1 + q + q^2 + q^3 + ... = 1/1 - q 1 + 2q + 3q^2 + 4q^3 + ... = 1/(1 - q)^2 Two g
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1 + q + q^2 + q^3 + ... = 1/1 - q 1 + 2q + 3q^2 + 4q^3 + ... = 1/(1 - q)^2 Two gunmen A and B are fighting in a duel. The first one, A, has a probability of hitting 75% and the other one B has 45% chances of hitting a target. In each round they shoot simultaneously at each other. The game is played by rounds. At each round the players shoot at each other simultaneously, hence their speed doesn't matter. The truel is over when ONLY ONE player is still alive or if both of them are dead. We assumed that each player has infinitely many bullets. What are the probabilities of A winning, B winning, and nobody winning? What is the expected number of rounds in a duel? What is the expected number of rounds given A won? What is the expected number of rounds given B won? What is the expected number of rounds given nobody won? Repeat part one but instead of given probabilities use p_A for probability that A hits a target, p_B for probability that B hits a target. Also use q_A and q_B for probability that A or B miss a target. Questions are the same for each group What are the probabilities of A winning, B winning, and nobody winning? What is the expected number of rounds in a duel? What is the expected number of rounds given A won? What is the expected number of rounds given B won? What is the expected number of rounds given nobody won?Explanation / Answer
Answer:
Probability of A hitting the target = 75% = 0.75
Probability of B hitting the target = 45% = 0.45
Probability of A winning: 1st shot: A will win if A hits B and B misses, therefore P = (0.75)*(1-0.45) = (0.75*0.55)
If both A and B misses in first shot then, IInd shot: A will win if A hits B and B misses, therfore P = (0.25*0.55)*(0.75*0.55) (combined probability of 1st miss shot and II shot)
If both A and B misses in Ist and IInd shot then, IIIrd shot: A will win if A hits B and B misses, therfore P = (0.25*0.55)*(0.25*0.55)(0.75*0.55) (combined probability of 1st, IInd miss shot and IIIrd shot)
This will go on till A wins the duel.
Therfore probability of A winning will form a geometric series of the form (0.75*0.55)*[1+r+r2+r3+...]
where r = (0.25*0.55)
Now the sum of infinite geometric series is 1/(1-r), replacing value of sum above:
(0.75*0.55)*[1/(1-0.25*0.55)] = 0.4125/(0.8625) = 0.4782
2. Similarliy Probability of B winning can also be calculated:
(0.45*0.25)*[1+r+r2+r3+...]
where r = (0.25*0.55)
= (0.45*0.25)*[1/(1-0.25*0.55)] = 0.1125/(0.8625) = 0.1304
3. Probability of nobody winning: [1-P(A wins)]*[1-P(B wins)] = [1- (0.4782)]*[(1-0.1304)] = 0.521*0.869 = 0.453
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