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CIE 9709 2022 June Paper 53 Q6

A Level / CIE / S1

CIE 9709 2022 June Paper 53 Paper · Question 6

题目

Problem

Sajid is practising for a long jump competition. He counts any jump that is longer than 6 m6\,\mathrm{m} as a success. On any day, the probability that he has a success with his first jump is 0.2. For any subsequent jump, the probability of a success is 0.3 if the previous jump was a success and 0.1 otherwise. Sajid makes three jumps.

(a) Draw a tree diagram to illustrate this information, showing all the probabilities.

(2)

(b) Find the probability that Sajid has exactly one success given that he has at least one success.

(5)

On another day, Sajid makes six jumps.

(c) Find the probability that only his first three jumps are successes or only his last three jumps are successes.

(3)

题目中文翻译

Sajid 正在为跳远比赛练习。他把任何超过 6 m6\,\mathrm{m} 的跳跃视为成功。无论哪一天，他第一次跳跃成功的概率均为 0.2。对于之后的每次跳跃，如果前一次跳跃成功，则本次成功的概率为 0.3；否则为 0.1。Sajid 跳跃三次。

(a) 画出树状图来表示这些信息，并标出所有概率。

(2)

(b) 已知 Sajid 至少成功一次，求他恰好成功一次的概率。

(5)

另一天，Sajid 跳跃六次。

(c) 求只有前三次跳跃成功或只有后三次跳跃成功的概率。

(3)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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第一次成功率0.2；之后各层只看上一跳成功还是失败，不能当独立同分布。

答题过程

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Use SS for success and FF for failure. On the first jump, the branches have probabilities P(S)=0.2\mathrm{P}(S)=0.2 and P(F)=0.8\mathrm{P}(F)=0.8.

For each of the second and third jumps:

  • From any SS branch, use 0.30.3 for SS and 0.70.7 for FF.
  • From any FF branch, use 0.10.1 for SS and 0.90.9 for FF.

Draw all three stages and label every outcome and branch probability.

(b)

解法一

思路

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恰一次成功有三条路径；至少一次成功用FFF的补集，最后作条件概率。

答题过程

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For exactly one success,

PathProbability
SFFSFF0.2(0.7)(0.9)=0.1260.2(0.7)(0.9)=0.126
FSFFSF0.8(0.1)(0.7)=0.0560.8(0.1)(0.7)=0.056
FFSFFS0.8(0.9)(0.1)=0.0720.8(0.9)(0.1)=0.072

Thus the probability of exactly one success is 0.126+0.056+0.072=0.2540.126+0.056+0.072=0.254.

The probability of at least one success is

1−P(FFF)=1−0.8(0.9)(0.9)=0.352.1-\mathrm{P}(FFF)=1-0.8(0.9)(0.9)=0.352.

Therefore

 P(exactly one∣at least one)= 0.2540.352=127176.\begin{align*} &\,\mathrm{P}(\text{exactly one}\mid\text{at least one})\\[4mm] =&\,\frac{0.254}{0.352}=\boxed{\frac{127}{176}}. \end{align*}

(c)

解法一

思路

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只有前三或后三成功，分别就是SSSFFF、FFFSSS，不包含其他路径。

答题过程

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The two required paths are disjoint:

P(SSSFFF)=0.2(0.3)2(0.7)(0.9)2=0.010206,\mathrm{P}(SSSFFF)=0.2(0.3)^2(0.7)(0.9)^2=0.010206, P(FFFSSS)=0.8(0.9)2(0.1)(0.3)2=0.005832.\mathrm{P}(FFFSSS)=0.8(0.9)^2(0.1)(0.3)^2=0.005832.

Hence

P=0.010206+0.005832=0.016038≈0.0160.\mathrm{P}=0.010206+0.005832=0.016038\approx\boxed{0.0160}.