Answer to Question #297944 in Statistics and Probability for Crizelda Salazar

Question

julia and tony play and game rock paper scissor assume that at each play they make their choices with equal probability 1/3 for each of the three moves and dependently of any previous play

A.)on any single play,what is a chance of a tie

B.) What is the chance that Julia wins at the first play

Let Random variable x-rock

The probability distribution is shown below

C.)Find P(x=1)

D.)Find P(x≥1)

Answer 1

Assume missing data is:

X - 0 1 2

Frequency (f) - 4 4 1

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

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) $P(x\ge1)=P(x=1)+P(x=2)=4/9+1/9=5/9$