题目

Problem

Three fair 6-sided dice, each with faces marked 1, 2, 3, 4, 5, 6, are thrown at the same time repeatedly. The score on each throw is the sum of the numbers on the uppermost faces.

(a) Find the probability that a score of 17 or more is first obtained on the 6th throw.

(3)

(b) Find the probability that a score of 17 or more is obtained in fewer than 8 throws.

(2)

题目中文翻译

反复同时掷三枚均匀的六面骰子，每枚骰子的各面均标有 1、2、3、4、5、6。每次的得分为三个朝上面的数字之和。

(a) 求在第 6 次掷骰时首次得到至少 17 分的概率。

(3)

(b) 求在少于 8 次掷骰中得到至少 17 分的概率。

(2)

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

三骰216个等概率结果，和17有三种(5,6,6)次序，和18只有(6,6,6)。首次成功第6次前五次须失败。

答题过程

展开

There are $6^3=216$ equally likely ordered outcomes. Scores of at least 17 occur on $(6,6,6)$ and the three orders of $(5,6,6)$ . Thus

p=\frac{1+3}{216}=\frac1{54}.

The required probability is

(1-p)^5p=\biggl(\frac{53}{54}\biggr)^5\frac1{54} =0.016866182\ldots\approx\boxed{0.0169}.

(b)

解法一

思路

展开

少于8次即前7次至少一次成功，取七次都失败的补集。

答题过程

展开

\mathrm{P}=1-\biggl(\frac{53}{54}\biggr)^7 =0.122646185\ldots\approx\boxed{0.123}.

解法二

思路

展开

也可以按首次成功在第1至第7次分类并求等比和。

答题过程

展开

Writing $q=\frac{53}{54}$ and $p=\frac1{54}$ ,

\begin{align*} \mathrm{P}=&\,p+qp+q^2p+\cdots+q^6p\\[4mm] =&\,p\frac{1-q^7}{1-q}\\[4mm] =&\,0.122646185\ldots\approx\boxed{0.123}. \end{align*}

CIE 9709 2022 November Paper 52 Q3

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

解法二

思路

答题过程