题目

Problem

In a certain area in the Arctic the probability that it snows on any given day is $0.7$ , independent of all other days.

(a) Find the probability that in a week (7 days) it snows on at least five days.

(3)

A week in which it snows on at least five days out of seven is called a ‘white’ week.

(b) Find the probability that in three randomly chosen weeks at least one is a white week.

(2)

In a different area in the Arctic, the probability that a week is a white week is $0.8$ .

(5)

题目中文翻译

在北极的某个地区，任意一天下雪的概率为 $0.7$ ，且与其他各天相互独立。

(a) 求一周（7 天）中至少有五天下雪的概率。

(3)

如果一周七天中至少有五天下雪，则称这一周为“白色周”。

(b) 求随机选取的三周中至少有一周是白色周的概率。

(2)

在北极的另一个地区，一周是白色周的概率为 $0.8$ 。

(5)

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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一周下雪天数服从二项分布，至少5天即5、6、7天三项。

答题过程

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Let $D\sim\mathrm{B}(7,0.7)$ . Then

\begin{align*} \mathrm{P}(D\geq5) =&\,\binom75(0.7)^5(0.3)^2\\[4mm] &\,\hspace{2pt}+7(0.7)^6(0.3)+(0.7)^7\\[4mm] =&\,0.6470695\approx\boxed{0.647}. \end{align*}

(b)

解法一

思路

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承接(a)的准确概率，取三周全都非白色周的补集。

答题过程

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From (a), let $p=0.6470695$ be the probability of a white week. The required probability is

1-(1-p)^3=0.956038998\ldots\approx\boxed{0.956}.

解法二

思路

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也可以用三周中恰1、2、3周为白色周的二项和。

答题过程

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Using $p=0.6470695$ from (a),

\begin{align*} \mathrm{P}=&\,3p(1-p)^2+3p^2(1-p)+p^3\\[4mm] =&\,0.956038998\ldots\approx\boxed{0.956}. \end{align*}

(c)

解法一

思路

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另一地区必须改用0.8。少于47即至多46，正态近似修正边界为46.5。

答题过程

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Let $W\sim\mathrm{B}(60,0.8)$ . Since $np=48$ and $n(1-p)=12$ are both greater than 5, use $Y\sim\mathrm{N}(48,9.6)$ .

With a continuity correction,

\begin{align*} \mathrm{P}(W<47) =&\,\mathrm{P}(W\leq46)\\[4mm] \approx&\,\mathrm{P}(Y<46.5)\\[4mm] =&\,\Phi\biggl(\frac{46.5-48}{\sqrt{9.6}}\biggr)\\[4mm] =&\,0.314149\ldots\\[4mm] \approx&\,\boxed{0.314}. \end{align*}

CIE 9709 2024 June Paper 51 Q5

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

(c)

解法一

思路

答题过程