Jim and Renee will play one game of Rock, Paper, Scissors.

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.
_________________

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!
_________________

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.
_________________

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.
_________________

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:
_________________

Probability of winning = (3C1*1C1) / (3C1*3C1) = 1/3
_________________

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

_______________________
Topics merged. Thank you.
_________________

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
_________________

Since we know that rocks>scissors, and that scissors > paper there is TWO scenarios in which Jim can win. One is that Jim picks rock and Renee picks anything other than rocks, two is that Jim picks scissors and Renee does NOT pick rock or scissors:

P(Scenario 1) = 1/3 probability that Jim picking rock TIMES 2/3 probability that Renee picks not rock = 2/9
P(Scenario 2) = 1/3 probability that Jim picking scissors TIMES 1/3 probability that Renee picks paper = 1/9

P(Scenario 1) OR P(Scenario 2) = 2/9 + 1/9 = 3/9 = 1/3

We have 3 situations in which Jim can win:

1. Rock beats Scissors
If Jim has to win, he has to pick the option 'Rock' (out of 3 options) and Renee has to pick 'Scissors' (of out 3 options)
\(\frac{1}{3}*\frac{1}{3}=\frac{1}{9}\)

2. Scissors beats Paper
Again, Jim has to pick 'Scissors' (out of 3 options) and Renee has to pick 'Paper' (out of 3 options)
\(\frac{1}{3}*\frac{1}{3}=\frac{1}{9}\)

3. Paper beats Rock
Jim has to pick 'Paper' (out of 3 options) and Renee has to pick 'Rock' (out of 3 options)
\(\frac{1}{3}*\frac{1}{3}=\frac{1}{9}\)

Those are all possible combinations that will lead to a win for Jim. All other combinations will let him lose or end in a tie.
To get the probability, we need to add all options, because we need only one of the possibility (not all of them).
\(\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}=\frac{1}{3}\)

Answer (E)

Regardless of what sign Jim throws, Renee could throw that one sign that would beat it, there is one sign that would tie, and one sign that would lose which means Renee is equally likely to throw any one of the three signs.

P(Jim will win) = 1/3.

Lets say that Jim throws a Rock sign. This means that he will win only if Renee throws a Scissors sign. Therefore, it is one in three chance that Renee will do so.

P(Renee will win) = 1/3
P(Tie) = 1/3

Probability that Jim will win = 1/3

Therefore correct answer would be (E).

Bunuel is this approach correct?

Favourable Outcomes: 3 (Outcomes that make Jim the winner)

Outcome 1: Rock - Scissors : 1/3 x 1/3 = 1/9 (they are independent event, each with a probability of 1/3), so multiple for both together
Outcome 2: Scissors - Paper: 1/3 x 1/3 = 1/9
Outcome 3: Paper - Rock: 1/3 x 1/3: 1/9

Total probability (sum of all the outcomes) = 1/9+1/9+1/9 = 3/9=1/3 (E)