There are 3*3=9 combinations (cases) possible. In 3 cases there will be a tie (rock-rock, 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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please explain me where did you get this from the question stimulus? How do you decide that there is tie situation? Please read the Stimulus again!

The Official Answer is wrong!
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We are told that Jim and Renee have an equal chance of choosing any one of the hand signs, so if they both choose Paper for example there will be a tie. How else?

So, the official answer is not wrong.
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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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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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Probability of winning = (3C1*1C1) / (3C1*3C1) = 1/3
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OFFICIAL EXPLANATION:


No matter what sign Jim throws, there is one sign Renee could throw that would beat it, one that would tie, and one that would lose. Renee is equally likely to throw any one of the three signs. Therefore, the probability that Jim will win is 1/3.

For example, Jim could throw a Rock sign. He will win only if Renee throws a Scissors sign. There is a one in three chance that Renee will do so.

In fact:
Probability that Renee will win = 1/3
Probability of a tie = 1/3
Probability that Jim will win = 1/3

The correct answer is E.




Please give kudos if this was helpful in any way!

https://gmatclub.com/forum/jim-and-rene ... fl=similar
absolutely the same question from manhattan gmat, Bunuel

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Topics merged. Thank you.
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Total no of combination : 3*3 = 9
Now no of ways jenne or any one can win is only 3 : If one get like this : Rock beats Scissors, Scissors beat Paper, and Paper beats Rock
Other cases either loose or tie.

total favourable cases/ total cases : 3/9 => 1/3

Answer E


Kudos !!

Solution:

We can make a table that tabulate the results from Jim’s point of view:



We see that Jim wins 3 out of 9 times; therefore, the probability he wins is 3/9 = 1/3.

Alternate Solution:

Since each player has three choices of hand signs, the total number of pairs of hand signs the two players can show is 3 x 3 = 9. Of these 9 possibilities, Jim wins in 3 of them (which are S-P, P-R and R-S where the first letter is Jim’s pick). Thus, the probability that Jim wins is 3/9 = 1/3.

Answer: E
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