Question 648024: "A student has to sit for an examination consisting of 3 questions selected randomly from a list of 100 questions. To pass, he should answer all the three questions. What is the probability that the student will pass the examination, if he knows the answer to 90 questions on the list."
I have asked this question before and i got the answer as .729 but its the wrong answer. I have also solved it the same but correct answer is .798 and i don't know how to get this one...?????????
Found 2 solutions by Theo, asd123456:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the answer that i get is .7265306122
90/100 * 89/99 * 88/98
3 questions are selected randomly from a list of 100.
that means without replacement.
all 3 have to be among the 90 that he knows.
on the first draw, the probability of drawing one that he knows is 90/100.
on the second draw, the probability of drawing one that he knows is 89/99.
on the third draw, the probability of drawing one that he knows is 88/98.
the probability of drawing all 3 is 90/100 * 89/99 * 88/98.
this is no different than drawing 3 balls of the same color from an urn that contains 90 white and 10 other colors.
the probability of drawing all 3 white without replacement is 90/100 * 89/99 * 88/98.
i have no idea where .798 comes from.
since the student must know the answer to all 3 questions, then the probability that the student will pass the exam is the same as the probability the student will get 3 questions that he knows the answer to.
that's my take.
assume only 3 questions and he draws 2 out of 3 and knows the answer to 2 out of 3.
the probability that he will pass the test will be:
2/3 * 1/2 = 2/6 = 1/3
here's how it works out.
assume the questions are a, b, and 1
he knows the answer to a and b but doesn't know the answer to 1.
the possible combinations of 2 that can be gotten from 3 questions is as follows:
ab
a1
b1
order is not important since it doesn't matter which comes first or which comes second.
there's 3 possible combinations, only 1 of which contains a and b.
probability is therefore 1/3 that he will pass the test if he needs to be able to answer both questions correctly.
same concept applies to 100 questions, 90 of which he knows the answer to.
that's my take.
i could be wrong, but i'm pretty sure i'm right if i understand the problem correctly.
Answer by asd123456(1)  (Show Source):
You can put this solution on YOUR website! Yaah....i also thought that answer should be this way. But we got the question in test and prof has given the correct answer as .798. Also searched it on Google, this question is from "probability through problems" book...and not getting its soft or hard copy to check the answer....and prof has not explained it yet but so eager to know how it could be .798....
well, its Ok..if you also think that it should be .726.
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