Answer to Question #289861 in Statistics and Probability for Liann

Question

Julia and Tony play the hand game rock-paper-scissors. Assume that, at each play they make their choices with equal probability 1/3 for each three moves, independently of any previous play.

a. On any single olay what is yhw chance of a tie?

b. What is the chance that Jula wins at the first play?

Let Random variable X-rock

The probability distribution is shown below

X - 0 1 2

Frequency (f) - 4 4 1

P(X)- 4/9 4/9 1/9

c. Find P(x=1)

d. Find P(x > 1)

Answer 1

Solution:

(1)

(a) P(a tie) = P(both players choose the same choice)

=P({Rock,rock}, {scissors, scissors}, {paper, paper})

=3/9 = 1/3 (since total no. of cases is 9)

(b) P(Julia wins the first play)

=6/9 (since there are 6 possibilities where tie do not occur)

=2/3

(2)

(c) $P(x=1)=4/9$

(d) $P(x>1)=P(x=2)=1/9$