Probability Q
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- prachi18oct
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A quick way to solve this question is to first recognize that they will tie 1/3 of the time. For example, if Jim picks Rock, there's a 1/3 chance Renee will also picks Rock. Similarly, if Jim picks Paper, there's a 1/3 chance Renee will also picks Paper etc.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
So, if there's a 1/3 chance they tie, there must be a 2/3 chance they don't tie (i.e., someone wins).
At this point, we should recognize that, if they do not tie, each person has an equal chance of winning (unless one of them is a mind-reader ).
In other words, we can expect Jim to win 1/2 of the games that are not ties.
So, Jim should win 1/3 of all games (since 1/2 of 2/3 is 1/3)
The answer is E
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Brent
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Another option is to use the basic probability formula along with some counting.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
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
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Brent
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Here's another way to look at it.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?
A) 5/6
B) 2/3
C) 1/2
D) 5/12
E) 1/3
OA to follow.
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
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Brent
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There are actually 9 possible outcomes.prachi18oct wrote: Favourable outcomes when Jim wins will be RS, RP, SP, PR ( the first pick is for JIM)
or
Now total outcomes will be RS, RP,SP,PR,SR,PR,PS,RP = 8
so probability that Jim wins should be 1/2 ?
let's list them in the form: Jim's outcome - Renee's outcome:
R-R
R-P
R-S
P-R
P-P
P-S
S-R
S-P
S-S
Since 3 of the 9 outcomes are favorable to Jim winning, the probability is 3/9 = [spoiler]1/3[/spoiler]
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Brent
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You have some duplicates in your list.prachi18oct wrote: Now total outcomes will be RS, RP,SP,PR,SR,PR,PS,RP = 8
You have also missed all of the ties (in which neither person wins): RR, SS, PP
Also, we have...
Also note that RP is NOT a winning outcome for Jim.Favourable outcomes when Jim wins will be RS, RP, SP, PR ( the first pick is for JIM)
Cheers,
Brent
- prachi18oct
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Thankyou Brent for these great multiple approaches and pointing out my error.
Actually I assumed that there are only single hand sign for each type and hence did not count any duplicates.
Thanks for correcting me !
Actually I assumed that there are only single hand sign for each type and hence did not count any duplicates.
Thanks for correcting me !
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There are actually 9 possible outcomes. let's list them in the form: Jim's - Renee's :
R-R
R-P
R-S
P-R
P-P
P-S
S-R
S-P
S-S
Winning Outcomes - R-S , S-P , P-R
Since 3 of the 9 outcomes are favorable to Jim winning, the probability is 3/9 = 1/3
R-R
R-P
R-S
P-R
P-P
P-S
S-R
S-P
S-S
Winning Outcomes - R-S , S-P , P-R
Since 3 of the 9 outcomes are favorable to Jim winning, the probability is 3/9 = 1/3