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Math Help - Probability problem.

  1. #1
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    Probability problem.

    Please, help me! ( I think my difficulties with it because of my poor English a cause- result? or intersection of 2 events? )
    Ann estimates that there is a 60% chance that she will get a raise if she asks her boss.
    a) If there is a 40% chance that Ann will, in fact, ask her boss for a raise, what is the probability that she will get a raise because she asked a one?
    b) If there is a 10% chance that Ann's boss will give her a raise before she even asks for one, what is the probability that Ann will get a raise without asking for one?

    I see that it's about conditional probability, but can't get the answer.
    Thank you!
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  2. #2
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    Re: Probability problem.

    My guess is that

    Prob(raise|ask)=.6*.4=.24

    Prob(raise|not ask)=.10
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  3. #3
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    Re: Probability problem.

    Why is Prob(raise|not ask)=.10 ?
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  4. #4
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    Re: Probability problem.

    Part 2 of your original post says it!
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  5. #5
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    Re: Probability problem.

    Quote Originally Posted by Romanka View Post
    Ann estimates that there is a 60% chance that she will get a raise if she asks her boss.
    a) If there is a 40% chance that Ann will, in fact, ask her boss for a raise, what is the probability that she will get a raise because she asked a one?
    b) If there is a 10% chance that Ann's boss will give her a raise before she even asks for one, what is the probability that Ann will get a raise without asking for one?
    I see that it's about conditional probability, but can't get the answer.
    Frankly I doubt you have received sound replies yet.
    We have no way to know that the events are independent!
    If they are independent, then conditional probability is a moot question.
    Basically, I find this a deeply flawed question. It may be a translation problem.
    We need more clarification on what the question actually says.
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  6. #6
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    Re: Probability problem.

    Well they are dependent by definition, whether or not she asks for the raise affects the probability that she gets one, according to the given values. I do agree that the wording is misleading so you might disagree with this..

    The "because she asked for one" can be interpreted as, "what is the probability that she asked for a raise AND that she received one." In this case, it would be P(ask)*P(get raise|ask), where. So it would be 0.4 * 0.6 = 0.24.

    The second question is worded confusingly, because it says "before she even asks." This implies that she might be considering asking for a raise but her boss might give her one before she even had the chance to ask.

    If we interpret it literally... the answer is simply 10%.. the question is basically a rephrase of the very first statement in that problem.. "there is a 10% chance she will get a raise before she asks for one".. this implies that the chance that she does get a raise without asking for one is indeed simply 10%.. but then, it might be a problem with the wording.
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  7. #7
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    Re: Probability problem.

    That is the original question from the test (not my translation). I thought I can't clarify it for myself and understand because of my English, but you are agree with me- it's "a deeply flawed question" ((
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  8. #8
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    Re: Probability problem.

    Quote Originally Posted by SworD View Post
    it might be a problem with the wording.
    the only explanation that I have
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  9. #9
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    Re: Probability problem.

    Well I interpreted the wording literally and gave you answers, is there any way you can verify?
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  10. #10
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Probability problem.

    Another interpretation of the problem can result in:
    a=ask
    na=not ask
    r=raise

    P(r\cap a)=.40

    P(r\cap \text{na})=.10

    P(r|a)=.40/.50=.80

    P(r|\text{na})=.10/.50=.20
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  11. #11
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    Re: Probability problem.

    Thank you, all!!!
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