题目

Problem

On any given day, Cooper either wears a blue jumper or he wears a green jumper or he does not wear a jumper. The probability that he wears a blue jumper is 0.6 and the probability that he wears a green jumper is 0.3. Whether Cooper wears a jumper of either colour, or does not wear a jumper, on any day is independent of his choice on any other day.

(a) Find the probability, that in a week (7 days), Cooper wears a blue jumper on at least 5 days.

(3)

(b) Use a suitable approximation to find the probability that, in any 150-day period, Cooper does not wear a jumper on fewer than 22 days.

(5)

题目中文翻译

在任意一天，Cooper 要么穿蓝色毛衣，要么穿绿色毛衣，要么不穿毛衣。他穿蓝色毛衣的概率为 0.6，穿绿色毛衣的概率为 0.3。Cooper 在任何一天选择穿哪种颜色的毛衣或不穿毛衣，都与其他任何一天的选择相互独立。

(a) 求在一周（7 天）内，Cooper 至少有 5 天穿蓝色毛衣的概率。

(3)

(b) 利用适当的近似方法，求在任意连续 150 天内，Cooper 不穿毛衣的天数少于 22 的概率。

(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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每天是否穿蓝毛衣是独立、成功概率固定的试验，7天内的蓝毛衣天数服从二项分布。至少5天即5、6、7天。

答题过程

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Let $X$ be the number of days on which Cooper wears a blue jumper. Then $X\sim\mathrm{B}(7,0.6)$ , so

\begin{align*} \mathrm{P}(X\geq5) =&\,\binom75(0.6)^5(0.4)^2\\[4mm] &\,\hspace{2pt}+7(0.6)^6(0.4)+(0.6)^7\\[4mm] =&\,0.419904\\[4mm] \approx&\,\boxed{0.420}. \end{align*}

(b)

解法一

思路

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不穿毛衣的概率为0.1，用150次二项分布的正态近似。少于22天为21天及以下，连续性修正边界是21.5。

答题过程

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The probability of not wearing a jumper on any day is

p=1-0.6-0.3=0.1.

Let $X\sim\mathrm{B}(150,0.1)$ be the number of days with no jumper. Since $np=15$ and $n(1-p)=135$ are both greater than 5, use

Y\sim\mathrm{N}(15,13.5),

where $13.5=150(0.1)(0.9)$ is the variance. With a continuity correction,

\begin{align*} \mathrm{P}(X<22)=&\,\mathrm{P}(X\leq21)\\[4mm] \approx&\,\mathrm{P}(Y<21.5)\\[4mm] =&\,\Phi\biggl(\frac{21.5-15}{\sqrt{13.5}}\biggr)\\[4mm] =&\,\Phi(1.769075\ldots)\\[4mm] =&\,0.961559\ldots\\[4mm] \approx&\,\boxed{0.962}. \end{align*}

CIE 9709 2025 November 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