Results 1 to 5 of 5

Math Help - Independence Question - Can anyone check my solution please?

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    69

    Question Independence Question - Can anyone check my solution please?

    Your computer has been acting very strangely lately, and you suspect that it might have a virus on it. Unfortunately, all 15 of the different virus detection programs you own are outdated. You know that if your computer does have a virus, each of the programs, independently of the others, has a 0.77 chance of believing that your computer as infected, and a 0.23 chance of thinking your computer is fine. On the other hand, if your computer does not have a virus, each program has a 0.93 chance of believing that your computer is fine, and a 0.07 chance of wrongly thinking your computer is infected. Given that your computer has a 0.64 chance of being infected with some virus, and given that you will believe your virus protection programs only if 11 or more of them agree, find the probability that your detection programs will lead you to the right answer.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Mar 2009
    Posts
    69

    Question

    I did the following calculation for this question:

    P(more than 10 program says infectious/infectious): i= from 11 to 15
    Comb(15,i)*0.77^i*0.23^(15-i)
    P(more than 10 program says not infectious/not infectious):
    i= from 11 to 15
    Comb(15,i)*0.93^i*0.07^(15-i)

    so;

    1365 * (0.77^11)*(0.23^4) = 0.215497369

    455 * (0.77^12)*(0.23^3) = 0.240482571

    105 * (0.77^13)*(0.23^2) = 0.185790883

    15 * (0.77^14)*(0.23^1) = 0.0888565092

    1* (0.77^15) * 1 = 0.0198317426

    and

    1365 * (0.93^11) * (0.07^4) = 0.0147515361

    455 * (0.93^12) * (0.07^3) = 0.0653282312

    105 * (0.93^13) * (0.07^2) = 0.200292049

    15 * (0.93^14) * 0.07 = 0.380146135

    1* (0.93^15) * 1 = 0.336700862

    Then, I found out;

    P(more than 10 program says infectious/infectious):
    0.215497369 + 0.240482571 + 0.185790883 + 0.0888565092 + 0.0198317426 =

    0.750459075

    P(more than 10 program says not infectious/not infectious):
    0.0147515361 + 0.0653282312 + 0.200292049 + 0.380146135 + 0.336700862 =

    0.997218813


    Finally, I multiplied these 2 probabilities and get the result 0.750459075 but it says "incorrect". What did I do wrong...?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by essedra View Post
    Your computer has been acting very strangely lately, and you suspect that it might have a virus on it. Unfortunately, all 15 of the different virus detection programs you own are outdated. You know that if your computer does have a virus, each of the programs, independently of the others, has a 0.77 chance of believing that your computer as infected, and a 0.23 chance of thinking your computer is fine. On the other hand, if your computer does not have a virus, each program has a 0.93 chance of believing that your computer is fine, and a 0.07 chance of wrongly thinking your computer is infected. Given that your computer has a 0.64 chance of being infected with some virus, and given that you will believe your virus protection programs only if 11 or more of them agree, find the probability that your detection programs will lead you to the right answer.
    Let X be the random variable 'number of virus detection programs that say that the computer has a virus' when there is a virus.
    X ~ Binomial(n = 15, p = 0.77).

    Let Y be the random variable 'number of virus detection programs that say that the computer has no virus' when there is no virus.
    Y ~ Binomial(n = 15, p = 0.93).

    Pr(11 or more detection programs say there is a virus | there is a virus) = \Pr( X \geq 11).

    Pr(11 or more detection programs say there is no virus | there is no virus) = \Pr(Y \geq 11).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2009
    Posts
    69

    Unhappy

    Quote Originally Posted by mr fantastic View Post
    Let X be the random variable 'number of virus detection programs that say that the computer has a virus' when there is a virus.
    X ~ Binomial(n = 15, p = 0.77).

    Let Y be the random variable 'number of virus detection programs that say that the computer has no virus' when there is no virus.
    Y ~ Binomial(n = 15, p = 0.93).

    Pr(11 or more detection programs say there is a virus | there is a virus) = \Pr( X \geq 11).

    Pr(11 or more detection programs say there is no virus | there is no virus) = \Pr(Y \geq 11).
    Yes, I did exactly the same, as you can see above but the answers were incorrect it said.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by essedra View Post
    Yes, I did exactly the same, as you can see above but the answers were incorrect it said.
    Well then, it will depend on what is meant by "find the probability that your detection programs will lead you to the right answer" .... Perhaps whoever wrote the question has a different interpretation than the one in this thread.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: March 10th 2011, 09:12 AM
  2. inclusion-exclusion question check solution help
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: May 16th 2010, 07:16 AM
  3. could some check solution please
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: February 17th 2010, 11:46 PM
  4. Check my limits Solution
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: October 7th 2009, 08:58 AM
  5. L. Independence and existence of solution
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: August 27th 2009, 08:45 PM

Search Tags


/mathhelpforum @mathhelpforum