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Stats Question

  1. Sep 11, 2015 #1
    1. The problem statement, all variables and given/known data

    Relative frequencies of amount purchased and method of payment

    Cash Credit Debit
    $<20 .09 .03 .04
    $20-$100 .05 .21 .18
    >$100 .03 .23 .14


    (a) What proportion of purchases are paid for in cash?

    (b) Given that a purchase is for more than $100, what is the probability that it is paid for by credit?

    (c) Are payment by credit and amount > $100 independent events?

    2. Relevant equations

    Conditional probability, independence, etc.

    3. The attempt at a solution

    (a) .09 + .05 + .03 = .17

    (b) P(credit | >$100) = P(credit >$100)/P(>$100) = .23/.40 = .575

    (c) Independent if P(credit >$100) = P(credit)P(>$100) = .23.


    P(credit)P(>$100) = 0.47*0.40 = .188. not independent.


    Does that look right?
     
    Last edited: Sep 11, 2015
  2. jcsd
  3. Sep 11, 2015 #2

    andrewkirk

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    It looks right to me.
     
  4. Sep 11, 2015 #3
    Thanks. I initially overcomplicated it and then thought about it a bit more carefully.

    I have another question. Is this a true relation P(~E | F) + P(E | F) = 1?
     
  5. Sep 11, 2015 #4
    Actually, I had a typo in there. It should be right now.
     
  6. Sep 11, 2015 #5

    andrewkirk

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    You can reason that one out using the measures:
    $$P(E| F)\equiv\frac{P(E\cap F)}{P(F)}$$
    So
    $$P(E| F)+P(\sim E| F)\equiv\frac{P(E\cap F)}{P(F)}+\frac{P((\sim E)\cap F)}{P(F)}
    =\frac{P(E\cap F)+P((\sim E)\cap F)}{P(F)}
    $$

    The two sets in the numerator are disjoint, so you can use the rule for the probability/measure of the union of two disjoint sets to convert the numerator to a single P(something).
     
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