1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Another prob question

  1. Oct 9, 2006 #1

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    Sol:

    [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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Another prob question
  1. Another question (Replies: 3)

  2. Another question (Replies: 3)

Loading...