Thread: Probability question

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

Originally Posted by charles2

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

You should give exact answers:

b) 2/3.

c) Pr(both 1's) = (1/3)(1/3) = ...., Pr(both 2's) = (1/3)(1/3) = ...., Pr(both 3's) = (1/3)(1/3) = .... Therefore Pr(same number with two spins) = ....

You should carefully think about where these answers have come from ....

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

Originally Posted by charles2

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?

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 ....

Originally Posted by mr fantastic

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 ....

Ahh. I don't think I understood the question. I was thinking that it was asking what the likelihood that both spins land on 1 or 2, therefore I was excluding any outcome with 3.