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Probability problem

  1. Oct 1, 2004 #1

    I have an exercise and i want to know if i solved it correct.Can you help?

    Three people A,B,C throw a coin (First is A,second comes B and finally C)
    The first that will bring head wins and the game stops.

    The exercise wants:
    a)The sample space
    b)Determinate the following events in the sample space
    i)A={A WINS}
    ii)B={B WINS)
    iii)(A union B)complement
    a)sample space{H,TH,TTH,TTT}
    bii)P(B)={TH}=P(Acomplement intersection B)=1/4
    biii)(A union B)complement=Acomplement intersection Bcomplement=

    is it corrent?
  2. jcsd
  3. Oct 1, 2004 #2
    a) I think the sample space has countable infinite number of elements.
    what if all A, B, C fail in their first trials? They again start with A. So ur sample space is extended to TTTH, TTTTH......etc,

    bi)A can win in the first round with prob 1/2
    or in the 2nd round with prob 1/16
    or in the 3rd round with prob 1/128

    Therefore P(A) = 1/2+1/16+1/128+....... = 4/7

    bii) Smilarly P(B) = 1/4+1/32+.......= 2/7
    biii)A and B r mutually exclusive. P(A union B) = P(A) + P(B) = 6/7

    P(A union B)complement = 1-6/7 = 1/7.

    Aliter: P(A union B) means either A wins or B wins. So P(A union B)complement means neither A nor B wins ==> P(C wins) = 1/7(find out as above).
  4. Oct 1, 2004 #3
    You have right.
    But can you explain me how did you find
    "or in the 2nd round with prob 1/16
    or in the 3rd round with prob 1/128"

    i know that is:1/2*1/2*1/2*1/2

    but can you explain me why we multiply the probablilities plz?
  5. Oct 1, 2004 #4
    because they r independent. that means whether B gets a H/T doesnt depend on the previous outcome(A' toss). similarly with C and so on........
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