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Part 1: Gunman X and gunman Y are in a duel. Let's say that [#permalink]
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23 Jul 2003, 01:08
This topic is locked. If you want to discuss this question please repost it in the respective forum. Part 1:
Gunman X and gunman Y are in a duel. Let's say that gunman X will kill his target with one shot with probability P, and gunman Y will kill his target with one shot with probability Q. If gunman X shoots first, and they alternate shots until somebody dies, what is the probability that gunman X will survive?
Useful identity: if 0 < r < 1, then a*(1 + r + r^2 + r^3 + ... + r^infinity) converges to: a / (1  r).
(A) P / (P + Q)
(B) P / ((1  P)(1  Q))
(C) P / (1  (1  P)(1  Q))
(D) P / (1  P * Q)
(E) P * (1  Q)(1  P)
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AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993



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X kills Y by doing one shot  P or P*[(1P)(1Q)]^0
X kills Y by doing two shots P*[(1P)(1Q)]^1
X kills Y by doing three shots P*[(1P)(1Q)]^2
.............
X kills Y by doing N shots  P*[(1P)(1Q)]^(N1)
.............
X kills Y by doing an infinite number of shots  P*[(1P)(1Q)]^inf
We need to sum all the foregoing probabilities by employing the hint formula and take P as a common: r = (1P)(1Q)
P/(1+(1P)(1Q))
no such option, but I think it is C.
As written C is wrong, its denominator is less than 1, making the option be more than 1. Absurd.
I insist that the answer be P/(1+(1P)(1Q))



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stolyar wrote: X kills Y by doing one shot  P or P*[(1P)(1Q)]^0 X kills Y by doing two shots P*[(1P)(1Q)]^1 X kills Y by doing three shots P*[(1P)(1Q)]^2 .............
X kills Y by doing N shots  P*[(1P)(1Q)]^(N1) .............
X kills Y by doing an infinite number of shots  P*[(1P)(1Q)]^inf
We need to sum all the foregoing probabilities by employing the hint formula and take P as a common: r = (1P)(1Q)
P/(1+(1P)(1Q))
no such option, but I think it is C. As written C is wrong, its denominator is less than 1, making the option be more than 1. Absurd.
I insist that the answer be P/(1+(1P)(1Q))
Why is it absurd that the denominator be less than 1?
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AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993



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because the numerator is less than 1.
albeit... 0.9/0.99=0.90909... is OK.
Even so, I think there is a typo in C.



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stolyar wrote: because the numerator is less than 1. albeit... 0.9/0.99=0.90909... is OK.
Even so, I think there is a typo in C.
The identity = a / (1  r) which matches C.
Think about it. a * (1 + r + r^2 + ...) is "a" times (1 + an infinite series of positive numbers). Which means the result > a.
a / (1  r) must be > a, so the denominator must be < 1.
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AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993



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my bad
if 0 < r < 1, then a*(1 + r + r^2 + r^3 + ... + r^infinity) converges to: a / (1  r). A minus and not a plus! But there was a plus  I remember. A hard day...
So, r=(1P)(1Q)
and X=P/(1(1P)(1Q))
C again.



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stolyar wrote: my bad
if 0 < r < 1, then a*(1 + r + r^2 + r^3 + ... + r^infinity) converges to: a / (1  r). A minus and not a plus! But there was a plus  I remember. A hard day...
So, r=(1P)(1Q)
and X=P/(1(1P)(1Q))
C again.
Very nicely done and a fairly clear explanation (let me know if you want a more detail explanation). Now try part 2!
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AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993



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Challenge Problem: Probability Part 1: [#permalink]
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23 Jul 2003, 04:54
I agree with Akamaibrah  it should be C



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Re: Challenge Problem: Probability Part 1: [#permalink]
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23 Jul 2003, 21:16
AkamaiBrah wrote: Part 1: Gunman X and gunman Y are in a duel. Let's say that gunman X will kill his target with one shot with probability P, and gunman Y will kill his target with one shot with probability Q. If gunman X shoots first, and they alternate shots until somebody dies, what is the probability that gunman X will survive?
Useful identity: if 0 < r < 1, then a*(1 + r + r^2 + r^3 + ... + r^infinity) converges to: a / (1  r).
(A) P / (P + Q) (B) P / ((1  P)(1  Q)) (C) P / (1  (1  P)(1  Q)) (D) P / (1  P * Q) (E) P * (1  Q)(1  P)
I am not very sure on how I should interpret the question .
The point that I am trying to make is that , I am not very sure that u can "ADD" all the distinct probabalities.. Each probability is complete in itself.
The questions might need more information to get an answere. Like if I were to ask what is the probabality that X survives after "n" shots , assuming that he kills Y on the n'th shot .. and that means he will survive.
Then question is complete .
else if I were to ask u what is the probability that I withdraw a " King" from a pack of cards.
You might want to ask me how many cards do I withdraw since in each case the probabality changes .. but u cannot add those probabalities to get to an answere.......
So probabality that he survives should actually not be the "sum" of his probabality that he survives shots till infinity.......



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Re: Challenge Problem: Probability Part 1: [#permalink]
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23 Jul 2003, 21:45
Quote: I am not very sure on how I should interpret the question . The point that I am trying to make is that , I am not very sure that u can "ADD" all the distinct probabalities.. Each probability is complete in itself. The questions might need more information to get an answere. Like if I were to ask what is the probabality that X survives after "n" shots , assuming that he kills Y on the n'th shot .. and that means he will survive.
Then question is complete . else if I were to ask u what is the probability that I withdraw a " King" from a pack of cards.
You might want to ask me how many cards do I withdraw since in each case the probabality changes .. but u cannot add those probabalities to get to an answere.......
So probabality that he survives should actually not be the "sum" of his probabality that he survives shots till infinity
The probability of "something happening" OR "something else happening" is the sum of the probability of the two events, so long as the two events are distinct or mutually exclusive (i.e., cannot happen at the same time).
For example, if I ask you: If I deal 4 cards out of a deck, what is the probability that the Ace of Spades will be one of those cards.
Well, the Ace can be either the 1st card OR the 2nd card OR the 3rd card OR or the 4th card. Since it cannot be any two of the previous events at the same time, we can add the probabilities together.
P(1st card) = 1/52
P(2nd card) = 51/52 * 1/51 = 1/52
P(3rd card) = 51/52 * 50/51 * 1/50 = 1/52
P(4th card) = 51/52 * 50/52 * 49/50 * 1/49 = 1/52
Hence, the total probability of all of the above is 4/52.
Note, if you were to extend this to this entire deck, the probability of drawing the Ace of Spades is 52 * 1/52 = 1 or the obvious conclusion that it is certain that we will draw the Ace of Spades if we examine every card in the deck.
Computing the probability of "duels" is a similar type of problem. Player X can die in either the first round OR the second round OR the third round OR the fourth round OR .... the 1000th round. In this case, There is always a chance for each person to die in a particular round if he has survived to that point.
As you correctly point out, we can calculate the exact probability that the person will die in a SPECIFIC round. However, if we want to calculate the probabilty that he will die IN GENERAL, we must add up all of the individual probabilities that he will die in each specific round. This probability gets geometrically smaller and smaller as the rounds progress, and the total probabilty (the sum of all of the probabilities) always converges to a single number for each player.
Hope this made some sense.
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AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993




Re: Challenge Problem: Probability Part 1:
[#permalink]
23 Jul 2003, 21:45






