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Probability with replacement

  1. Oct 13, 2011 #1
    A coin purse contains 9 nickels and 1 quarter. Two coins are drawn WITH REPLACEMENT. What is the set S of this random experiment? What are the probabilities of the elementary events?

    i'm a little confused by the "with replacement" concept. i'm just wondering if i'm doing it correctly. i have the equation n-1+k choose n-1 so here's my work:

    set S = {two nickels, two quarters, one nickel and one quarter}
    total possible ways of selection: (10-1+2) choose (10-1) = 55
    P[two nickels] = (9-1+2) choose (9-1) / 55 = 45/55
    P[two quarters] = (2-1+2) choose (2-1) / 55 = 3/55
    P[one nickel and one quarter] = 1 - Pall other events = 1 - (45/55 + 3/55) = 7/55

    have i done the problem correctly?
     
  2. jcsd
  3. Oct 13, 2011 #2
    What is the probability of drawing a quarter for the first coin? With replacement, would the probability of drawing a quarter the second time be any different?

    The results of the coin drawings would be independent events, so what would the probability of drawing two quarters be, in terms of the probabilities of drawing a quarter for the first and second drawings?
     
  4. Oct 13, 2011 #3
    probability for drawing a quarter for the first coin = 1/10
    probability of drawing a quarter for the second coin = 1/10
    so P[drawing 2 quarters] = 1/100 ?

    do i follow the same approach for the probability of each of the elementary events? like for 1 nickels..
    probability of drawing 1 nickel first = 9/10
    probability of drawing 1 nickel second = 9/10
    so P[drawing 2 nickels] = 81/100

    and then for 1 nickel and 1 quarter
    probability of drawing 1 nickel = 9/10
    probability of drawing 1 quarter = 1/10
    P[1 nickel, 1 quarter] = 9/100

    but the probabilities do not add up to 1 so i am still confused :\
     
  5. Oct 13, 2011 #4

    Mark44

    Staff: Mentor

    Mod note: Moved from Engineering & CS section.
     
  6. Oct 13, 2011 #5
    There are two ways the two coins drawn would be a nickel and a quarter.
     
    Last edited: Oct 13, 2011
  7. Oct 13, 2011 #6
    Ahh, I see. This is because order doesn't matter, correct? so P[1 nickel, 1 quarter] = 9/100 + 9/100 = 18/100 = 9/50
     
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