### Combination, Permutation and Probability in a game

Question Sample Titled 'Combination, Permutation and Probability in a game'

In each round of a game, stones are dropped randomly one by one into one of the six holes which are placed as a circular ring as shown in the figure. Suppose that each stone has an equal chance to fall into any one of the holes. Each hole can hold at most ${6}$ stones.

holes

A player in the game can choose any one of the following options.

 Option 1: Three stones are dropped one by one. If three stones fall into the same hole, the player can get ${120}$ dollars. If the three stones fall into three different hole, the player gets ${90}$ dollars. Otherwise, the player gets no dollars. Option 2: Four stones are dropped one by one. If four stones fall into the same holes, the player can get ${270}$ dollars. If three of them fall into the same hole, the player can get ${120}$ dollars. If four stones fall into four adjacent holes, the player gets ${120}$ dollars. Otherwise, the player gets no dollars.

 (a) Find the expected dollars got if a player selects option 1. (3 marks) (b) Which option should a player select in order to maximize the expected dollars got? Explain your answer. (4 marks) (c) Michelle selects the option with higher expected dollars got and she plays this game for three rounds. She thinks that her chance of getting no dollars is less than ${0.07}$. Do you agree? Explain your answer. (2 marks)

 (a) ${P}{\left(\text{same hole}\right)}$ $={1}\times\dfrac{{1}}{{6}}\times\dfrac{{1}}{{6}}=\dfrac{{1}}{{36}}$ 1A ${P}{\left(\text{3 in different holes}\right)}$ $={1}\times\dfrac{{5}}{{6}}\times\dfrac{{4}}{{6}}=\dfrac{{5}}{{9}}$ 1A Expected dollars $=\dfrac{{1}}{{36}}\times{120}+\dfrac{{5}}{{9}}\times{90}$  $=\dfrac{{160}}{{3}}$ 1A  (b) For option 2,  ${P}{\left(\text{same hole}\right)}$ $={1}\times\dfrac{{1}}{{6}}\times\dfrac{{1}}{{6}}\times\dfrac{{1}}{{6}}=\dfrac{{1}}{{216}}$ ${P}{\left(\text{3 in the same hole}\right)}$ $={{C}_{{3}}^{{4}}}{\left({1}\times\dfrac{{5}}{{6}}\times\dfrac{{1}}{{6}}\times\dfrac{{1}}{{6}}\right)}$ 1A  $=\dfrac{{5}}{{54}}$ ${P}{\left(\text{4 adjacent holes}\right)}$ $={6}{{P}_{{4}}^{{4}}}{\left(\dfrac{{1}}{{6}}\right)}^{{4}}$ 1A  $=\dfrac{{1}}{{9}}$ Expected dollars $=\dfrac{{1}}{{216}}\times{270}+\dfrac{{5}}{{54}}\times{120}+\dfrac{{1}}{{9}}\times{120}$ 1M  $=\dfrac{{925}}{{36}}$  $<\dfrac{{160}}{{3}}$ The expected dollars of option 1 is higher. Thus, option 1 should be selected. 1A (c) For option 1,  ${P}{\left(\text{no dollars}\right)}$ $={1}-\dfrac{{1}}{{36}}-\dfrac{{5}}{{9}}$  $=\dfrac{{5}}{{12}}$ ${P}{\left(\text{no dollars in 3 rounds}\right)}$ $={\left(\dfrac{{5}}{{12}}\right)}^{{3}}$ 1M  $={0.07233796296296294}$  $>{0.07}$ Thus, Michelle is disagreed. 1A

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