

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Type: Question • Score: 2 • Views: 2,818 • Replies: 9

[+3] - engineer - 01/30/2012I'm not sure how to answer number 1. The mean number of correct answer from the first half is P1(n/2) and the mean for the second half is P2(n/2) so the middle of the distribution for the entire...

@Sheldon1,

I'm not sure how to answer number 1. The mean number of correct answer from the first half is P1(n/2) and the mean for the second half is P2(n/2) so the middle of the distribution for the entire test is (P1+P2)(n/2) but describing the distribution is going to be hard. Each half forms a normal distribution where the standard deviation is sqrt[ (P)(1-P)(n/2) ], but if P1 and P2 are radically different, these two distributions wouldn't even overlap. The sum of two normal distributions is not easily described.

Number 2 is straightforward. The probability the bill is in the book is 1/3. The probability that he will miss the bill is (1-Pi). The probability it is in the book and he will miss it is (1-Pi)/3

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@engineer,

Hmm...I think the first problem is binome distribution. So we have:
X:B(n/2,P1) and Y:B(n/2,P2), and from that Z=X+Y.
...that`s what I have so far, but I don`t know further...

For the second one...hmmm...i don`t have clue...I understand the part vith 1/3 (`couse there are 3 books)...but I don`t understand the rest...is there some formula that you uesd or it can be written a bit detailed?

Btw, tnx for hleping, it means me very much!

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@Sheldon1,

It's been a long time since I took stats (although I use them daily) so I'm not sure of your nomenclature but I think you are on the right track. The distribution is going to be a combination of two normal curves with the tighter distribution rising out of the middle on the wider one.

The logic on the second one runs like this:
- If the probability of finding the bill in book i (assuming it is in that book) is Pi, then the probability of not finding the book is (1-Pi). Two things have to happen to get your answer. The bill has to be in the book and Donald must fail to find it so the probability is the product of those two probabilities (1/3)(1-Pi).

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@Sheldon1,

Quote:

2) Donald Duck had hidden $100 bill in one of three books that he currently reads, but he forgot in which one. Pi is the probability that he will find the bill by quickly browsing the i-th book (i = 1,2,3) if the bill really is in that book (it is possible Pi <1).Let`s suppose that Donald Duck quickly browsed through the book and didn`t found the bill. What is the probability that bill really is in that book?

It's only 3 books, so forget probability just take the books hold them upside down and shake. If there's any money to be found in there it'll happen a lot faster this way than by casual browsing. You may receive a failing grade for my answer; yet, having found the money will make you a winner.

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Scrooge McDuck would find money in all the books.

http://gnomeaggedon.files.wordpress.com/2009/07/scrooge-mcduck-christmas-carol.jpg

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What is the probability Donald Duck is alive, reading and has $100 to hide...

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@Ceili,

If Donald Duck had $100, he'd need to find someplace like a book to hide it, because he can't put it in his pocket. After all, HE HAS NO PANTS!

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@engineer,

I don't think it's quite that straightforward. I think this is a case of conditional probability and requires Bayes' theorem. Assuming the bill is equally likely to be in any of the three books:

Let A = The bill is in book i.
Let B = Donald browsed book i and didn't find the bill.

We want to know P(A) given B (which is P(A|B)).

Bayes' theorem:
P(A|B) = P(B|A) * P(A) / P(B)

P(A) = 1/3
P(B) = [(1-P1) + (1-P2) + (1-P3)] / 3
= 1 - (P1+P2+P3)/3
P(B|A) = 1-Pi

Therefore,
P(A|B) = (1-Pi) * (1/3) / [1 - (P1+P2+P3)/3]

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@joefromchicago,

The McDuck clan uses the books to hide money, not find it.

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Student and Donald Duck

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