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)=$\displaystyle (\frac{1}{36}\times \frac{5}{36})+(\frac{2}{36}\times \frac{6}{36})+(\frac{1}{36}\times \frac{6}{36})=0.0177$

P(neither wins, i.e. score difference less than 4)=$\displaystyle 1-0.0177=0.982$

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 $\displaystyle ^{15}C_4=1365$ 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, $\displaystyle ^{10}C_1\times^5C_3+^{10}C_2\times^5C_2+^{10}C_3\t imes^5C_1=1150$

So, P(not all on same line)=$\displaystyle \frac{1150}{1365}=\frac{230}{273}$

answer is $\displaystyle \frac{89}{91}$

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)=$\displaystyle (\frac{2}{3})^3=\frac{8}{27}$

P(A wins in 4 sets)=$\displaystyle ^4C_1\times (\frac{2}{3})^3\times\frac{1}{3}=\frac{32}{81}$

P(A wins in 5 matches)=$\displaystyle ^5C_2\times (\frac{2}{3})^3\times (\frac{1}{3})^2=\frac{80}{243}$

P(A wins)= the sum of the above probabilities, but the sum is greater than 1.

Answer is $\displaystyle \frac{64}{81}$

Thanks for the help.