[问答题]
Jack and John have decided to play a g
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4dht ry);9ghg1q rmw9y will be rolling a die seven times. O
dhhm9 qrygr)1 g;4t9w ne roll of a die is considered as one round of the game. On each round, John agrees to pay Jack 4 dollars if 1 or 2 is rolled, Jack agrees to pay John 2 dollars if 3,4,5 or 6 is rolled, and who is paid wins the round. In the end, who earns money wins the game.
1. Show that the probability that Jack wins exactly two rounds is $\frac{224}{729}$ .
2. 1. Explain why the total number of outcomes for the results of the seven rounds is 128 .
2. Expand $(1+y)^{7}$ and choose a suitable value of y to prove that
$128=\binom{7}{0}+\binom{7}{1}+\binom{7}{2}+\binom{7}{3}+\binom{7}{4}+\binom{7}{5}+\binom{7}{6}+\binom{7}{7}$ .
3. Give a meaning of the equality above in the context of the seven rounds.
3. 1. Find the expected amount of money earned by each player in the game.
2. Who is expected to win the game?
3. Is this game fair? Justify your answer.
4. Jack and John have decided to play the game again.
1. Find an expression for the probability that John wins five rounds on the first game and two rounds on the second game. Give your answer in the form
$\binom{7}{r}^{2}\left[\frac{1}{3}\right]^{s}\left[\frac{2}{3}\right]^{t}$
where the values of r, s and t are to be found.
2. Use your answer to (d) (i) and seven similar expressions to write down the probability that John wins a total of seven rounds over two games as the sum of eight probabilities.
3. Hence prove that
$\binom{14}{7}=\sum_{k=0}^{7}\binom{7}{k}^{2}$
5. Now Jack and John roll a die 12 times. Let A denote the number of rounds Jack wins. The expected value of A can be written as
$\mathrm{E}[A]=\sum_{r=0}^{12} r\binom{12}{r}\left[\frac{a^{12-r}}{b^{12}}\right]$
1. Find the value of a and b .
2. Differentiate the expansion of $(1+y)^{12}$ to prove that the expected number of rolls Jack wins is 4 .
