Results 1 to 11 of 11

Math Help - just to check a simple probability problem

  1. #1
    Newbie
    Joined
    May 2011
    Posts
    19

    just to check a simple probability problem

    I worked this simple problem and am not sure if it is correct and I have another 19 questions similar to this therefore I need to be sure of what I am doing.

    The problem broken in simple terms..
    company paced order for 2 products
    event E => 1st product out of stock Pr(E)=0.3
    event F => 2nd product out of stock Pr(F)=0.2
    P(E or F) = 0.2

    a) Pr( both out of stock) => (Pr(E)*(Pr(F)) = (0.3*0.2)= 0.6

    b) are e and f independent => I think they are independent since e has nothing to do with f. (I am not sure if this is the correct reason). But if they were independent why should they have an intersection. Pr(EandF)

    c) p(F given E) => P(F|E) = P(F and E) / P(E) = 0.6/0.3 and here I noticed I have worked wrong since Pr can never be greater than 1


    well , should I reduce p(e or f)=0.4 thus
    ans a ) 0.3*0.2-0.4 ??


    Thanks in advance
    Last edited by terence; December 1st 2011 at 05:59 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Re: just to check a simple probability problem

    Quote Originally Posted by terence View Post
    I worked this simple problem and am not sure if it is correct and I have another 19 questions similar to this therefore I need to be sure of what I am doing.

    The problem broken in simple terms..
    company paced order for 2 products
    event E => 1st product out of stock Pr(E)=0.3
    event F => 2nd product out of stock Pr(F)=0.2
    P(E or F) = 0.2

    a) Pr( both out of stock) => (Pr(E)*(Pr(F)) = (0.3*0.2)= 0.6

    b) are e and f independent => I think they are independent since e has nothing to do with f. (I am not sure if this is the correct reason). But if they were independent why should they have an intersection. Pr(EandF)

    c) p(F given E) => P(F|E) = P(F and E) / P(E) = 0.6/0.3 and here I noticed I have worked wrong since Pr can never be greater than 1


    well , should I reduce p(e or f)=0.4 thus
    ans a ) 0.3*0.2-0.4 ??


    Thanks in advance
    For the first part remember that even if events are not indepentant that

    P(A \cup B) = P(A)+P(B)-P(A \cap B)

    You everything except the last one.

    For the 2nd one you are correct two events are independant if

    P(A)P(B)=P(A \cap B) use your answer from the first part to answer this.

    For part 3 just use the correct answer from part 1.

    I hope this clears it up.

    TES
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2011
    Posts
    19

    Re: just to check a simple probability problem

    Quote Originally Posted by TheEmptySet View Post
    For the first part remember that even if events are not indepentant that

    P(A \cup B) = P(A)+P(B)-P(A \cap B)
    Oh this was my misconception then, i thought that this was for dependent events only.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Re: just to check a simple probability problem

    Quote Originally Posted by terence View Post
    Oh this was my misconception then, i thought that this was for dependent events only.

    Thanks
    Remember if events are independant we have that

    P(A \cap B)=0

    Then the formula becomes

    P(A \cup B)=P(A)+P(B)-0=P(A)+P(B)

    This is the formula is they are independant!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,708
    Thanks
    1638
    Awards
    1

    Re: just to check a simple probability problem

    Quote Originally Posted by TheEmptySet View Post
    Remember if events are independant we have that \color{red}P(A \cap B)=0
    That is not correct.
    We have P(A \cap B)=0 if the events are mutually exclusive.
    Independent events are not necessary mutually exclusive.

    If the events are independent then P(A\cap B)=P(A)\cdot P(B)~.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    May 2011
    Posts
    19

    Re: just to check a simple probability problem

    But for the problem P (A intersection B ) is not equal to 0. Therefore if I can understand well ,it can be confirmed that events are mutually exclusive. At the same time though events are independent.

    If events are dependent, they are still multiplied but but P(B) will have a different value than the original one. (conditional probability should be taken care of then.)

    Am i correct ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,708
    Thanks
    1638
    Awards
    1

    Re: just to check a simple probability problem

    Quote Originally Posted by terence View Post
    I worked this simple problem and am not sure if it is correct and I have another 19 questions similar to this therefore I need to be sure of what I am doing.
    The problem broken in simple terms..
    company paced order for 2 products
    event E => 1st product out of stock Pr(E)=0.3
    event F => 2nd product out of stock Pr(F)=0.2
    P(E or F) = 0.2
    This question contains an internal error.
    It is impossible for \color{red}P(E\cup F)=0.2.
    So either you copied it incorrectly or it is a faulty question.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    May 2011
    Posts
    19

    Re: just to check a simple probability problem

    it is 0.4 sorry, I typed incorrectly. I have worked the question out. a) 0.3*0.2-0.4 = 0.2 b)indipendent c) 0.2/0.3 =2/3

    but i still have the following doubts..

    "But for the problem P (A intersection B ) is not equal to 0. Therefore if I can understand well ,it can be confirmed that events are mutually exclusive. At the same time though events are independent.

    If events are dependent, they are still multiplied but but P(B) will have a different value than the original one. (conditional probability should be taken care of then.)

    Am i correct ?"
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,708
    Thanks
    1638
    Awards
    1

    Re: just to check a simple probability problem

    Quote Originally Posted by terence View Post
    it is 0.4 sorry, I typed incorrectly.
    a) \begin{align*} P(E\cap F) &= P(E)+P(F)-P(E\cup F)\\ &= 0.3+0.2-0.4 \\ &=0.1    \end{align*}.

    b) 0.06=P(A)P(B)\ne P(A\cap B)=0.1 Therefore, they are not independent.

    c) P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{0.1}{0.3}=\frac{1}{3}
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    May 2011
    Posts
    19

    Re: just to check a simple probability problem

    Quote Originally Posted by Plato View Post
    a) \begin{align*} P(E\cap F) &= P(E)+P(F)-P(E\cup F)\\ &= 0.3+0.2-0.4 \\ &=0.1    \end{align*}.


    Isn't it like this \begin{align*} P(E\cap F) &= P(E)*P(F)-P(E\cup F)\end{align*}?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,708
    Thanks
    1638
    Awards
    1

    Re: just to check a simple probability problem

    Quote Originally Posted by terence View Post
    Isn't it like this \begin{align*} P(E\cap F) &= {\color{red}P(E)*P(F)}-P(E\cup F)\end{align*}?
    That is simply and completely wrong.

    This is the rule:
    P(A\cup B)=P(A)+P(B)-P(A\cap B) thus P(A\cap B)=P(A)+P(B)-P(A\cup B).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Probability Problem... (check, please!)
    Posted in the Statistics Forum
    Replies: 3
    Last Post: April 4th 2011, 06:44 PM
  2. Simple Probability Problem
    Posted in the Statistics Forum
    Replies: 3
    Last Post: April 6th 2010, 06:14 AM
  3. Replies: 4
    Last Post: January 31st 2010, 09:49 AM
  4. Replies: 3
    Last Post: April 13th 2009, 06:51 PM
  5. [SOLVED] Probability problem check please
    Posted in the Statistics Forum
    Replies: 10
    Last Post: January 15th 2009, 12:26 AM

Search Tags


/mathhelpforum @mathhelpforum