Results 1 to 13 of 13
Like Tree2Thanks
  • 1 Post By Plato
  • 1 Post By Jame

Math Help - Probability problems

  1. #1
    Senior Member Paze's Avatar
    Joined
    Nov 2012
    From
    Iceland
    Posts
    379
    Thanks
    19

    Probability problems

    A test contains 10 questions and each question has 4 options to answer in with 1 being the correct answer.
    What is the probability of getting at least 4 questions right if answers are randomly selected?

    What is the probability of getting between 2 to 7 questions right (3,4,5,6) if answers are randomly selected?

    I'm completely stranded in probability. I'm trying to get it in the P and C formats but it won't work. Help is very much appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,708
    Thanks
    1638
    Awards
    1

    Re: Probability problems

    Quote Originally Posted by Paze View Post
    [I]A test contains 10 questions and each question has 4 options to answer in with 1 being the correct answer.
    What is the probability of getting at least 4 questions right if answers are randomly selected?

    What is the probability of getting between 2 to 7 questions right (3,4,5,6) if answers are randomly selected?


    The probability of getting exactly k correct is:
    \mathcal{P}(X=k)=\binom{10}{k}(.25)^k(.75)^{10-k}

    Now just express these as sums.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Paze's Avatar
    Joined
    Nov 2012
    From
    Iceland
    Posts
    379
    Thanks
    19

    Re: Probability problems

    Thank you. Is this the correct answer to my first problem?

    ((10*9*8*7*6*5*4)/7!)/4^10 - Wolfram|Alpha

    And this for my 2nd?

    ((10*9*8*7)/4!&#41 ;/4^10 - Wolfram|Alpha
    Last edited by Paze; March 12th 2013 at 08:18 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,708
    Thanks
    1638
    Awards
    1

    Re: Probability problems

    Quote Originally Posted by Paze View Post
    Thank you. Is this the correct answer to my first problem?
    ((10*9*8*7*6*5*4)/7!)/4^10 - Wolfram|Alpha
    NO!
    The answer to the first is \sum\limits_{k = 4}^{10} {\mathcal{P}(X=k)}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Paze's Avatar
    Joined
    Nov 2012
    From
    Iceland
    Posts
    379
    Thanks
    19

    Re: Probability problems

    Quote Originally Posted by Plato View Post
    NO!
    The answer to the first is \sum\limits_{k = 4}^{10} {\mathcal{P}(X=k)}
    Hm, I didn't learn it with sums

    Is that the only way to do it?

    I feel like this is the same problem: https://www.khanacademy.org/math/pro...-in-five-flips

    Only my problem contains 4 possibilities instead of 2 and I have more 'flips'.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Feb 2011
    Posts
    83
    Thanks
    2

    Re: Probability problems

    For the first, you break it down into cases and use the addition principle.

    Case 1: Exactly 4 questions right
    -Pick 4 questions to be right. The other 6 are wrong. What is the probability of this?

    Case 2: Exactly 5 questions right
    and so on.

    The 2nd problem uses cases and addition principle too, however the cases are different.

    This isn't straight up counting per se , but independent trails that are either a success with probability .25 or a failure with probability .75
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,708
    Thanks
    1638
    Awards
    1

    Re: Probability problems

    Quote Originally Posted by Paze View Post
    I feel like this is the same problem: https://www.khanacademy.org/math/pro...-in-five-flips
    Did you read reply #2? That shows you exactly how to do these.

    Now if you are foolish enough to use KhanAcademy for serious study, then you get what you deserve.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member Paze's Avatar
    Joined
    Nov 2012
    From
    Iceland
    Posts
    379
    Thanks
    19

    Re: Probability problems

    Quote Originally Posted by Plato View Post
    Did you read reply #2? That shows you exactly how to do these.

    Now if you are foolish enough to use KhanAcademy for serious study, then you get what you deserve.
    I did read your reply but I don't even understand it. I'm learning to solve these problems using factorials. I think your way of solving it is more like the binomial theorem while I am asking for pascals triangle..I want to learn the basics before I simply apply a formula to solve it.

    If you can either prove your formula or direct me to a proof, I would gladly look into using that method instead.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,708
    Thanks
    1638
    Awards
    1

    Re: Probability problems

    Quote Originally Posted by Paze View Post
    I did read your reply but I don't even understand it. I'm learning to solve these problems using factorials. I think your way of solving it is more like the binomial theorem while I am asking for pascals triangle..I want to learn the basics before I simply apply a formula to solve it.
    If you can either prove your formula or direct me to a proof, I would gladly look into using that method instead.
    Do you know about binomial coefficients?
    \binom{N}{k}=\frac{N!}{k!(N-k)!}, the number of combinations of N objects taken k at a time?

    It is also the number of ways the arrange N 1's and k 0's in a string.

