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Math Help - The last few questions

  1. #1
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    The last few questions

    1. In a game of chance each player throws two unbiased dice and scores the difference of the two numbers shown on the dice. two players compete in the game. one wins if he scores at least 4 or more than his opponent. find the probability that neither wins.
    The possible scores for a throw are {5,4,3,2,1,0}
    Combinations of scores to win are (5,1); (4,0); (5,0).
    P(score 5)=1/36
    P(score 4)=2/36
    P(score 1)=5/36
    P(score 0)=6/36
    P(score difference greater than 4)= (\frac{1}{36}\times \frac{5}{36})+(\frac{2}{36}\times \frac{6}{36})+(\frac{1}{36}\times \frac{6}{36})=0.0177
    P(neither wins, i.e. score difference less than 4)= 1-0.0177=0.982
    answer is supposed to be 0.914, I can't find anywhere i have gone wrong, so i need someone to help me see if i have.

    2. Six lines are drawn in a plane and produced to give all their points of intersection. If no three lines are concurrent and no two lines are parallel, show that there are 15 points of intersection. Find the probability that if four lines are chosen at random, they are not all on one of the given lines.
    I drew the lines up, with 15 points of intersection and found that there are 5 points of intersection on each line.
    There are ^{15}C_4=1365 ways to choose 4 out of 15.
    if they are not all on the same line, then they can be 3 on one and 1 on any of 5 other lines, 2 on one, or 1 on one.
    So I have, ^{10}C_1\times^5C_3+^{10}C_2\times^5C_2+^{10}C_3\t  imes^5C_1=1150
    So, P(not all on same line)= \frac{1150}{1365}=\frac{230}{273}
    answer is \frac{89}{91}

    3. A tennis player A has a probability of 2/3 of winning a set against a player B. A match is won by the player who first wins three sets. Find the probability that A wins the match.
    There can be 3, 4, 5, matches
    P(A wins in 3 sets)= (\frac{2}{3})^3=\frac{8}{27}
    P(A wins in 4 sets)= ^4C_1\times (\frac{2}{3})^3\times\frac{1}{3}=\frac{32}{81}
    P(A wins in 5 matches)= ^5C_2\times (\frac{2}{3})^3\times (\frac{1}{3})^2=\frac{80}{243}
    P(A wins)= the sum of the above probabilities, but the sum is greater than 1.
    Answer is \frac{64}{81}
    Thanks for the help.
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  2. #2
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    Quote Originally Posted by arze View Post
    1. In a game of chance each player throws two unbiased dice and scores the difference of the two numbers shown on the dice. two players compete in the game. one wins if he scores at least 4 or more than his opponent. find the probability that neither wins.
    The possible scores for a throw are {5,4,3,2,1,0}
    Combinations of scores to win are (5,1); (4,0); (5,0).
    P(score 5)=1/36
    P(score 4)=2/36
    P(score 1)=5/36
    P(score 0)=6/36
    P(score difference greater than 4)= (\frac{1}{36}\times \frac{5}{36})+(\frac{2}{36}\times \frac{6}{36})+(\frac{1}{36}\times \frac{6}{36})=0.0177
    P(neither wins, i.e. score difference less than 4)= 1-0.0177=0.982
    answer is supposed to be 0.914, I can't find anywhere i have gone wrong, so i need someone to help me see if i have.
    There are 2 ways to score a five (6,1) and (1,6) so P(5)=\frac{2}{36}

    There are 4 ways to score a four (6,2), (2,6), (5,1), (1,5) so P(4)=\frac{4}{36}

    There are 10 ways to score a one
    (2,1), (1,2), (3,2), (2,3), (4,3), (3,4), (5,4), (4,5), (6,5), (5,6) so P(4)=\frac{10}{36}

    Using these, and doubling since either can win, the answer comes out.
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  3. #3
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    Quote Originally Posted by arze View Post
    2. Six lines are drawn in a plane and produced to give all their points of intersection. If no three lines are concurrent and no two lines are parallel, show that there are 15 points of intersection. Find the probability that if four points of intersection are chosen at random, they are not all on one of the given lines.
    I drew the lines up, with 15 points of intersection and found that there are 5 points of intersection on each line.
    There are ^{15}C_4=1365 ways to choose 4 out of 15.
    if they are not all on the same line, then they can be 3 on one and 1 on any of 5 other lines, 2 on one, or 1 on one.
    So I have, ^{10}C_1\times^5C_3+^{10}C_2\times^5C_2+^{10}C_3\t  imes^5C_1=1150
    So, P(not all on same line)= \frac{1150}{1365}=\frac{230}{273}
    answer is \frac{89}{91}
    For 1 line... no points of intersection
    For 2 lines... one point of intersection
    For 3 lines... an additional 2, giving 1+2 = 3 points of intersection
    For 4 lines... an additional 3, giving 1+2+3 = 6 points of intersection
    For 5 lines... 1+2+3+4 = 10
    For 6 lines... 1+2+3+4+5 = 15 points of intersection.

    It is much simpler to consider that as there are 5 points of intersection per line in this situation, you only need select 4 from these 5 from any one line.
    Then multiply by the number of lines to find the total number of ways there can be 4 points of intersection on a line.

    \frac{6\binom{5}{4}}{\binom{15}{4}}=\frac{6(5)}{\f  rac{15(14)13(12)}{4(3)2}}=\frac{30}{15(7)13}

    =\frac{6}{273}=\frac{2}{91}

    Hence the probabilty that the 4 points are not on any line is 1-\frac{2}{91}=\frac{91}{91}-\frac{2}{91}=\frac{89}{91}
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  4. #4
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    Quote Originally Posted by arze View Post
    3. A tennis player A has a probability of 2/3 of winning a set against a player B. A match is won by the player who first wins three sets. Find the probability that A wins the match.
    There can be 3, 4, 5, matches
    P(A wins in 3 sets)= (\frac{2}{3})^3=\frac{8}{27}
    P(A wins in 4 sets)= ^4C_1\times (\frac{2}{3})^3\times\frac{1}{3}=\frac{32}{81}
    P(A wins in 5 matches)= ^5C_2\times (\frac{2}{3})^3\times (\frac{1}{3})^2=\frac{80}{243}
    P(A wins)= the sum of the above probabilities, but the sum is greater than 1.
    Answer is \frac{64}{81}
    Thanks for the help.
    For the 3rd one, arze

    you are overcounting.

    There are only \binom{3}{1}=3 ways that player A can win in 4 sets....
    if player B wins the 1st, 2nd or 3rd set.
    B can't win the 4th as the game would have been over.
    Similarly, for the 5-set match, Player B cannot win the 5th set,
    he can win 2 of the first 4 but A must win the 5th.

    Hence \frac{2^3}{3^3}+\binom{3}{1}\frac{2^3}{3^3}\frac{1  }{3}+\binom{4}{2}\frac{2^3}{3^3}\frac{1^2}{3^2}[MATh]=\frac{8}{27}+\frac{8}{27}+\frac{6}{9}\frac{8}{27} =(1+1+\frac{6}{9})\frac{8}{27}[/tex]

    =\frac{24}{9}\frac{8}{27}=\frac{64}{81}
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