Results 1 to 5 of 5

Math Help - Probability question

  1. #1
    Newbie
    Joined
    May 2010
    From
    Portland, OR
    Posts
    8

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by charles2 View Post
    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 ....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2010
    From
    Portland, OR
    Posts
    8
    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
    Last edited by charles2; June 6th 2010 at 04:26 PM. Reason: I understand B
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by charles2 View Post
    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 ....
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2010
    From
    Portland, OR
    Posts
    8
    Quote Originally Posted by mr fantastic View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: February 10th 2013, 02:11 PM
  2. Probability question..
    Posted in the Statistics Forum
    Replies: 0
    Last Post: January 21st 2011, 05:04 PM
  3. Probability question involving (A given B type question )
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: November 9th 2009, 09:08 AM
  4. Probability question
    Posted in the Statistics Forum
    Replies: 1
    Last Post: September 27th 2007, 10:35 PM
  5. A probability question and an Expectations question
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: January 29th 2006, 06:13 PM

Search Tags


/mathhelpforum @mathhelpforum