Results 1 to 6 of 6

Math Help - compute probability

  1. #1
    Newbie
    Joined
    Aug 2007
    Posts
    3

    compute probability

    I need to work this math question, using the binomial probability formula,can someone help me figure it out. A 10-question multiple choice exam is given, and each question has five possible answers. Pascal takes his exam and guesses at every question. Use the binomial orobability formula to find the probability(to 5 deciminal places) that:
    a) he gets exactly 2 questions correct.
    b)he gets no questions correct.
    c) he gets at least 1 question correct(use the information from part (b) to answer this part.
    d) he gets at least 9 questions correct.
    e) Without using the binomial probability formula, determine the probability that he gets exactly 2 questions correct.
    f) Compare your answers to parts (a) and (f). If they are not the same, explain why.
    Is there a math wiz out there, that could possibly help me with this problem, or try to walk me through it,I would really appreciate it, Thanks Toni
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by toni_45miss@hotmail.com View Post
    I need to work this math question, using the binomial probability formula,can someone help me figure it out. A 10-question multiple choice exam is given, and each question has five possible answers. Pascal takes his exam and guesses at every question. Use the binomial orobability formula to find the probability(to 5 deciminal places) that:
    a) he gets exactly 2 questions correct.
    b)he gets no questions correct.
    c) he gets at least 1 question correct(use the information from part (b) to answer this part.
    d) he gets at least 9 questions correct.
    e) Without using the binomial probability formula, determine the probability that he gets exactly 2 questions correct.
    f) Compare your answers to parts (a) and (f). If they are not the same, explain why.
    Is there a math wiz out there, that could possibly help me with this problem, or try to walk me through it,I would really appreciate it, Thanks Toni
    A binomialy distributed random variable with N trials and probability of a favourable outcome on a single trial p has prbability of n favourable outcomes in N trials of:

    <br />
b(n;N,p) = \frac{N!}{(N-n)!n!}~p^n(1-p)^{n-1}<br />

    where the exclamation mark denotes the factorial function.

    In this question N=10 and p=1/5, so for part (a) we have:

    <br />
p(n=2) = b(2;10,0.2) = \frac{10!}{8!2!}~0.2^2~0.8^{8}=45 ~0.2^2~0.8^{8} \approx 0.30199<br />

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2007
    Posts
    3

    To Captain Black man you are good, you sure your not in my math class? So do I use 0'

    Man you are great, are you sure you are not in my math class? So do I use all 0's for (b),or not. This type of math has lost me
    Quote Originally Posted by CaptainBlack View Post
    A binomialy distributed random variable with N trials and probability of a favourable outcome on a single trial p has prbability of n favourable outcomes in N trials of:

    <br />
b(n;N,p) = \frac{N!}{(N-n)!n!}~p^n(1-p)^{n-1}<br />

    where the exclamation mark denotes the factorial function.

    In this question N=10 and p=1/5, so for part (a) we have:

    <br />
p(n=2) = b(2;10,0.2) = \frac{10!}{8!2!}~0.2^2~0.8^{8}=45 ~0.2^2~0.8^{8} \approx 0.30199<br />

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,683
    Thanks
    615
    Hello, toni!

    I will assume you know the Binomial Probability Formula.

    We know that: . \begin{array}{ccccc}P(\text{right}) & = & \frac{1}{5} & = & 0.2 \\ P(\text{wrong}) & = & \frac{4}{5} & = & 0.8\end{array}


    A 10-question multiple choice exam is given, and each question has 5 possible answers.
    Pascal takes his exam and guesses at every question.
    Use the binomial orobability formula to find the probability(to 5 deciminal places) that:

    a) he gets exactly 2 questions correct.
    P(\text{2 right})\;=\; {10\choose2}(0.2)^2(0.8)^8


    b) he gets no questions correct.
    P(\text{0 right}) \;=\;{10\choose0}(0.2)^0(0.8)^{10}


    c) he gets at least 1 question correct
    (use the information from part (b) to answer this part.
    The opposite of "at least one right" is "none right".

    Hence: . P(\text{at least 1 right}) \;=\;1 - P(\text{0 right})



    d) he gets at least 9 questions correct.
    "At least 9 right" means: 9 right or 10 right.


    P(\text{at least 9 right}) \;=\;P(\text{9 right}) + P(\text{10 right})

    . . . . . . . . . . . . = \;{10\choose9}(0.2)^9(0.8)^1 + {10\choose10}(0.2)^{10}(0.8)^0


    I'll let you do the arithmetic . . .

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by toni_45miss@hotmail.com View Post
    I need to work this math question, using the binomial probability formula,can someone help me figure it out. A 10-question multiple choice exam is given, and each question has five possible answers. Pascal takes his exam and guesses at every question. Use the binomial orobability formula to find the probability(to 5 deciminal places) that:
    a) he gets exactly 2 questions correct.
    b)he gets no questions correct.
    c) he gets at least 1 question correct(use the information from part (b) to answer this part.
    d) he gets at least 9 questions correct.
    e) Without using the binomial probability formula, determine the probability that he gets exactly 2 questions correct.
    f) Compare your answers to parts (a) and (f). If they are not the same, explain why.
    Is there a math wiz out there, that could possibly help me with this problem, or try to walk me through it,I would really appreciate it, Thanks Toni
    Quote Originally Posted by CaptainBlack View Post
    A binomialy distributed random variable with N trials and probability of a favourable outcome on a single trial p has prbability of n favourable outcomes in N trials of:

    <br />
b(n;N,p) = \frac{N!}{(N-n)!n!}~p^n(1-p)^{n-1}<br />

    where the exclamation mark denotes the factorial function.

    In this question N=10 and p=1/5, so for part (a) we have:

    <br />
p(n=2) = b(2;10,0.2) = \frac{10!}{8!2!}~0.2^2~0.8^{8}=45 ~0.2^2~0.8^{8} \approx 0.30199<br />

    RonL
    For part b) we want n=0, so:

    <br />
p(n=0) = b(0;10,0.2) = \frac{10!}{10!~0!}~0.2^0~0.8^{10}=0.8^{10} \approx 0.10737<br />

    Note: 0!=1

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Aug 2007
    Posts
    3

    Math,math,math!!!

    Yes, puzzeled I am, but you guys are explaining it to me in a way I am kinda getting it, thank you all for all of the help, Toni Hopefully someday I can return the favor.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. how to compute this probability?
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: January 16th 2012, 04:54 AM
  2. how to compute this probability?
    Posted in the Statistics Forum
    Replies: 8
    Last Post: January 2nd 2012, 11:55 AM
  3. Compute each probability of Y
    Posted in the Statistics Forum
    Replies: 1
    Last Post: September 17th 2009, 03:15 AM
  4. Compute the probability of each
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: May 4th 2009, 04:04 PM
  5. Compute the probability of the min. of some RV.
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 20th 2008, 02:49 AM

Search Tags


/mathhelpforum @mathhelpforum