rishijhawar 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?
5/6
2/3
1/2
5/12
1/3
OA to follow.
Another option is to use the basic probability formula along with some counting.
So P(Jim wins) =
[# of outcomes where Jim wins] /
[total number of possible outcomes]
Let's begin with the denominator (it's always best to begin with the denominator - you'll see why shortly)
If Jim can choose a hand sign in 3 ways and Renee can choose a hand sign in 3 ways,
the total number of possible outcomes equals
9 (3 x 3 = 9)
At this point, we know that P(Jim wins) =
[# of outcomes where Jim wins] /
9
This means we can eliminate answer choices A, C, and D since it is impossible for
??/
9 to simplify to equal 5/6, 1/2 or 5/12
We're now left with 2/3 and 1/3
From here, we could apply some logic. Since each person has an equal probability of winning, the answer cannot be 2/3 since 2/3 + 2/3 is greater than 1. So, the answer must be
E
Alternatively, we could list all of the outcomes where Jim wins and then count them.
Aside: In many cases, listing and counting outcomes is a LOT of work. However, since the total number of outcomes is 9, I know that the number of outcomes where Jim wins will be less than 9, so it might be quite easy to just list them.
Here are the outcomes where Jim wins:
1. J=Rock, R=Scissors
2. J=Scissors, R=Paper
3. J=Paper, R=Rock
So, the
[# of outcomes where Jim wins] =
3, which means P(Jim wins) =
3 /
9
= 1/3
=
E
Cheers,
Brent
