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Another prob question

  1. Oct 9, 2006 #1


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    I find the questions of this week are more difficult than usual. I am in dire need of assistance!

    A and B fight in a duel. They pick up their guns and fire once at each other. A kills B with probability [itex]p_A[/itex] and B kills A with probability [itex]p_B[/itex]. If no one is killed, they repeat the process. What is....

    a) The probability that A does not die


    [itex]\Omega[/itex]: Every possible outcome of the duel.

    [itex]\Omega = \left\{ (A), (B), (AB), (d,A), (d,B), (d,AB), (d,d,A),..., (d,d,d,d,...)\right\}[/itex]

    where A means "A wins", B measn "B wins", AB means "A and B both die" and "d" means a draw.

    E: A does not die.

    [itex]E_i[/itex]: The duel lasts i rounds and end up with the death of B and the survival of A.

    O: Nobody ever dies, i.e. it is the case of perpetual draws.

    [tex]E=\bigcup_{i=1}^{\infty}E_i \cup O[/tex]

    Since these are all disjoint sets, P(E) is the sum of the probabilities, and we have

    [tex]P(E_i) = [(1-p_A)(1-p_B)]^{i-1}p_A(1-p_B) \equiv r^{i-1}p_A(1-p_B)[/tex]

    (I made the hypothesis of independance btw the shots, i.e. P(A and B miss) = P(A misses)P(B misses))

    [tex]P(O)=\lim_{i\rightarrow \infty}[(1-p_A)(1-p_B)]^{i-1}=0[/tex]

    [tex]\therefore P(E) = p_A(1-p_B)\sum_{i=0}^{\infty}r^i= \frac{p_A(1-p_B)}{1-r}[/tex]
    Last edited: Oct 10, 2006
  2. jcsd
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