Here's the question:
Suppose that a wheel, partitioned into three equal quadrants (labeled 1, 2, and 3), is spun 2 times.
a) List all the outcomes in the Sample Space.
Code:1,1 1,2 1,3 2,1 2,2 2,3 3,1 3,2 3,3
b) What is the probability that each spin lands on either the number "1" or the number "2"?
With this question, would I simply be counting the outcomes with only "1's" and "2's" and divided the total number of outcomes? For example, I counted "1,1" "1,2" "2,1" "2,2" = 4 and divide by 9 to get .44
Is this correct?
c) What is the probability that both spins land on the same number?
Each number has a .33 probability of being landed on, so I assume I multiply the probability twice to get .1089
b) I don't understand how the answer is 2/3 since there are 4 ways to get a 1 or a 2 on both spins. If I am going to find the probability that both spins land on either 1 or 2, wouldn't I divide that amount with the total number of possible outcomes?
edit: I think I understand. There is a .66 (2/3) probability that a spin lands on either 1 or a 2 and I then multiply the probability to get .44
c) p(both land on 1)=(1/3)(1/3)= .1089
On the first spin you can get a 1 by spinning (1, 1), (1, 2), (1, 3). Similar logic for getting a 2. By my counting, 6 of the 9 possibilities satisfy the requirement.
Of course, you could always just use symmetry: On a single spin 2 of the 3 sections of the wheel are favourable. Therefore ....