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IAL 2023 June S2 Q7

A Level / Edexcel / S2

IAL 2023 June Paper · Question 7

题目

Problem

A bakery sells muffins individually at an average rate of 8 muffins per hour.

(a) Find the probability that, in a randomly selected one-hour period, the bakery sells at least 4 but not more than 8 muffins.

(3)

A sample of 5 non-overlapping half-hour periods is selected at random.

(b) Find the probability that the bakery sells fewer than 3 muffins in exactly 2 of these periods.

(5)

Given that 4 muffins were sold in a one-hour period, (c) find the probability that more muffins were sold in the first 15 minutes than in the last 45 minutes.

(4)

(Total for Question 7 is 12 marks)

题目中文翻译

一家面包店平均每小时单独售出 8 个松饼。

(a) 求在随机选取的一小时内，面包店售出至少 4 个但不超过 8 个松饼的概率。

随机选取 5 个互不重叠的半小时区间。

(b) 求在这 5 个时间段中，恰好有 2 个时间段里面包店售出少于 3 个松饼的概率。

已知在一小时内售出了 4 个松饼， (c) 求前 15 分钟售出的松饼数多于最后 45 分钟售出的概率。

（第 7 题共 12 分）

解答

(a)

解法一

思路

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一小时内售出的松饼数可用均值为 8 的泊松分布建模。“至少 4 个但不超过 8 个”对应 4⩽R⩽84\leqslant R\leqslant8，使用两个累积概率之差即可由统计表求得。

答题过程

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Let RR be the number of muffins sold in one hour. Then

R∼Po⁡(8).R\sim\operatorname{Po}(8).

Using the Statistical Tables,

P(4⩽R⩽8)= P(R⩽8)−P(R⩽3)= 0.5925−0.0424= 0.5501.\begin{align*} P(4\leqslant R\leqslant8) =&\,P(R\leqslant8)-P(R\leqslant3) \\ =&\,0.5925-0.0424 \\ =&\,\boxed{0.5501}. \end{align*}

(b)

解法一

思路

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半小时内的平均销量为 4，所以先用 Po⁡(4)\operatorname{Po}(4) 求一个半小时内售出少于 3 个的概率。5 个互不重叠时段可视为独立，再用二项分布求恰好两个时段满足条件的概率。

答题过程

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Let HH be the number of muffins sold in one half-hour period. Then

H∼Po⁡(4).H\sim\operatorname{Po}(4).

Using the Statistical Tables,

p=P(H<3)=P(H⩽2)=0.2381.p=P(H<3)=P(H\leqslant2)=0.2381.

Let YY be the number of the five periods in which fewer than 3 muffins are sold. Since the periods do not overlap,

Y∼B⁡(5,0.2381).Y\sim\operatorname{B}(5,0.2381).

Therefore,

P(Y=2)= (52)(0.2381)2(1−0.2381)3= 0.25073…= 0.251(to 3 significant figures).\begin{align*} P(Y=2) =&\,\binom{5}{2}(0.2381)^2(1-0.2381)^3 \\ =&\,0.25073\ldots \\ =&\,\boxed{0.251}\quad\text{(to 3 significant figures)}. \end{align*}

(c)

解法一

思路

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把一小时拆成前 15 分钟和后 45 分钟，两段销量因泊松过程的独立增量而分别服从 Po⁡(2)\operatorname{Po}(2) 与 Po⁡(6)\operatorname{Po}(6)。已知总销量为 4，前段销量多于后段销量只可能是 (W,X)=(3,1)(W,X)=(3,1) 或 (4,0)(4,0)，再使用条件概率。

答题过程

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Let WW be the number sold in the first 15 minutes and XX the number sold in the last 45 minutes. Then

W∼Po⁡(2),X∼Po⁡(6),W\sim\operatorname{Po}(2), \qquad X\sim\operatorname{Po}(6),

and WW and XX are independent. Given that W+X=4W+X=4, the event W>XW>X occurs when (W,X)=(3,1)(W,X)=(3,1) or (4,0)(4,0). Therefore,

P(W=3)P(X=1)= e−2233!×e−661!= 8e−8,P(W=4)P(X=0)= e−2244!×e−6= 23e−8,P(R=4)= e−8844!= 5123e−8.\begin{align*} P(W=3)P(X=1) =&\,\frac{e^{-2}2^3}{3!}\times\frac{e^{-6}6}{1!} \\ =&\,8e^{-8}, \\ P(W=4)P(X=0) =&\,\frac{e^{-2}2^4}{4!}\times e^{-6} \\ =&\,\frac{2}{3}e^{-8}, \\ P(R=4) =&\,\frac{e^{-8}8^4}{4!} \\ =&\,\frac{512}{3}e^{-8}. \end{align*}

Hence,

P(W>X∣W+X=4)= 8e−8+23e−85123e−8= 13256.\begin{align*} P(W>X\mid W+X=4) =&\,\frac{8e^{-8}+\dfrac{2}{3}e^{-8}} {\dfrac{512}{3}e^{-8}} \\ =&\,\boxed{\frac{13}{256}}. \end{align*}

解法二

思路

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官方替代路线使用泊松过程的条件分布：已知一小时总共售出 4 个，每个松饼落在前 15 分钟的条件概率为 1560=14\dfrac{15}{60}=\dfrac14。因此前 15 分钟的销量服从二项分布，再求其大于 2 的概率。

答题过程

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Conditionally on 4 muffins being sold in the hour, let FF be the number sold in the first 15 minutes. Then

F∼B⁡(4,14).F\sim\operatorname{B}\left(4,\frac{1}{4}\right).

More muffins are sold in the first 15 minutes than in the last 45 minutes if F>4−FF>4-F, so F>2F>2. Hence,

P(F>2)= P(F=3)+P(F=4)= (43)(14)3(34) +(14)4= 12256+1256= 13256.\begin{align*} P(F>2) =&\,P(F=3)+P(F=4) \\ =&\,\binom{4}{3}\left(\frac14\right)^3\left(\frac34\right) \\ &\,\hspace{2pt}+\left(\frac14\right)^4 \\ =&\,\frac{12}{256}+\frac{1}{256} \\ =&\,\boxed{\frac{13}{256}}. \end{align*}