SOLUTION: In a duel between X and Y both duelists have 2 bullets. The first one to shoot is X, with a chance to hit of 30%, afterwards if X missed Y shoots with a chance to hit of 60%, after

Question 1045311: In a duel between X and Y both duelists have 2 bullets. The first one to shoot is X, with a chance to hit of 30%, afterwards if X missed Y shoots with a chance to hit of 60%, afterwards if Y missed X shoots with a chance to hit of 40% and finally if X missed Y shoots with a chance to hit of 50%. What is the probability that Y will hit X?
Found 3 solutions by ikleyn, Boreal, robertb:
Answer by ikleyn(52900)   (Show Source): You can put this solution on YOUR website!
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In a duel between X and Y both duelists have 2 bullets. The first one to shoot is X, with a chance to hit of 30%,
afterwards if X missed Y shoots with a chance to hit of 60%,
afterwards if Y missed X shoots with a chance to hit of 40% and
finally if X missed Y shoots with a chance to hit of 50%.
What is the probability that Y will hit X?
~~~~~~~~~~~~~~~~~~~~~~~~

I think it is

(1-0.3)*0.6 + (1-0.3)*0.4*(1-0.4)*0.5 = 0.504.

Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
X hits----0.3
X misses-0.7------Y hits 0.6
-------------------Y misses 0.4----X hits 0.4
-------------------X misses 0.6---Y hits 0.4
X hits Y:0.3
Y hits X:0.7*0.6 (0.42)
X hits: 0.3*0.4*0.4=0.048
Y hits: 0.7*0.4*0.6*0.4=0.0672
Y hits on the first attempt: 0.42
Y hits on the second attempt 0.0672
Answer is Probability of 0.4872 that Y will hit X.
--------------------------------
Probability X will hit Y is 0.3+0.7*0.4*0.4=0.412
The remainder of 0.111 is that neither hits the other.
Answer by robertb(5830)   (Show Source): You can put this solution on YOUR website!
We go by the assumption that once a duelist hits his opponent, then the hit opponent cannot hit back anymore, and the duel ends.
With that said, there are only 5 possible elements to this sample space, which are the following mutually exclusive events:
{X hit},
{X no-hit, Y hit},
{X no-hit, Y no-hit, X hit}
{X no-hit, Y no-hit, X no-hit, Y hit}
{X no-hit, Y no-hit, X no-hit, Y no-hit}.
This listing can also be illustrated using a TREE DIAGRAM, with the proper labeling of probabilities being indicated at every respective
branch.
We are interested in cases 2 and 4 above, where Y ends up hitting X.
===> probability is 0.7*0.6 + 0.7*0.4*0.6*0.5 = .
N.B.:
The probability that X hits Y is done similarly (represented by cases 1 and 3), and is
0.3 + 0.7*0.4*0.4 = 0.412.
The probability that neither of them hits the other during the whole process is 0.7*0.4*0.6*0.5 = 0.084. (Both of them walk off unscathed.)
Checking the probabilities for all cases gives 0.504 + 0.412 + 0.084 = 1.
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