Results 1 to 4 of 4

Math Help - help with two basic probability problems.

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    37

    help with two basic probability problems.

    Problem 1:
    There are 12 balls in an urn, 4 of which are white and 8 black. Three blndfold players, A, B, C draw a ball in turn, first A, thn B, then C. The winner is the one who first draws a white ball. Assuming that each (black) ball isreplaced after being drawn, find the ratio of the chances of the three players.

    Problem 2:
    There are 40 cards, 10 of each suit. A wagers B that he will draw four cards ad get one of each suit. What are the fair amount of wagers of each?

    For problem 2, I know the answer (from the back of the book) is A:1000 vs. B:8139. which implies that the probability P = 1000/(1000+8139)=1000/9139. Also the number of combiantions of 4 cards out of 40 is
    40 C 4 = 91390. Then the question that remains is how do i find (out of these 91390 different combinations) that there are 10^4=10000 combiantions that contain each card of a different suit?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2009
    Posts
    277
    Thanks
    2

    Suggestions

    1. Until someone wins, isn't the probability on any turn 4/12 that that person will win (because the black balls are replaced)? If so, then the probability anyone will win is the probability that he'll win on that turn and that everyone has lost before.

    2. A brute force way to figure out that probability is to consider the probability that you'll have drawn 2 cards from the same suit, or 3 cards, or 4 cards and subtract from the # ways of drawing 4 cards in general.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2009
    Posts
    37
    Quote Originally Posted by qmech View Post
    1. Until someone wins, isn't the probability on any turn 4/12 that that person will win (because the black balls are replaced)? If so, then the probability anyone will win is the probability that he'll win on that turn and that everyone has lost before.

    2. A brute force way to figure out that probability is to consider the probability that you'll have drawn 2 cards from the same suit, or 3 cards, or 4 cards and subtract from the # ways of drawing 4 cards in general.
    on problem 2, the probability of drawing 4 cards of the same suit is (10/40)(9/39)(8/38)(7/37) = 21/9139. The probability of getting the complement (that is not getting 4 cards of the same suit) is (9139-21)/9139 or 1-(21/9139) which is equal to 9118/9139 which is not the correct answer. For instance you can draw 3 cards of the same suit a the fourth one of a different suit.
    If my interpretation of your solution is not correct? Then what do you exaclty mean by "the # ways of drawing 4 cards in general". You also say "substract" something from this number (the # ways of drawing 4 cards in general), but it is not clea what to substract
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by santiagos11 View Post
    Problem 1:
    There are 12 balls in an urn, 4 of which are white and 8 black. Three blndfold players, A, B, C draw a ball in turn, first A, thn B, then C. The winner is the one who first draws a white ball. Assuming that each (black) ball isreplaced after being drawn, find the ratio of the chances of the three players.

    Problem 2:
    There are 40 cards, 10 of each suit. A wagers B that he will draw four cards ad get one of each suit. What are the fair amount of wagers of each?

    For problem 2, I know the answer (from the back of the book) is A:1000 vs. B:8139. which implies that the probability P = 1000/(1000+8139)=1000/9139. Also the number of combiantions of 4 cards out of 40 is
    40 C 4 = 91390. Then the question that remains is how do i find (out of these 91390 different combinations) that there are 10^4=10000 combiantions that contain each card of a different suit?
    There are 10^4 ways to choose 4 cards each of a different suit because there are 10 ways to choose the club, 10 ways to choose the spade, 10 ways to choose the heart, and 10 ways to choose the diamond.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: March 1st 2011, 04:39 PM
  2. Some basic combinatorics problems
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 27th 2009, 10:19 AM
  3. basic trigonometry problems, help?
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: February 22nd 2009, 11:41 PM
  4. basic calc two problems! thanks for the help!
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 10th 2007, 09:50 PM
  5. basic calculus problems
    Posted in the Calculus Forum
    Replies: 7
    Last Post: January 18th 2007, 02:31 PM

Search Tags


/mathhelpforum @mathhelpforum