a little probability problem

by Jeremiah Mburuburu Tue May 01, 2012 11:22 am

a professor of political science gives his students 20 questions, and assures them that the final exam will consist of 8 of them. all the questions are equally important, and the professor has no preference for any of them. each question requires long study. one student prepares for 14 of them. what is the probability that the student has prepared for 6 or more of the 8 questions that appear on the final?

by Petrichor Tue May 01, 2012 11:49 am

14.56%

by chameli Tue May 01, 2012 11:50 am

female student?

chances that the professor has secretly handed her the chosen 8 are 100%

she prepares the other 6 just to throw her friends off

the two are "probably " married and with child by now

by Jeremiah Mburuburu Tue May 01, 2012 12:07 pm

atcg wrote:14.56%

your answer is different from mine.

you might check if the sum of P(student has prepared for k of 8 Qs on the final) for k = 0, 1, 2,...,8 according to your calculation is equal to 1.

by Jeremiah Mburuburu Tue May 01, 2012 12:45 pm

chameli wrote:female student?

chances that the professor has secretly handed her the chosen 8 are 100%

she prepares the other 6 just to throw her friends off

the two are "probably " married and with child by now

yes, the student was female. that's why the chances that he revealed the 8 chosen Qs to her are 0; the prof was gay.

by chameli Tue May 01, 2012 12:49 pm

naaah u just made that up

otoh are most statistics and maths professors gay and on the lam ?

by MulaiAzhagi Tue May 01, 2012 1:30 pm

===> Does it mean that the numbers 20 and 14 given are irrelevant?

by Jeremiah Mburuburu Tue May 01, 2012 1:58 pm

MulaiAzhagi wrote:===> Does it mean that the numbers 20 and 14 given are irrelevant?

were you asking me?

by Petrichor Tue May 01, 2012 2:28 pm

7.75%, final answer.

by Jeremiah Mburuburu Tue May 01, 2012 2:36 pm

atcg wrote:7.75%, final answer.

that answer is also different from mine.

by MulaiAzhagi Tue May 01, 2012 2:41 pm

Jeremiah Mburuburu wrote:
MulaiAzhagi wrote:===> Does it mean that the numbers 20 and 14 given are irrelevant?
were you asking me?

Yes.

I was going down the following path. Looks like it is not right.

let p denote the probability of a question showing up on the final exam
let q denote the probability of a question not showing up on the final exam

p = 8/14 = 0.57
q = 1-p = 1 - 0.57 =0.43

This is a binomial distribution

The probability of exactly 6 out of 14 questions showing up P(r=6) = Cn,r * p**r * q**(n-r)
= C14,6 * (0.57)**6 * (0.43)**(14-6)
The probability of exactly 7 out of 14 questions showing up P(r=7) = Cn,r * p**r * q**(n-r)
= C14,7 * (0.57)**7 * (0.43)**(14-7)
The probability of exactly 8 out of 14 questions showing up P(r= Cool

= Cn,r * p**r * q**(n-r)
= C14,8 * (0.57)**8 * (0.43)**(14- Cool

P(r >= 6) = P(r=6) + P(r=7) + P(r= Cool

by Jeremiah Mburuburu Tue May 01, 2012 3:40 pm

MulaiAzhagi wrote:
Jeremiah Mburuburu wrote:
MulaiAzhagi wrote:===> Does it mean that the numbers 20 and 14 given are irrelevant?
were you asking me?

Yes.

