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Math Help - 2 quick questions on probability!

  1. #1
    Junior Member
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    Apr 2006
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    2 quick questions on probability!

    1. The probabilities that a batch of 4 computers will contain 1, 2, 3, and 4 defective computers are 0.6274, 0.3102, 0.0575, and 0.001 respectively.

    a. Set up a probability distribution to describe this situation.
    b. Based on these probabilities how many defective computers would you expect to find in this batch?
    c. How much would this value reasonably be expected to fluctuate?
    d. What’s the probability of finding at most 2 defective computers in this batch?

    2. Based on past records, a car insurance company has determined that 9% of all drivers were involved in a car accident last year.

    a. If 12 drivers were randomly selected, what is the probability of getting 0 who were involved in a car accident last year?
    b. If 12 drivers were randomly selected, what is the probability of getting 3 or more who were involved in a car accident last year?

    Thanks!!
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  2. #2
    Member
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    Mar 2008
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    1. a. The probability distribution is more or less given already. There are 5 possible outcomes of the experiment which tests the computers for defects.

    E_0: 0 computers are defective
    E_1: 1 computer is defective
    E_2: 2 computers are defective
    E_3: 3 computers are defective
    E_4: 4 computers are defective

    You define a probability distribution P:\Omega \rightarrow [0,1] such that  \sum P(E_i) = 1, i=0,1,2,3,4 . Your probability space is the set of all events of some experiment, so it might be more precise: computer 1 has the defect, computer 2 has the defect, computer 3 has the defect, computer 4 has the defect. Then the events above are some subsets (elements of the power set of the 4 outcomes).

    Since you are given probabilities of the events E_0 ... E_4 your probability distribution is simply P(E_0), P(E_1), P(E_2), P(E_3), P(E_4).

    b. Given a probability distribution you can find expectation. It is given by \mu = \sum E_i P(E_i)

    c. I suspect this is asking you to find the standard deviation of the distribution. It is given by \sum (\mu - E_i)P(E_i) (or the variance which is just this number squared)

    d. Think about this as the probability of finding exactly 0 defective computers plus the probability of finding exactly 1 defective computer plus the probability of finding exactly 2 defective computers.

    2. a. Assuming that the probability of each random person having been in an accident is independent of the probabilities of any others being in an accident this should be easy. Likewise for part b, just phrase the question as the union of disjoint events.
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