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3/27/05 2:11 AM
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tower1
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Edited: 27-Mar-05
Member Since: 02/24/2003
Posts: 378
 
I think I'm losing hair over this question and I need your help before I go bald, not that there's anything wrong with that:) "A multiple-choice examination has 20 questions each having five possible answers, only one of which is correct. Suppose one of the students who takes the examination answers the questions by guessing. What is the probability that he answers at least 15 correctly? What is the probability he answers exactly 8 correctly? How do I do this?
3/27/05 3:56 PM
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Dogbert
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Edited: 27-Mar-05
Member Since: 01/01/2001
Posts: 14755
The probability that a certain answer is answered correctly is P=.2 The probability that this question is answered wrongly is Q=1-P=.8 There are [n!]/[(n-r)!] ways to get a sample of size r out of a population with n elements. I guess you've learned the formula already. So we have [20!/12!] ways to get exactly 8 out of 20 right. The probability of each of these cases is Q8*P12. So the answer to the second problem is [20!/12!]*Q8*P12. For the first question you simply calculate the same problem for 20, 19, 18, 17, 16, 15 being right and add these probabilities. I don't have an easier way at hand.
3/27/05 6:27 PM
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Andrew Yao
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Edited: 27-Mar-05 06:28 PM
Member Since: 01/01/2001
Posts: 2736
Just to add more detail at each step: C(20,n) is the number of ways to pick any n correct answers out of 20 questions. The first thing you ask yourself when trying to figure out the number of ways to do something is to ask "does order matter, and is there replacement?" See http://en.wikipedia.org/wiki/Combinations_and_permutations You know to use combinations in this case because there is no order(we don't care if I get the first one right and the second one wrong, or vice versa) and no replacement(after you've gotten a question right or wrong, it's not considered again). Now we know how many ways you can get n correct, and we need to find the probability of any one of these occuring. The probability that a given way of choosing exactly n correct answers will occur randomly is .2^n * .8^(20-n). You can think of it like this. What if there were only 2 questions instead of 20? Then the possible outcomes are {correct, correct} , {correct, incorrect}, {incorrect, correct}, and {incorrect, incorrect}. The probabilities of these are .2*.2, .2*.8, .8*.2, and .8*.8, respectively. In general, the probability of a given outcome would be .2^(number correct) * .8^(number incorrect) So now we know there are C(20, n) ways to pick exactly n correct answers, and the probability of a given one of these ways occuring is .2^n * .8^(20-n). The probability that any one of the ways will occur is the sum of the probabilities that a given one of the ways will occur. = C(20, n) * .2^n * .8^(20-n)
3/27/05 8:23 PM
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tower1
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Edited: 27-Mar-05
Member Since: 02/24/2003
Posts: 379
Thanks alot guys... That was very helpful. I'm sure I'll have some more questions soon. Just curious, is anyone familiar with continuous random variables and exponential continuous random variables?

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