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CIE 9709 2022 June Paper 53 Q4

A Level / CIE / S1

CIE 9709 2022 June Paper 53 Paper · Question 4

题目

Problem

Ramesh throws an ordinary fair 6-sided die.

(a) Find the probability that he obtains a 4 for the first time on his 8th throw.

(1)

(b) Find the probability that it takes no more than 5 throws for Ramesh to obtain a 4.

(2)

Ramesh now repeatedly throws two ordinary fair 6-sided dice at the same time. Each time he adds the two numbers that he obtains.

(c) For 10 randomly chosen throws of the two dice, find the probability that Ramesh obtains a total of less than 4 on at least three throws.

(4)

题目中文翻译

Ramesh 投掷一枚普通的均匀六面骰子。

(a) 求他在第 8 次投掷时才首次得到 4 的概率。

(1)

(b) 求 Ramesh 得到 4 所需的投掷次数不超过 5 的概率。

(2)

现在 Ramesh 重复同时投掷两枚普通的均匀六面骰子，每次将所得的两个数字相加。

(c) 对于随机选取的 10 次双骰投掷，求 Ramesh 至少有三次得到的数字之和小于 4 的概率。

(4)

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

展开

前七次都不是4，第八次才是4。

答题过程

展开 P(first 4 on throw 8)= (56)716= 0.046513608…≈0.0465.\begin{align*} \mathrm{P}(\text{first 4 on throw 8}) =&\,\biggl(\frac56\biggr)^7\frac16\\[4mm] =&\,0.046513608\ldots\approx\boxed{0.0465}. \end{align*}

(b)

解法一

思路

展开

不超过5次出现首个4，取前五次完全没出现4的补集。

答题过程

展开 P(first 4 by throw 5)= 1−(56)5= 46517776≈0.598 (3 s.f.).\begin{align*} \mathrm{P}(\text{first 4 by throw 5}) =&\,1-\biggl(\frac56\biggr)^5\\[4mm] =&\,\frac{4651}{7776}\approx\boxed{0.598}\text{ (3 s.f.)}. \end{align*}

解法二

思路

展开

也可加总首次出现于第1至5次的五个互斥事件。

答题过程

展开

If TT is the throw on which the first 4 appears, then

P(T≤5)= ∑r=15(56)r−116= 161−(5/6)51−5/6= 46517776≈0.598 (3 s.f.).\begin{align*} \mathrm{P}(T\leq5)=&\,\sum_{r=1}^{5}\biggl(\frac56\biggr)^{r-1}\frac16\\[4mm] =&\,\frac16\frac{1-(5/6)^5}{1-5/6}\\[4mm] =&\,\frac{4651}{7776}\approx\boxed{0.598}\text{ (3 s.f.)}. \end{align*}

(c)

解法一

思路

展开

小于4的和只有2、3，对应三个有序结果；十次成功数取二项补集。

答题过程

展开

The pairs giving a total below 4 are (1,1)(1,1), (1,2)(1,2) and (2,1)(2,1), so the success probability is 3/36=1/123/36=1/12.

For ten independent throws, let Y∼B(10,1/12)Y\sim\mathrm{B}(10,1/12). Then

P(Y≥3)= 1−[(1112)10+10112(1112)9 +(102)(112)2(1112)8]= 0.044483694…≈0.0445.\begin{align*} \mathrm{P}(Y\geq3)=&\,1-\biggl[\biggl(\frac{11}{12}\biggr)^{10} +10\frac1{12}\biggl(\frac{11}{12}\biggr)^9\\[4mm] &\,\hspace{2pt}+\binom{10}{2}\biggl(\frac1{12}\biggr)^2 \biggl(\frac{11}{12}\biggr)^8\biggr]\\[4mm] =&\,0.044483694\ldots\approx\boxed{0.0445}. \end{align*}