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Homework Help: Probability Question homework

  1. May 3, 2014 #1
    1. The problem statement, all variables and given/known data
    Five companies (A,B,C,D and E) that make electrical relays compete each year to be the sole supplier of relays to a major automobile manufacturer. The auto company's records show that the probabilities of choosing a company to be the sole supplier are

    Supplier chain: A B C D E
    Probability: .20 .25 .15 .30 .10

    A) Suppose that supplier E goes out of business this year, leaving the remaining four companies to compete with one another. What are the new probabilities of companies A,B,C and D being chosen as the sole supplier this year?
    B) Suppose the auto company narrows the choice of suppliers to companies A and C. What is the probability that company A is chosen this year?

    2. Relevant equations
    I'm truly not sure where to start on this. I feel like E' = .9 is relevant to find each. Like P(A|E') but I'm having trouble figuring out where to go.

    3. The attempt at a solution
    P(A|E') = P(A and E')/P(E') = P(A)*P(E')/P(E') = P(A) .....yeah I'm obviously confused ;)
  2. jcsd
  3. May 3, 2014 #2
    A) For Part A, E' is now 0.9 (let us call this the total that all probabilities can be as of know) so to finf the new probability of (A) you would P(A)/E' = 0.2/0.9 = 0.22222 = 0.22
    Do the same for B,C,D, and E and you should have a new probability whose sum is (1)

    B)for Part B, the total is now 0.5 (0.2 + 0.3) so for A the probability is 0.2/0.5 = 0.4 and C is 0.6, so the chance that A is chosen is 0.4
  4. May 3, 2014 #3
    Ummm Farouk C = 0.15 and not 0.3, so actually the total would be 0.35....
  5. May 3, 2014 #4

    Ray Vickson

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    When you apply conditional probability arguments, you are essentially assuming that the relative odds remain the same after E fails. In other words: before E fails we have
    [tex] \frac{P(B)}{P(A)} = \frac{.25}{.20}, \:
    \frac{P(C)}{P(A)} = \frac{.15}{.25}, \: \frac{P(D)}{P(A)} = \frac{.30}{.20} [/tex]
    Assume these same ratios hold after E fails, but now we must also have ##P(A) + P(B) + P(C) + P(D) = 1##. Can you find the new Ps? They ought to be the same as ##P(A|E')##, etc.

    If you draw a Venn diagram, the events A, B, C, D and E are mutually exclusive (no overlap) and together give the whole sample space. So, what is ##A \cap E'##? Just look at the diagram to see. That allows an easy computation of ##P(A|E') = P(A \cap E')/P(E')##.
  6. May 3, 2014 #5


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    So you are really asking how E's 10% of the market is to be divided among the others. I would think there should be more information but, assuming that the division is to be proportional to each chain's current percentage. The entire "market" of A, B, C, and D was .20+ .25+ .15+ .30= .90 so A's new share is .20/.90= .2222...., B's new share is .25/.90= .2777..., etc.
    Pretty much the same thing: the total is now just .20+ .15= .35. A's share would be .20/.35= .571428.... and B's share would be .15/.35= .428571....

    Don't just memorize formulas! Learn concepts and think!
  7. May 3, 2014 #6
    Yea My bad :)) Misread it but still the same idea right?
    Edit: It would be 0.2/(0.2 + 0.15) = 0.5714 ??
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