# Math Help - help with a problem of probability combinations

1. ## help with a problem of probability combinations

my first post:
let's see. I was solving a problem of probability combinations, but I had a little problem in dealing with a question, here is the problem:

"a student must meet a test that consists of 3 questions randomly selected from a list of 100 questions (each question has the same probability of being selected). to approve the review he needs to answer at least two questions correctly. What is the probability that the student approve the test if he only knows the answers to 90 questions on the list?"

I started getting the following:

(i) (100 combination 3) = 161700 ........... number of possible questions in groups of 3
(ii) (90 combination 3) = 117480 ............ number of possible questions in groups of 3 students know

really did not know which is the total probability, if 1 / 100 (for each question) or 1 / (100 combination 3) (for each group of questions).

then: the 10 questions that the student does not know how it affects the probability?

very grateful in advance, a response

2. The entire space consists of all possible combinations of choosing three questions out of 100. There are however, 90C3 and 90C2*10 possible combinations that will allow him to pass the test so:

Total # of possible 3-question combinations: 100 choose 3
Total # of possible 3-question combinations that student knows all answers: 90C3
Total # of possible 3-question combinations that students knows two answers and doesn't know the last one: (90 choose 2)X10

Probability he will be given a 3-question combinaiton he knows: $\frac{90C3+90C2*10}{100C3}$

This assumes that the test is premade, and its simply a matter of selecting one of the premade tests. If he is actually picking questions then its much easier:

$[\frac{90}{100}\frac{89}{99}\frac{10}{98}]3+[\frac{90}{100}\frac{89}{99}\frac{88}{98}]3$

3. Originally Posted by ANDS!
The entire space consists of all possible combinations of choosing three questions out of 100. There are however, 90C3 and 90C2*10 possible combinations that will allow him to pass the test so:

Total # of possible 3-question combinations: 100 choose 3
Total # of possible 3-question combinations that student knows all answers: 90C3
Total # of possible 3-question combinations that students knows two answers and doesn't know the last one: (90 choose 2)X10

Probability he will be given a 3-question combinaiton he knows: $\frac{90C3+90C2*10}{100C3}$

This assumes that the test is premade, and its simply a matter of selecting one of the premade tests. If he is actually picking questions then its much easier:

$\frac{90}{100}\frac{89}{99}+\frac{90}{100}\frac{89 }{99}\frac{88}{98}$
I understand that 1-q = 90C2 ... but why " *10"?

4. There are ten questions he does not know - therefore, if you have 90C2, you need to multiply that number by 10 to get the overall number of 3-question tests he can get where he knows two of the answers (90C2) and doesn't know the last one (10 of those).

5. Originally Posted by ANDS!
There are ten questions he does not know - therefore, if you have 90C2, you need to multiply that number by 10 to get the overall number of 3-question tests he can get where he knows two of the answers (90C2) and doesn't know the last one (10 of those).
lol. is true, did not I think. thank you very much

6. Argh. My mistake on the second "scenario": It should actually be a binomial, with $(100C3)(\frac{90}{100})^2(\frac{10}{100})+(100C3)( \frac{90}{100})^3(\frac{10}{100})^0$

7. Hello, killertapia!

Welcome aboard!

A student takes a test that has 3 questions randomly selected from a list of 100 questions
(each question has the same probability of being selected).
To pass the test he needs to answer at least two questions correctly.
What is the probability that the student passes the test
if he knows the answers to omnly 90 questions on the list?
I got the same answer as ANDS!

What is the probability that he fails the test?

There are: $_{100}C_{3} \:=\:161,\!700$ possible tests.

How many of these tests contain no answers he knows?
. . There are: . $_{10}C_3 \:=\:120$ such tests.

How many of these tests have exactly one answer he knows?
. . There are: . $\left(_{90}C_1\right) \left(_{10}C_2\right) \:=\:4,\!050$ such tests.

Hence, he fails on: . $120 + 4,\!050 \:=\:4,\!170$ of the tests.

And: . $P(\text{fail}) \:=\:\frac{4,\!170}{161,\!700} \:=\:\frac{139}{5390}$

Therefore: . $P(\text{pass}) \;=\;1 - \frac{139}{5390} \;=\;\frac{5251}{5390}$

8. Originally Posted by ANDS!
Argh. My mistake on the second "scenario": It should actually be a binomial, with $(100C3)(\frac{90}{100})^2(\frac{10}{100})+(100C3)( \frac{90}{100})^3(\frac{10}{100})^0$
i'm sorry but, Are you sure that is well? I mean I know how is the binomial function, but with this data I get a very large number
on an other hand thanks for the other scenario, i get a correct answer