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Math Help - Probability that X is appointed as a manager

  1. #1
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    Probability that X is appointed as a manager

    Question:
    The chance of three persons Mr X, Y and Z becoming manager of a company are 4:2:3. The probability that bonus scheme shall be introduced if X, Y and Z become managers are 0.3,0.5,0.8 respectively. Find the probability that the bonus scheme will be introduced. What is the probability that X is appointed as a manager?

    Answer:

    P(S) = Probability of bonus being introduced.
    P(X) = Probability of Mr X being Manager.
    P(Y) = Probability of Mr Y being Manager.
    P(Z) = Probability of Mr Z being Manager.
    P(S/X) = Probability of scheme being introduced when X becomes manager.
    P(S/Y) = Probability of scheme being introduced when Y becomes manager.
    P(S/Z) = Probability of scheme being introduced when Z becomes manager.

    Probability of bonus scheme being introduced
    P(S) = P(X).P(S/X) + P(Y).P(S/Y) + P(Z).P(S/Z) = 4(0.3) + 2(0.5) + 3(0.8) = 4.6

    Probability that X is appointed as manager
    P(S/X) = \frac{4(0.3)}{4(0.3)+2(0.5)+3(0.8)} = 0.26

    Is this Right or Wrong solution......Please let me now ........
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  2. #2
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    Quote Originally Posted by Stats View Post
    Question:
    The chance of three persons Mr X, Y and Z becoming manager of a company are 4:2:3. The probability that bonus scheme shall be introduced if X, Y and Z become managers are 0.3,0.5,0.8 respectively. Find the probability that the bonus scheme will be introduced. What is the probability that X is appointed as a manager?

    Answer:

    P(S) = Probability of bonus being introduced.
    P(X) = Probability of Mr X being Manager.
    P(Y) = Probability of Mr Y being Manager.
    P(Z) = Probability of Mr Z being Manager.
    P(S/X) = Probability of scheme being introduced when X becomes manager.
    P(S/Y) = Probability of scheme being introduced when Y becomes manager.
    P(S/Z) = Probability of scheme being introduced when Z becomes manager.

    Probability of bonus scheme being introduced
    P(S) = P(X).P(S/X) + P(Y).P(S/Y) + P(Z).P(S/Z) = 4(0.3) + 2(0.5) + 3(0.8) = 4.6

    Probability that X is appointed as manager
    P(S/X) = \frac{4(0.3)}{4(0.3)+2(0.5)+3(0.8)} = 0.26

    Is this Right or Wrong solution......Please let me now ........
    First off - probability answers will always be between 0 and 1, so your P(S) is wrong.

    Are these the only candidates for the position of manager?
    Assuming they are then..
    Turn the ratio into a fraction/decimal/percentage/whatever you like working with probability in. So 4+2+3=9. So X has 4/9ths of a chance to be manager, Y 2/9, Z 3/9.

    So the probability of X becoming manager is 4/9 or 44.44 %

    Then your method for finding the probability of the bonus seems to be correct, but you have to change the P(X), P(Y), and P(Z) in a similar fashion to what we've just done here..
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  3. #3
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    Quote Originally Posted by Unenlightened View Post
    First off - probability answers will always be between 0 and 1, so your P(S) is wrong.

    Are these the only candidates for the position of manager?
    Assuming they are then..
    Turn the ratio into a fraction/decimal/percentage/whatever you like working with probability in. So 4+2+3=9. So X has 4/9ths of a chance to be manager, Y 2/9, Z 3/9.

    So the probability of X becoming manager is 4/9 or 44.44 %

    Then your method for finding the probability of the bonus seems to be correct, but you have to change the P(X), P(Y), and P(Z) in a similar fashion to what we've just done here..

    Please can u give me a proper solution of P(S)
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  4. #4
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    Why don't you give it a shot yourself first?

    Use the same formula you used in your first attempt. But the proper values of P(X), P(Y), and P(Z).
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  5. #5
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    Is this right?

    Quote Originally Posted by Unenlightened View Post
    Why don't you give it a shot yourself first?

    Use the same formula you used in your first attempt. But the proper values of P(X), P(Y), and P(Z).
    P(S) = \frac{4}{9}(0.3)+ \frac{2}{9}(0.5)+ \frac{3}{9}(0.8)

    =0.46 (Is this right?)
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  6. #6
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    Quote Originally Posted by Stats View Post
    P(S) = \frac{4}{9}(0.3)+ \frac{2}{9}(0.5)+ \frac{3}{9}(0.8)

    =0.46 (Is this right?)
    It's right up until where you wrote down 0.46.
    4.6/9 is the answer, not 4.6/10..
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  7. #7
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    What about this?

    Quote Originally Posted by Unenlightened View Post
    It's right up until where you wrote down 0.46.
    4.6/9 is the answer, not 4.6/10..

    And P(S/X) = 1.2/0.46 = 2.6

    Is this right ?
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  8. #8
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    P(s/x) =

    P(S/X) = Probability of scheme being introduced when X becomes manager. = P(X) * P(S)
    = 4/9 * 0.3
    =0.133333


    Rember stats, Probability of happening of some thing will be never be greater than one.....
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