Results 1 to 9 of 9

Math Help - Probabilty problem

  1. #1
    Junior Member
    Joined
    Apr 2009
    Posts
    44

    Smile Probabilty problem

    A heavy-equipment salesperson can contact either one or two customers per day with probability 1/3 and 2/3 ,respectively .Each contact will result in either no sale or a 50,000$ sale ,with the probabilities 0.9 and 0.1 ,respectively .Give the probability distribution for daily sales .Find the mean and standard deviation of the daily sales .


    I am completely confused with this Qn because I was unable to apply E(X) and V(X) here .
    Can somebody explain me in detail ?


    What is the meaning of E(X+Y) provided that X and Y are two random variables ?
    What is the meaning of E(X-Y) ?

    Thank You
    Last edited by K A D C Dilshan; May 18th 2009 at 10:32 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    Let X=$ in sales per day.
    It seems X can be 0, 50k or 100k based on 0,1 or 2 sales.
    You need to partition this, i.e.

     P(X=0)=P(X=0, 2 customers) +P(X=0, 1 customer)

     = P(X=0|2 customers) P(2 customers) + P(X=0|1 customer) P(1 customer)

     =(.9)(.9)(2/3) + (.9)(1/3)

    Since  P(X=0)+P(X=50k)+P(X=100k)=1

    I'll only compute  P(X=100k)=P(X=100k, 2 customers)

    =P(X=100k| 2 customers)P(2 customers)=(.1)(.1)(2/3)

    Then you compute E(X) by its definition and V(X)=E(X^2)-(EX)^2.

    And we need to assume independence between the customers too, when the salesperson see two of them.
    Last edited by matheagle; May 18th 2009 at 11:53 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Firstly, you need to find which values a daily sales can take.

    In the case where there is only one customer :
    0 or 50000
    In the case where there are two customers :
    0 (both contacts result in no sale), 50000 (exactly one of the contacts), 100000 (both contacts)

    So finally, the possible values are :
    0,50000,100000

    Now I have to ask you... do you know about conditional expectation ?

    Otherwise, you'll have to calculate the different probabilities associated to each value.
    For example for value 0 :
    \mathbb{P}(X=0)=\mathbb{P}(X=0|N=1)\mathbb{P}(N=1)  +\mathbb{P}(X=0|N=2)\mathbb{P}(N=2)= 0.9/3+0.9^2\times 2/3

    What is the meaning of E(X+Y) provided that X and Y are two random variables ?
    What is the meaning of E(X-Y) ?
    Imagine you throw two dices. X represents the number on the first dice, Y on the second. What's the average of the sum of numbers ? That's E(X+Y).
    What's the average for the difference of numbers ? That's E(X-Y).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Apr 2009
    Posts
    44
    can you please moo show me the steps to achieve the answers ? ..
    are those 0.9 and 0.1 are conditional probabilities ?
    ifso is this right ?
    0 -0.9
    50000 -1/30
    100000-1/15

    what is conditional expectation ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    I have the conditionals above
    P(AB)=P(A|B)P(B) after I partitioned this via the number of customers.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by K A D C Dilshan View Post
    can you please moo show me the steps to achieve the answers ? ..
    are those 0.9 and 0.1 are conditional probabilities ?
    ifso is this right ?
    0 -0.9
    50000 -1/30
    100000-1/15

    what is conditional expectation ?
    0.9 just means that there is no sale.

    In other words, if you have N=1, what's the probability that the number of sales is 0 ? It's 0.9
    If you have N=2, what's the probability that the number of sales is 0 ? It's 0.9x0.9, because you need two contacts to give no sale.

    Conditional probability...that's not for now if you've never heard of it
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Apr 2009
    Posts
    44
    please moo show the steps to solve the problem ?
    and how to prove E(X+Y)=E(X)+E(Y)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    E(X+Y) =\int\int (x+y)f(x,y)dxdy

    = \int\int xf(x,y)dxdy +\int\int yf(x,y)dxdy=E(X)+E(Y)

    which also works in the discrete and mixed settings.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Apr 2009
    Posts
    44

    Smile

    I am really sorry everybody ..I couldn't see your all answers as you all have used latex code here ..sorry for making same reply again again ..thank you eagle and moo giving me the idea of conditional expected values ..

    I have another problem
    how to perform the above proof for a discreet distribution ..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Probabilty problem
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 21st 2010, 09:04 PM
  2. problem involving probabilty
    Posted in the Statistics Forum
    Replies: 1
    Last Post: January 18th 2010, 11:11 AM
  3. Probabilty for a fair game problem w. binomials
    Posted in the Statistics Forum
    Replies: 6
    Last Post: November 22nd 2009, 12:23 PM
  4. Probabilty problem
    Posted in the Statistics Forum
    Replies: 0
    Last Post: May 15th 2009, 07:09 PM
  5. NCAA Probabilty Problem: HELP!!!!!!!
    Posted in the Advanced Statistics Forum
    Replies: 8
    Last Post: March 28th 2008, 11:54 AM

Search Tags


/mathhelpforum @mathhelpforum