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Thread: Probability practise

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    Probability practise




    An electrical system is set up such that power can get from the source to the switch if both parts A and B are working, if part C is working or if all 3 parts are working. The probability that any one part fails is 12% and all parts fail independently. Let X be the number of parts that fail.
    a. Find the probability distribution of X
    b. Find the probability that the system does not work.
    c. How many parts are expected to fail?
    d. If there are 10 of these machines, how many total parts are expected to fail?



    My answers
    a,X=0,1,2,3..and their probabilities
    p(x=0)=1-(0.12*0.12*0.12),p(x=1)=0.12*0.12*0.92and so on..
    b. a fails b fails c fails=0.12*0.12*0.12
    c.N*expected mean
    10*expected mean
    let me know if I aam correct.thanks

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    Re: Probability practise

    For part A) can you elaborate a little more? Is there a specific distribution you know that this follows? I'm just wondering because your P(X=1) is wrong.

    B) The problem is more complicated than you realize. It's possible for the system to fail in more situations than just all 3 parts failing.

    C) You're probably right but what do you mean by N and what do you mean by expected mean? It's relatively simple so why don't you just tell us the numeric answer. Same with D.

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    Re: Probability practise

    part a, not sure what you are asking, I just thought there can be these 4 possibilities,like no part fails 1 fails 2 fail,3 fail..
    and p(x=1), should be 0.12*.82*0.82...sorry about 0.92..was a typo.
    part b, so is it like, if part a and part b work or part c(only works) the system works..and if part a and b fail or one of them fails or c fails?
    and i am mean to find an expected value for mean and then just multiple with an N, dont know how can I come up with numeric in this part?

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    Re: Probability practise

    Quote Originally Posted by SadieKhan View Post
    p(x=1), should be 0.12*.82*0.82
    This is still not correct. To convince yourself of this why don't you find the probabilities for X=0,1,2,3 using your method. Do they add up to 1? Since those are the only options then P(X=0)+P(X=1)+P(X=2)+P(X=3) = 1 should be true.

    Also:
    Quote Originally Posted by SadieKhan View Post
    p(x=0)=1-(0.12*0.12*0.12)
    This isn't correct either.

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    Re: Probability practise

    Quote Originally Posted by Dason View Post
    This is still not correct. To convince yourself of this why don't you find the probabilities for X=0,1,2,3 using your method. Do they add up to 1? Since those are the only options then P(X=0)+P(X=1)+P(X=2)+P(X=3) = 1 should be true.
    got it..
    will try and report back and as for not giving you the numeric value for part D, I have to find all the probabilities first then multiply by x before summing up.
    Hadn't worked that far yet

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    Re: Probability practise

    I have figured it out.

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    Re: Probability practise

    What did you end up getting? Did you get that I was alluding to the binomial distribution in my first post?

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    Re: Probability practise

    X Probability P(x)
    0 (0.88)(0.88)(0.88) 0.681472




    1
    (0.12)(0.88)(0.88) 0.278784

    2 (0.12)(0.12)(0.88) 0.038016
    3 (0.12)(0.12)(0.12) 0.001728


    b. Find the probability that the system does not work.
    p(system doesn’t work)=P(A and C fail and B works)+P(B and C fail and A works)+P(A fails B and C fail)
    =(0.12)(0.12)(0.88)+(0.12)(0.12(0.88)+(0.12)(0.12)(0.12)
    =0.038016+0.038016+0.001728
    =0.07776



    c. How many parts are expected to fail?
    x p(x) xp(x)
    0 0.681472
    0
    1 0.278784
    0.278784

    2 0.038016 0.076032

    3 0.001728 0.005184


    Expectation of x=0.36
    How many parts are expected to fail?
    =3*0.36=1.08.
    Hence one part is expected to fail.
    d. If there are 10 of these machines, how many total parts are expected to fail?
    =10*0.36=3.6..or we may say almost 4 parts are expected to fail.

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    Re: Probability practise

    right...........?

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    Re: Probability practise

    Quote Originally Posted by SadieKhan View Post
    X Probability P(x)
    0 (0.88)(0.88)(0.88) 0.681472
    1 (0.12)(0.88)(0.88) 0.278784
    2 (0.12)(0.12)(0.88) 0.038016
    3 (0.12)(0.12)(0.12) 0.001728
    You get the right probabilities but you're missing things. What that chart really should look like is:
    Code: 
    X	Probability	P(x)
    0	1*(0.88)(0.88)(0.88)	0.681472
    1	3*(0.12)(0.88)(0.88)	0.278784
    2	3*(0.12)(0.12)(0.88)	0.038016
    3	1*(0.12)(0.12)(0.12)	0.001728
    Can you spot the difference?
    b. Find the probability that the system does not work.
    p(system doesn’t work)=P(A and C fail and B works)+P(B and C fail and A works)+P(A fails B and C fail)
    =(0.12)(0.12)(0.88)+(0.12)(0.12(0.88)+(0.12)(0.12)(0.12)
    =0.038016+0.038016+0.001728
    =0.07776
    And I mentioned you having the wrong things before because it seems to crop up here. Do your calculation again. Notice that
    (.12)*(.12)*(.88) does not equal .038016.
    c. How many parts are expected to fail?
    x p(x) xp(x)
    0 0.681472
    0
    1 0.278784
    0.278784

    2 0.038016 0.076032

    3 0.001728 0.005184


    Expectation of x=0.36
    You should stop right here. I don't know what you're doing after this is and this is the correct answer.
    How many parts are expected to fail?
    =3*0.36=1.08.
    Hence one part is expected to fail.
    You should be getting 3*(.12)=0.36 as the expected number to fail. When I asked about the binomial distribution this is partly why. If you learn some basic distributions then these questions become quite easy. This process follows a binomial distribution. The expected value of a binomial is n*p so in this case 3*(.12) = .36.
    d. If there are 10 of these machines, how many total parts are expected to fail?
    =10*0.36=3.6
    Once again just stop right here. When talking about the expected value NEVER round to the nearest integer. Nobody cares. Expected value is a mathematical property and it doesn't have to be an integer.
    ..or we may say almost 4 parts are expected to fail.

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    Re: Probability practise

    I couldnt understand why you put, 1, 3,3,1 in first part?

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    Re: Probability practise

    Rest is cool..and I thought for once in my life that I did it all correct:P

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    Re: Probability practise

    Quote Originally Posted by SadieKhan View Post
    I couldnt understand why you put, 1, 3,3,1 in first part?
    It might help if you wrote out all 8 ways things could happen.
    Let F be that that item failed, S be that it didn't fail then the 8 options are:

    1) A F, B F, C F
    2) A F, B F, C S
    3) A F, B S, C F
    4) A F, B S, C S
    5) A S, B F, C F
    6) A S, B F, C S
    7) A S, B S, C F
    8) A S, B S, C S

    Now the probability that one fails is the sum of the probabilities where only one fails. How many ways do one item fail?

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    Re: Probability practise

    3.........,got it

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    Re: Probability practise


    and I forgot i did ..
    =Power(0.12,3):|

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