Expected points: games of dropping balls

Question Sample Titled 'Expected points: games of dropping balls'

In each round of a game, balls are dropped randomly one by one into one of the holes which are arranged in ring pattern as shown in the figure. Suppose that each ball has an equal chance to fall into any one of the hole. Each hole can hold at most ${4}$ balls.

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

 Option 1: Three balls are dropped one by one. If three balls fall into the same hole, the player can get ${50}$ points. If the three balls fall into three different holes, the player gets ${10}$ points. Otherwise, the player gets no points. Option 2: Four balls are dropped one by one. If four balls fall into the same hole, the player can get ${120}$ points. If ${3}$ of them fall into the same hole, the player can get ${50}$ points. If four balls fall into four adjacent holes, the player gets ${40}$ points. Otherwise, the player gets no points.

 (a) Find the expected points got if Lucas selects option 1. (3 marks) (b) Which option should a player select in order to maximize the expected points got? Explain your answer. (4 marks) (c) Lucas selects the option with higher expected points got and he plays this game for two rounds. He thinks that his chance of getting no points is less than ${0.2}$ . Do you agree? Explain your answer. (2 marks)

 (a) ${P}{(}$same hole${)}$ $={1}\times\dfrac{{1}}{{7}}\times\dfrac{{1}}{{7}}=\dfrac{{1}}{{49}}$ 1A ${P}{\left({3}\right.}$ different holes${)}$ $={1}\times\dfrac{{6}}{{7}}\times\dfrac{{5}}{{7}}=\dfrac{{30}}{{49}}$ 1A Expected points $=\dfrac{{1}}{{49}}\times{50}+\dfrac{{30}}{{49}}\times{10}$  $=\dfrac{{50}}{{7}}$ 1A  (b) For option 2, ${P}{(}$same hole${)}$ $={1}\times\dfrac{{1}}{{7}}\times\dfrac{{1}}{{7}}\times\dfrac{{1}}{{7}}$  $=\dfrac{{1}}{{343}}$ ${P}{(}$3 in the same hole${)}$ $={1}\times\dfrac{{6}}{{7}}\times\dfrac{{1}}{{7}}\times\dfrac{{1}}{{7}}\times{{C}_{{1}}^{{4}}}$  $=\dfrac{{24}}{{343}}$ 1A ${P}{(}$4 adjacent holes${)}$ $={1}\times\dfrac{{1}}{{7}}\times\dfrac{{1}}{{7}}\times\dfrac{{1}}{{7}}\times{4}!\times{7}$  $=\dfrac{{24}}{{343}}$ 1A Expected points $=\dfrac{{1}}{{343}}\times{120}+\dfrac{{24}}{{343}}\times{50}+\dfrac{{24}}{{343}}\times{40}$ 1M  $=\dfrac{{2280}}{{343}}<\dfrac{{50}}{{7}}$ $\therefore$Option 1 should be selected. 1A  (c) Select option 1. ${P}{(}$no points${)}$ $={\left({1}-\dfrac{{1}}{{49}}-\dfrac{{30}}{{49}}\right)}^{{2}}$ 1M  $=\dfrac{{324}}{{2401}}<{0.2}$ 1A $\therefore$He is correct.

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