here's my take at it:


to find P(jim wins), lets figure out all of his winning situations:

(1) Jim chooses rock; renee chooses scissors (note their decisions are independent)

(2) Jim chooses paper, renee chooses rock

(3) Jim chooses scissors, renee chooses paper

So P(Jim wins) = P(1) + P(2) + P(3).


the probability for (1) is P(jim chooses rock)*P(renee chooses scissors) <== (independent events allows you to multiply)
= 1/3*1/3=1/9

(2) and (3) yield the same result.


1/9*3 = 1/3 = answer

best of luck!

pkw209 wrote:Explanation please!

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

Answer = [spoiler]1/3[/spoiler]

let rock, paper, scissors be R,P, and S.


now, each Jim and Renee can have one of these three hand signs. So, total number of cases = 3 * 3 = 9

(RP,PR,RS,SR,SP,PS,SS,RR,PP)

we have three cases where Jim wins. therefore, probab= 3/9 = 1/3.

Jim has 3 options rock , paper or scissor

If he chooses either one , for him to win Reene will be left with only one option for jim to win

Hence 1/3 * 1/3

Now jim has 3 choices

Hence 1/3 *1/3 * 3

Answer = 1/3
E