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bmwhype2
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prob - guesing 25 Feb 2008, 10:11
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A student has answered 7 of the 10 questions on the true-false test correctly and has decided to guess randomly on the rest. If all of the questions are equally weighted, what is the probability of the student receiving a score of 80% or more on the test?
(A) 0.125
(B) 0.5
(C) 0.7
(D) 0.8
(E) 0.875



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prob - guesing 25 Feb 2008, 10:14
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bmwhype2
A student has answered 7 of the 10 questions on the true-false test correctly and has decided to guess randomly on the rest. If all of the questions are equally weighted, what is the probability of the student receiving a score of 80% or more on the test?
(A) 0.125
(B) 0.5
(C) 0.7
(D) 0.8
(E) 0.875
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E: he has to answer at least one question out of three remaining correctly. 1 - (1/2)^3 = 7/8

(1/2)^3 - probability that he will mess up on each of 3 remaining questions.
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prob - guesing 25 Feb 2008, 20:52
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maratikus
bmwhype2
A student has answered 7 of the 10 questions on the true-false test correctly and has decided to guess randomly on the rest. If all of the questions are equally weighted, what is the probability of the student receiving a score of 80% or more on the test?
(A) 0.125
(B) 0.5
(C) 0.7
(D) 0.8
(E) 0.875

E: he has to answer at least one question out of three remaining correctly. 1 - (1/2)^3 = 7/8

(1/2)^3 - probability that he will mess up on each of 3 remaining questions.
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very impressive. OA is E
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prob - guesing 25 Feb 2008, 21:12
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where am i messing up here:

we need to find out the probability of him getting 1 right, 2 right or 3 right.

probability of 1 right and 2 wrong is (1/2)*(1/2)*(1/2) right ?
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prob - guesing 25 Feb 2008, 21:16
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its easier if u use the 1-affirmative shortcut for "at least one" questions

the q is asking for "at least one correct."
therefore, we want 1-P(none correct)

p(correct) = 1/2
p(wrong) = 1/2


1 - 3 wrong
1 - (1/2)^3
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prob - guesing 26 Feb 2008, 01:04
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E.

So he answered 7/10 correctly, so they're asking what's the chance that he answers at least 1 right on the last 3?

This can be also reformulated to:

1 - Pr(0 questions right).

1 - (0.5)(0.5)(0.5) = 1 - 1/2^3 = 8/8 - 1/8 = 7/8.

Thus E), 0.875.

Very good question to learn, the hard probability questions can be solved using this method.
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prob - guesing 26 Feb 2008, 06:05
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Even if one does not know probability, it is simple by means of the process of elimination...

Here, the chance is anyway 70%+...

That leaves us only C, D and E...

And when we are looking at an assumed likelihood, it would clearly be above 70%, with any of the three being right..

so, that means, we are left with D and E....

For those who dont know Probability, its a wild guess from here...

But, a wise guess nevertheless. and the chance that the guess may turn out to be right after the elimination process is 1 out of 2, vis a vis 1 out of 5 before the elimination process...
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prob - guesing 26 Feb 2008, 07:48
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bmwhype2
A student has answered 7 of the 10 questions on the true-false test correctly and has decided to guess randomly on the rest. If all of the questions are equally weighted, what is the probability of the student receiving a score of 80% or more on the test?
(A) 0.125
(B) 0.5
(C) 0.7
(D) 0.8
(E) 0.875
Show more

Simple all we need to do is answer what is the prob. of the student not getting an 80%. Then subtract 1. This method is much easier.

1/2*1/2*1/2 = 1/8 --> 1-1/8 = 7/8.

E
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prob - guesing 26 Feb 2008, 10:50
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maratikus
bmwhype2
A student has answered 7 of the 10 questions on the true-false test correctly and has decided to guess randomly on the rest. If all of the questions are equally weighted, what is the probability of the student receiving a score of 80% or more on the test?
(A) 0.125
(B) 0.5
(C) 0.7
(D) 0.8
(E) 0.875

E: he has to answer at least one question out of three remaining correctly. 1 - (1/2)^3 = 7/8

(1/2)^3 - probability that he will mess up on each of 3 remaining questions.
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E. exactly. 1 - (1/2)^3 = .875 or 87.5%
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prob - guesing 01 Mar 2008, 19:36
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1/2+1/4+1/8
Clear E.



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