I find the questions of this week are more difficult than usual. I am in dire need of assistance!(adsbygoogle = window.adsbygoogle || []).push({});

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]

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Another prob question

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**