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25 Apr 2019, 03:06
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
36% (01:50) correct
64%
(01:28)
wrong
based on 140
sessions
History
Date
Time
Result
Not Attempted Yet
Larry and Julius are playing a game, taking turns throwing a ball at a bottle sitting on a ledge. Larry throws first. The winner is the first person to knock the bottle off the ledge. At each turn the probability that a player knocks the bottle off the ledge is 1/2, independently of what has happened before. What is the probability that Larry wins the game?
(A) 1/2
(B) 3/5
(C) 2/3
(D) 3/4
(E) 4/5
(A) 1/2
(B) 3/5
(C) 2/3
(D) 3/4
(E) 4/5
27 Apr 2019, 02:25
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If Larry wins, he either wins on the first move, or the third move, or the fifth move, etc. Let represent "player wins", and represent "player loses". Then the events corresponding to Larry winning are W,LLW,LLLLW.....
Thus the probability of Larry winning is
1/2,1/8,1/32......
This is a infinite geometric series with ratio: 1/4
Formula to solve this is (1/1-r) where r is the common ratio
hence the answer is :2/3
Thus the probability of Larry winning is
1/2,1/8,1/32......
This is a infinite geometric series with ratio: 1/4
Formula to solve this is (1/1-r) where r is the common ratio
hence the answer is :2/3
General Discussion
27 Apr 2019, 04:09
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This question is a decision-tree problem.
Obviously, because Larry goes first, Larry clearly has the advantage if she luckily wins in the first turn. This game is an unfair game for Julius -> The probability that Larry wins must be higher than 1/2.
We have the following scenarios: (LW as Larry wins, LL as Larry loses, JW as Julius wins, JL as Julius loses)
LW
LL -> JW
--------JL -> LW
--------------- LL -> JW
-----------------------JL -> LW
------------------------------LL -> [...]
This will continue until there is one person wins the game or we can say it continues to infinitive.
The chance for Larry to win lies at 1st, 3rd, 5th, 7th turns -> odd turns
=> The probability now is: (0.5)+ (0.5)^3 + (0.5)^5 + (0.5)^7 + ... + (0.5)^n with n is an odd number
The question now asks us to count the sum of a sequence.
Applying the Geometric Progression, we have b1=0.5, r = (0.5)^2, the sum is 0.5/[1 - (0.5)^2] = 0.5/0.75 = 2/3
Obviously, because Larry goes first, Larry clearly has the advantage if she luckily wins in the first turn. This game is an unfair game for Julius -> The probability that Larry wins must be higher than 1/2.
We have the following scenarios: (LW as Larry wins, LL as Larry loses, JW as Julius wins, JL as Julius loses)
LW
LL -> JW
--------JL -> LW
--------------- LL -> JW
-----------------------JL -> LW
------------------------------LL -> [...]
This will continue until there is one person wins the game or we can say it continues to infinitive.
The chance for Larry to win lies at 1st, 3rd, 5th, 7th turns -> odd turns
=> The probability now is: (0.5)+ (0.5)^3 + (0.5)^5 + (0.5)^7 + ... + (0.5)^n with n is an odd number
The question now asks us to count the sum of a sequence.
Applying the Geometric Progression, we have b1=0.5, r = (0.5)^2, the sum is 0.5/[1 - (0.5)^2] = 0.5/0.75 = 2/3
