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IAL 2026 Jan S2 A Q7

A Level / Edexcel / S2

IAL 2026 Jan A Paper · Question 7

题目

Problem

Members of a conservation group record the number of sightings of a rare animal. The number of sightings follows a Poisson distribution with a rate of 1 every 2 months.

(a) Find the smallest value of nn such that the probability that there are at least nn sightings in 2 months is less than 0.05

(2)

(b) Find the smallest number of months, mm, such that the probability of no sightings in mm months is less than 0.05

(2)

(c) Find the probability that there is at least 1 sighting per month in each of 3 consecutive months.

(3)

(d) Find the probability that the number of sightings in an 8 month period is equal to the expected number of sightings for that period.

(2)

(e) Given that there were 4 sightings in a 4 month period, find the probability that there were more sightings in the last 2 months than in the first 2 months.

(3)

(Total for Question 7 is 12 marks)

题目中文翻译

一个保护组织的成员记录稀有动物的目击次数。目击次数服从泊松分布，速率为每 2 个月 1 次。

(a) 求最小的 nn 值，使得 2 个月内至少有 nn 次目击的概率小于 0.05。

(b) 求最小的月数 mm，使得 mm 个月内没有目击的概率小于 0.05。

(c) 求连续 3 个月中每个月至少有 1 次目击的概率。

(d) 求 8 个月期间目击次数等于该期间期望目击次数的概率。

(e) 已知 4 个月期间有 4 次目击，求最后 2 个月的目击次数多于前 2 个月的概率。

（第 7 题共 12 分）

解答

(a)

解法一

思路

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两个月的平均目击次数为 11，所以使用参数为 11 的泊松分布。要证明所得 nn 是最小值，需要同时检查相邻的两个尾概率：n=3n=3 尚未低于 0.050.05，而 n=4n=4 已低于 0.050.05。

答题过程

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Let XX be the number of sightings in 2 months. Then

X∼Po⁡(1).X\sim\operatorname{Po}(1).

Now,

P(X≥3)= 1−P(X≤2)= 0.0803,\begin{align*} P(X\geq3) =&\,1-P(X\leq2)\\ =&\,0.0803, \end{align*}

which is greater than 0.050.05. However,

P(X≥4)= 1−P(X≤3)= 0.0190,\begin{align*} P(X\geq4) =&\,1-P(X\leq3)\\ =&\,0.0190, \end{align*}

which is less than 0.050.05. Therefore, the smallest value is

n=4.\boxed{n=4}.

(b)

解法一

思路

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每个月的平均目击次数为 0.50.5，所以 mm 个月的泊松参数为 0.5m0.5m。没有目击的概率是 e−0.5me^{-0.5m}；解不等式后，再取满足条件的最小整数月数。

答题过程

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Let YY be the number of sightings in mm months. Then

Y∼Po⁡(0.5m).Y\sim\operatorname{Po}(0.5m).

Hence,

P(Y=0)=e−0.5m.P(Y=0)=e^{-0.5m}.

We require

e−0.5m<0.05.e^{-0.5m}<0.05.

Taking natural logarithms,

−0.5m< ln⁡0.05,m> −2ln⁡0.05= 5.991….\begin{align*} -0.5m <&\,\ln0.05,\\ m >&\,-2\ln0.05\\ =&\,5.991\ldots. \end{align*}

Therefore, the smallest whole number of months is

m=6.\boxed{m=6}.

(c)

解法一

思路

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单月目击次数服从参数为 0.50.5 的泊松分布。单月至少一次的概率为 1−P(0)1-P(0)；互不重叠月份的目击次数相互独立，因此连续三个月都满足条件的概率是单月概率的三次方。

答题过程

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Let WW be the number of sightings in one month. Then

W∼Po⁡(0.5).W\sim\operatorname{Po}(0.5).

For one month,

P(W≥1)=1−P(W=0)=1−e−0.5.P(W\geq1)=1-P(W=0)=1-e^{-0.5}.

The numbers of sightings in disjoint months are independent. Therefore, the required probability is

[P(W≥1)]3= (1−e−0.5)3= 0.060916…≈ 0.0609.\begin{align*} \big[P(W\geq1)\big]^3 =&\,\big(1-e^{-0.5}\big)^3\\ =&\,0.060916\ldots\\ \approx&\,\boxed{0.0609}. \end{align*}

(d)

解法一

思路

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八个月包含四个“两个月”时段，因此期望目击次数为 44，泊松参数也是 44。题目要求实际次数等于期望次数，所以计算随机变量恰好等于 44 的概率。

答题过程

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Let SS be the number of sightings in 8 months. Then

S∼Po⁡(4),E(S)=4.S\sim\operatorname{Po}(4), \qquad E(S)=4.

Therefore,

P(S=4)= e−4444!= 0.195366…≈ 0.195.\begin{align*} P(S=4) =&\,e^{-4}\frac{4^4}{4!}\\ =&\,0.195366\ldots\\ \approx&\,\boxed{0.195}. \end{align*}

(e)

解法一

思路

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前、后两个两个月时段的目击次数相互独立，并且都服从参数为 11 的泊松分布。在总次数为 44 的条件下，后两个月次数较多只可能对应 (0,4)(0,4) 或 (1,3)(1,3)；把这两个联合概率相加，再除以四个月共 44 次的概率。

答题过程

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Let UU and VV be the numbers of sightings in the first and last 2 months respectively. Then

U∼Po⁡(1),V∼Po⁡(1),U\sim\operatorname{Po}(1), \qquad V\sim\operatorname{Po}(1),

and UU and VV are independent. Also,

U+V∼Po⁡(2).U+V\sim\operatorname{Po}(2).

Given that U+V=4U+V=4, the event V>UV>U occurs when (U,V)=(0,4)(U,V)=(0,4) or (1,3)(1,3). Hence,

N= P(U=0,V=4)+ P(U=1,V=3)= e−24!+e−23!= 5e−224.\begin{align*} N =&\,P(U=0,V=4)\\ +&\,P(U=1,V=3)\\ =&\,\frac{e^{-2}}{4!}+\frac{e^{-2}}{3!}\\ =&\,\frac{5e^{-2}}{24}. \end{align*}

Also,

D= P(U+V=4)= e−2244!= 16e−224.\begin{align*} D =&\,P(U+V=4)\\ =&\,e^{-2}\frac{2^4}{4!}\\ =&\,\frac{16e^{-2}}{24}. \end{align*}

Therefore,

P(V>U∣U+V=4)= ND= 516.\begin{align*} P(V>U\mid U+V=4) =&\,\frac{N}{D}\\ =&\,\boxed{\frac{5}{16}}. \end{align*}

解法二

思路

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已知四个月总共有 44 次目击后，每次目击落在前两个月或后两个月的条件概率均为 12\frac12。因此前两个月的次数服从二项分布；后两个月次数更多等价于前两个月至多有 11 次。

答题过程

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Conditional on there being 4 sightings in total, let ZZ be the number occurring in the first 2 months. Since the two periods have equal lengths,

Z∼B⁡(4,12).Z\sim\operatorname{B}\bigg(4,\frac12\bigg).

The last 2 months contain more sightings when

4−Z>Z,4-Z>Z,

so Z≤1Z\leq1. Therefore,

P(Z≤1)= P(Z=0)+P(Z=1)= (12)4+4(12)4= 516.\begin{align*} P(Z\leq1) =&\,P(Z=0)+P(Z=1)\\ =&\,\bigg(\frac12\bigg)^4 +4\bigg(\frac12\bigg)^4\\ =&\,\boxed{\frac{5}{16}}. \end{align*}