题目
Problem
Sally plays a game in which she can either win or lose.
A turn consists of up to 3 games. On each turn Sally plays the game up to 3 times. If she wins the first 2 games or loses the first 2 games, then she will not play the 3rd game.
- The probability that Sally wins the first game in a turn is 0.7
- If Sally wins a game the probability that she wins the next game is 0.6
- If Sally loses a game the probability that she wins the next game is 0.2
(a) Use this information to complete the tree diagram on page 3
Tree diagram
(3)
(b) Find the probability that Sally wins the first 2 games in a turn.
(2)
(c) Find the probability that Sally wins exactly 2 games in a turn.
(2)
Given that Sally wins 2 games in a turn,
(d) find the probability that she won the first 2 games.
(2)
Given that Sally won the first game in a turn,
(e) find the probability that she won 2 games.
(2)
解答
(a)
解法一
思路
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树图每一组分支的概率要相加为 1。如果上一局赢了，下一局赢的概率是 0.6，输的概率就是 0.4；如果上一局输了，下一局赢的概率是 0.2，输的概率就是 0.8。
答题过程
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The missing probabilities are:
P(lose first game)=1−0.7=0.3.
After a win:
P(win next)=0.6,P(lose next)=0.4.
After a loss:
P(win next)=0.2,P(lose next)=0.8.
The same conditional probabilities are used on the third-game branches when a third game is played.
(b)
解法一
思路
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前两局都赢，就是沿着树图的 WW 路线相乘。
答题过程
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P(wins first 2 games)==0.7(0.6)0.42.
(c)
解法一
思路
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赢恰好 2 局有三种可能：WW，WLW，LWW。注意 WW 后不会玩第三局，所以它本身就是一种“赢 2 局”的情况。
答题过程
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P(exactly 2 wins)====P(WW)+P(WLW)+P(LWW)0.7(0.6)+0.7(0.4)(0.2)+0.3(0.2)(0.6)0.42+0.056+0.0360.512.
(d)
解法一
思路
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这是条件概率。分母是“赢 2 局”的总概率，分子是“前两局都赢”的概率。
答题过程
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P(WW∣exactly 2 wins)===P(exactly 2 wins)P(WW)0.5120.420.8203125.
Therefore the required probability is
0.820
to 3 significant figures.
(e)
解法一
思路
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已知第一局赢了。接下来要赢 2 局，可以是第二局直接赢，也可以第二局输、第三局再赢。
答题过程
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Given that Sally won the first game,
P(wins 2 games∣won first game)===P(W on second)+P(L on second and W on third)0.6+0.4(0.2)0.68.