I was going down the following path. Looks like it is not right.

let p denote the probability of a question showing up on the final exam
let q denote the probability of a question not showing up on the final exam

p = 8/14 = 0.57
q = 1-p = 1 - 0.57 =0.43

This is a binomial distribution

The probability of exactly 6 out of 14 questions showing up P(r=6) = Cn,r * p**r * q**(n-r)
= C14,6 * (0.57)**6 * (0.43)**(14-6)
The probability of exactly 7 out of 14 questions showing up P(r=7) = Cn,r * p**r * q**(n-r)
= C14,7 * (0.57)**7 * (0.43)**(14-7)
The probability of exactly 8 out of 14 questions showing up P(r= = Cn,r * p**r * q**(n-r)
= C14,8 * (0.57)**8 * (0.43)**(14-

P(r >= 6) = P(r=6) + P(r=7) + P(r=

check whether the experiment in your mind is a bernoulli process, consisting of a sequence of identical, independent, dichotomous trials to which a binomial distribution applies.

by indophile Tue May 01, 2012 4:45 pm

by Guest Tue May 01, 2012 8:53 pm

~0.545

by Kris Tue May 01, 2012 11:06 pm

Jeremiah Mburuburu wrote:a professor of political science gives his students 20 questions, and assures them that the final exam will consist of 8 of them. all the questions are equally important, and the professor has no preference for any of them. each question requires long study. one student prepares for 14 of them. what is the probability that the student has prepared for 6 or more of the 8 questions that appear on the final?

>>>91%

Probably of 1 question wrong is 0.3; probably of a 2nd question wrong is 0.3. P(2 wrong) = 0,09.

1 less 0.09 is 91%.

by Jeremiah Mburuburu Wed May 02, 2012 12:39 am

blabberwock wrote:~0.545

absolutely!

by Jeremiah Mburuburu Wed May 02, 2012 1:32 am

Jeremiah Mburuburu wrote:a professor of political science gives his students 20 questions, and assures them that the final exam will consist of 8 of them. all the questions are equally important, and the professor has no preference for any of them. each question requires long study. one student prepares for 14 of them. what is the probability that the student has prepared for 6 or more of the 8 questions that appear on the final?

the answer is 0.5449. blabberwock is right.

the general expression for the P(the student has prepared for exactly k Qs that appear on the final) =

(14Ck)(6C(8-k)) / (20C8)

the denominator represents the total no. of possible selections of 8Qs the prof can make from the complete set of 20, and the numerator, the no. of selections favourable to the student that are among the total. (the favourable selctions are those in which exactly k of the 14 Qs the student studied for appear.)

the 1st factor in the numerator is the no. of ways of selecting k of the 14 Qs the student has prepared for, and the 2nd factor, the no. of ways of selecting the remaining 8-k of the 20-14 = 6 Qs the student hasn't prepared for.

this fraction is the correct probability because every selection of 8 Qs of 20 the prof could possibly make is of equal probability.

the probabilities for k = 6, 7, and 8 are, respectively, 0.3576, 0.1635, and 0.0238, for a sum of 0.5449, the answer.

by Petrichor Wed May 02, 2012 9:06 am

If you prepare 70%, you have better than even chance of scoring 75% - seems counter-intuitive.

by indophile Wed May 02, 2012 9:23 am

I was looking for a simple solution. It went like this -

Only 8 of the 20 will be on the exam. So 12 of the 20 will not be on the exam.

If the kid studies 13, then there will be at least 1 question she can answer

If she studies 14, then there will be at least 2 questions she can answer

That is, If she studies 14, correct answers for 2 are definite.

The question then is - out of the remaining 12 questions, what is the probability that she will be lucky to get at least 4 on the test (to make it at least 6 correct answers)

4/12 = 1/3.

So if she studies 14, she can answer correctly at least 6 with a 1/3 probability (At least 6 is taken to mean 6 or more as asked in the question).

by Sponsored content

a little probability problem

a little probability problem

Re: a little probability problem

Re: a little probability problem

Re: a little probability problem

Re: a little probability problem

Re: a little probability problem

Re: a little probability problem

Re: a little probability problem

Re: a little probability problem

Re: a little probability problem

Re: a little probability problem

Re: a little probability problem

Re: a little probability problem

Re: a little probability problem

Re: a little probability problem

Re: a little probability problem

Re: a little probability problem

Re: a little probability problem

Re: a little probability problem

Re: a little probability problem