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Jim and Renee will play one game of Rock, Paper, Scissors.
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26 Nov 2010, 07:13
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64% (01:27) correct 36% (01:29) wrong based on 696 sessions
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Jim and Renee will play one game of Rock, Paper, Scissors. In this game, each will select and show a hand sign for one of the three items. Rock beats Scissors, Scissors beat Paper, and Paper beats Rock. Assuming that both Jim and Renee have an equal chance of choosing any one of the hand signs, what is the probability that Jim will win? A. 5/6 B. 2/3 C. 1/2 D. 5/12 E. 1/3
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Re: Jim and Renee will play one game of Rock, Paper, Scissors.
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26 Nov 2010, 07:23
anilnandyala wrote: Jim and Renee will play one game of Rock, Paper, Scissors. In this game, each will select and show a hand sign for one of the three items. Rock beats Scissors, Scissors beat Paper, and Paper beats Rock. Assuming that both Jim and Renee have an equal chance of choosing any one of the hand signs, what is the probability that Jim will wi
5/6
2/3
1/2
5/12
1/3 There are 3*3=9 combinations (cases) possible. In 3 cases there will be a tie (rockrock, scissors/scissors, paper/paper). Now, out of 6 cases left Jim and Renee have equal chances of winning, so in 3 cases Jim will win and in other 3 Renee will win. So the probability Jim winning the game is (favorable outcomes) / (total # of outcomes) = 3/9 = 1/3. Or: no matter what sign Renee will select Jim can select 1 sign to win, 1 sign to get tie and 1 sign to loose, so the probability Jim winning the game is 1/3. Answer: E.
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Re: Jim and Renee will play one game of Rock, Paper, Scissors.
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17 Aug 2015, 09:41
Bunuel wrote: anilnandyala wrote: Jim and Renee will play one game of Rock, Paper, Scissors. In this game, each will select and show a hand sign for one of the three items. Rock beats Scissors, Scissors beat Paper, and Paper beats Rock. Assuming that both Jim and Renee have an equal chance of choosing any one of the hand signs, what is the probability that Jim will wi
5/6
2/3
1/2
5/12
1/3 There are 3*3=9 combinations (cases) possible. In 3 cases there will be a tie (rockrock, scissors/scissors, paper/paper). Now, out of 6 cases left Jim and Renee have equal chances of winning, so in 3 cases Jim will win and in other 3 Renee will win. So the probability Jim winning the game is (favorable outcomes) / (total # of outcomes) = 3/9 = 1/3. Or: no matter what sign Renee will select Jim can select 1 sign to win, 1 sign to get tie and 1 sign to loose, so the probability Jim winning the game is 1/3. Answer: E. I got a bit confused reg tie. But now it is clear. Thanks Bunuel for the solution. Can u help me refer few more problems such as this? What can be a general strategy to deal with this kind of problems?



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Re: Jim and Renee will play one game of Rock, Paper, Scissors.
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17 Aug 2015, 10:31
Mechmeera wrote: Bunuel wrote: anilnandyala wrote: Jim and Renee will play one game of Rock, Paper, Scissors. In this game, each will select and show a hand sign for one of the three items. Rock beats Scissors, Scissors beat Paper, and Paper beats Rock. Assuming that both Jim and Renee have an equal chance of choosing any one of the hand signs, what is the probability that Jim will wi
5/6
2/3
1/2
5/12
1/3 There are 3*3=9 combinations (cases) possible. In 3 cases there will be a tie (rockrock, scissors/scissors, paper/paper). Now, out of 6 cases left Jim and Renee have equal chances of winning, so in 3 cases Jim will win and in other 3 Renee will win. So the probability Jim winning the game is (favorable outcomes) / (total # of outcomes) = 3/9 = 1/3. Or: no matter what sign Renee will select Jim can select 1 sign to win, 1 sign to get tie and 1 sign to loose, so the probability Jim winning the game is 1/3. Answer: E. I got a bit confused reg tie. But now it is clear. Thanks Bunuel for the solution. Can u help me refer few more problems such as this? What can be a general strategy to deal with this kind of problems? Was not able to find similar questions to this one but you can practice more questions on probability and combinations from the links below: Also, check articles on these topics in our Important Topics Directory. Hope it helps.
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Re: Jim and Renee will play one game of Rock, Paper, Scissors.
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27 Jul 2016, 06:44
Hello! First question here.
If for example they play 3 times and the question ask for the prob. that Jim will win the three times
Would it be 1/3*1/3*1/3=1/27?
Thanks!



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Re: Jim and Renee will play one game of Rock, Paper, Scissors.
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27 Jul 2016, 06:58
juanweiser wrote: Hello! First question here.
If for example they play 3 times and the question ask for the prob. that Jim will win the three times
Would it be 1/3*1/3*1/3=1/27?
Thanks! ____________ Yes, that's correct.'
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Re: Jim and Renee will play one game of Rock, Paper, Scissors.
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29 Aug 2016, 12:43
anilnandyala wrote: Jim and Renee will play one game of Rock, Paper, Scissors. In this game, each will select and show a hand sign for one of the three items. Rock beats Scissors, Scissors beat Paper, and Paper beats Rock. Assuming that both Jim and Renee have an equal chance of choosing any one of the hand signs, what is the probability that Jim will win?
A. 5/6 B. 2/3 C. 1/2 D. 5/12 E. 1/3 Here's another way to look at it. Let's say that Renee already knows what she is going to select. So, that part is done. At this point, what are the possible outcomes? a) Jim selects the one sign that allows him to win b) Jim selects the one sign that allows Renee to win c) Jim selects the one sign that results in a tie. Since each of these outcomes is equally likely, the probability of each outcome is 1/3 So, the probability that Jim wins is 1/3 Answer:
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Re: Jim and Renee will play one game of Rock, Paper, Scissors.
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02 Sep 2018, 20:32
Probability of winning = (3C1*1C1) / (3C1*3C1) = 1/3
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Re: Jim and Renee will play one game of Rock, Paper, Scissors.
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02 Sep 2018, 20:32






