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Math Help - Easy probability help please

  1. #1
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    Easy probability help please

    Can someone explain to me how to do these problems. The answers are already there, I just need to know how to do them for a test tomorrow. I missed last week of school.

    1. Sixty percent of the employees in a firm are engineers. Seventy percent of the employees in the firm have been employed for over 5 years. Fifty five percent of the non engineers have been with the firm for over 5 years..

    A.) What’s the probability that an individual is either not an engineer or has been
    employed for over 5 years? . .88

    B.)If an employee has been with the firm for over 5 years, what’s the probability that he/she is not an engineer? .31

    C)What is the probability that a selected employee is both an engineer and has been employed for more than 5 years? .48


    D.) Is being an engineer independent of being with the firm for over 5 years? Explain using statistics? P(A)=.6
    P(A│B)= .69 A & B not independent
    A= engineer; B= employed for over 5 yrs


    2. Forty percent of the individuals in a survey are females. Sixty percent of the females drive foreign cars, while 70 percent of the males drive domestic cars.

    A.) If an individual drives a foreign car what’s the probability that the individual is a female?
    .57

    B.) What’s the probability that an individual drives a domestic car?
    .58
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  2. #2
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    Re: Easy probability help please

    Hey thespacepope.

    Why don't you start off by expressing your thoughts on why you think the answer is the answer and how you think you get there even if you aren't 100% sure?
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  3. #3
    Newbie rudsdiey's Avatar
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    Re: Easy probability help please

    Question 1
    In this case, the question focus on 2 main variables, which are Engineers (E) versus Non-Engineers (NE), and Employed over 5 years (EO) versus Employed less 5 years (EL).

    Lets collect all the data for E, NE, EO and EL.
    E = 60 ; means NE = 40
    EO= 70 ; means EL = 30
    Use sample 100

    So you need find the probability of following variables;
    1. Non-engineers employed over 5 years (NEO)
    2. Engineers employed over 5 years (EEO)
    3. Non-engineers employed less 5 years (NEL)
    4. Engineers employed less 5 years (EEL)

    1. Non-engineers employed over 5 years (NEO)
    Given 55% of NE is NEO, means
    NEO = (55/100)*40 = 22
    This means that there are 22 non-engineers of 70 employees who employed over 5 years.

    2. Engineers employed over 5 years (EEO)
    EEO = EO – NEO = 70 – 22 = 48
    This means that there are 48 engineers of 70 employees who employed over 5 years.

    3. Non-engineers employed less 5 years (NEL)
    NEL = NE – NEO = 40 – 22 = 18
    This means that there are 18 non-engineers of 30 employees who employed less 5 years.

    4. Engineers employed less 5 years (EEL)
    EEL = EL – NEL = 30- 18 = 12
    This means that there are 12 engineers of 30 employees who employed less 5 years.


    Above information will help to answer the question.

    Cheers
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  4. #4
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    Re: Easy probability help please

    Wanting to delete a topic makes it seem as if you don't want your professor to find this topic. I'm not saying this is necessarily the case, but that's what how it is interpreted by many.
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