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Math Help - contracts/Tenders

  1. #1
    Newbie
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    Feb 2008
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    contracts/Tenders



    A company has decided to bid for a new contract which is up for tender. There are three other competing companies who could possibly take on the contract and it is thought there is
    a 60% chance that competitor A will bid,
    a 50% chance that competitor B will bid,
    and 40% chance that competitor C will bid.

    1. What is the probability that one other competitor will bid for the contract?
    (probable answer: 0.6*0.5*0.6 + 0.4*0.5*0.6 + 0.4*0.5*0.4 =0.38)

    2. Assuming that all companies who bid for the contract have an equal opportunity of winning it, What is the probability that:

    (a) you will win the contract?
    ( guessed answer:P(win)=p(loss)=0.5 let our company be Z.
    p(Z win,A loss,B Loss,C Lose) +(Z w, A L, B L)*3[because instead of AB there could be BC or CA & their prob. is the same] +( Z w, A L) *3[because instead of A there could be B or C] + (Z w) =0.9375)

    (b) Competitor A wins the contract?
    (is it similar to the a one above?)

    Given that you do eventually win the contract, what is the probability

    (c) no other companies bid for the contract?

    (d)exactly one other company bid for the contract?




    Any kind of help will be appreciated. Thanks.
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  2. #2
    Senior Member
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    Melbourne
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    1. What is the probability that one other competitor will bid for the contract?
    (probable answer: 0.6*0.5*0.6 + 0.4*0.5*0.6 + 0.4*0.5*0.4 =0.38)
    You have calculated the probability of exactly one competitor bidding correctly. I would have interpreted this question as being the probability of one or more competitor bidding, but that's just me.

    2. Assuming that all companies who bid for the contract have an equal opportunity of winning it, What is the probability that:

    (a) you will win the contract?
    ( guessed answer:P(win)=p(loss)=0.5 let our company be Z.
    p(Z win,A loss,B Loss,C Lose) +(Z w, A L, B L)*3[because instead of AB there could be BC or CA & their prob. is the same] +( Z w, A L) *3[because instead of A there could be B or C] + (Z w) =0.9375)
    There may be a simpler way to do this, but I did it this way:

    P(0 bidders) = .12
    P(1 bidder) = .38
    P(2 bidders) = .38
    P(3 bidders) = .12

    P(winning against 0 opponents) = 1
    P(winning against 1 opponent) = .5
    P(winning against 2 opponents) = 1/3
    P(winning against 3 opponents) = .25

    P(winning) = .12*1+.38*.5+.38*1/3 +.12*.25 = .4666...

    (b) Competitor A wins the contract?
    (is it similar to the a one above?)
    This is indeed similar to the previous part. Try it yourself using the same method, but instead of finding the probability of there being a certain number of competitors you will need the probability that A will be competing against a certain number of competitors.

    (c) no other companies bid for the contract?
    \frac {0.12}{.46666...} 0.12 is the probability of winning unopposed and .4666... is the probability of winning.


    (d)exactly one other company bid for the contract?
    similar to c). find the probability of winning with one opponent and divide by probability of winning.
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  3. #3
    Newbie
    Joined
    Feb 2008
    Posts
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    thanks

    thanks a lot for your help badgerigar . i got the whole answer now. Could you also help me with other probability question? It is fairly complicated one.

    the link is: http://www.mathhelpforum.com/math-he...chemicals.html

    Thanks again.
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