    We think of the 1's as a place where a success occurs and the 0's where a failure happens.
    If success has a probability of p then failure has probability of 1-p.

    So the probability of exactly k successes in N trials is \binom{N}{k}(p)^k(1-p)^{N-k}.

    Thus on a test of ten questions with four options, a student just guesses.
    The probability of exactly 4 successes in 10 is \binom{10}{4}(.25)^4(.75)^{6}.

    For at least four is
    \binom{10}{4}(.25)^4(.75)^{6}+\binom{10}{5}(.25)^4  (.75)^{6}+\binom{10}{7}(.25)^7(.75)^{3}+\binom{10}  {8}(.25)^8(.75)^{2}+\binom{10}{9}(.25)^9(.75)^{1}+  \binom{10}{10}(.25)^{10}(.75)^{0}.
    Thanks from Paze
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member Paze's Avatar
    Joined
    Nov 2012
    From
    Iceland
    Posts
    379
    Thanks
    19

    Re: Probability problems

    Quote Originally Posted by Plato View Post
    Do you know about binomial coefficients?
    \binom{N}{k}=\frac{N!}{k!(N-k)!}, the number of combinations of N objects taken k at a time?

    It is also the number of ways the arrange N 1's and k 0's in a string.

    We think of the 1's as a place where a success occurs and the 0's where a failure happens.
    If success has a probability of p then failure has probability of 1-p.

    So the probability of exactly k successes in N trials is \binom{N}{k}(p)^k(1-p)^{N-k}.

    Thus on a test of ten questions with four options, a student just guesses.
    The probability of exactly 4 successes in 10 is \binom{10}{4}(.25)^4(.75)^{6}.

    For at least four is
    \binom{10}{4}(.25)^4(.75)^{6}+\binom{10}{5}(.25)^4  (.75)^{6}+\binom{10}{7}(.25)^7(.75)^{3}+\binom{10}  {8}(.25)^8(.75)^{2}+\binom{10}{9}(.25)^9(.75)^{1}+  \binom{10}{10}(.25)^{10}(.75)^{0}.
    Thanks a lot for that. This is actually the kind of logic I have been looking for in probability.

    I have come as far as understanding how the 1's represent success and the 0's represent failure. I also understand that if the probability of success is P then the probability of failure is 1-P.

    I am trying to understand the formula \binom{N}{k}(p)^k(1-p)^{N-k}

    Essentially we are saying: \binom{N}{k}=\frac{N!}{k!(N-k)!} which represents what exactly? Is 10 my N and 4 my K? 4 answers (choices) with 10 events (iterations)? In that case does this binomial coefficient represent the total amount of successes and failures and the rest of the formula acts as some sort of filter to answer the rest of the question, e.g. probability of exactly 4 instead of 10..?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member Paze's Avatar
    Joined
    Nov 2012
    From
    Iceland
    Posts
    379
    Thanks
    19

    Re: Probability problems

    Why isn't this essentially the same as saying:

    I have a dice with 4 sides. I throw it 10 times. What is the probability of getting the same side exactly twice?

    I calculate it according to https://www.khanacademy.org/math/pro...-combinatorics

    and compare it to your answer in hopes of gaining some insight into your answer, but we get different answers!
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Feb 2011
    Posts
    83
    Thanks
    2

    Re: Probability problems

    In the dice example each side has an equally likely chance of occuring.

    With the test problem, we have two things in the sample space
    Right with probability .25
    Wrong with probability .75

    the outcomes are not equally likely

    EDIT: Also is the dice/coin problem, order is important.

    In the test problem, order DOES NOT matter.

    Getting questions 1,4,6,10 wrong is the same as getting questions 4,10,6,1 wrong. We're not ordering the questions. We just want to pick 4 to be right, and the other 6 wil be wrong.
    Last edited by Jame; March 13th 2013 at 08:50 PM.
    Thanks from Paze
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Senior Member Paze's Avatar
    Joined
    Nov 2012
    From
    Iceland
    Posts
    379
    Thanks
    19

    Re: Probability problems

    Okay, I think this question might have been a bit out of my reach. I will go back to some more basic probability questions and try this one later!

    Thanks for your help guys.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Further Probability problems
    Posted in the Statistics Forum
    Replies: 3
    Last Post: December 16th 2010, 07:57 PM
  2. PROBABILITY problems help please
    Posted in the Statistics Forum
    Replies: 3
    Last Post: October 2nd 2009, 12:48 PM
  3. Probability Problems
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: May 6th 2008, 09:53 PM
  4. Probability problems
    Posted in the Statistics Forum
    Replies: 2
    Last Post: May 1st 2008, 05:25 PM
  5. Probability Problems
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: October 5th 2006, 03:07 AM

Search Tags


/mathhelpforum @mathhelpforum