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IAL 2022 Oct S2 Q1

A Level / Edexcel / S2

IAL 2022 Oct Paper · Question 1

题目

Problem

Bhavna produces rolls of cloth. She knows that faults occur randomly in her cloth at a mean rate of 1.5 every 15 metres.

(a) Find the probability that in 15 metres of her cloth there are

(i) less than 3 faults,

(ii) at least 6 faults.

(3)

Each roll contains 100 metres of Bhavna’s cloth. She selects 15 rolls at random. (b) Find the probability that exactly 10 of these rolls each have fewer than 13 faults.

(4)

Bhavna decides to sell her cloth in pieces. Each piece of her cloth is 4 metres long. The cost to make each piece is £5.00 She sells each piece of her cloth that contains no faults for £7.40 She sells each piece of her cloth that contains faults for £2.00 (c) Find the expected profit that Bhavna will make on each piece of her cloth that she sells.

(4)

(Total for Question 1 is 11 marks)

题目中文翻译

Bhavna 生产布卷。她知道布上的瑕疵随机出现，平均速率为每 15 米 1.5 个。

(a) 求在 15 米布中有以下情况的概率：

(i) 少于 3 个瑕疵，

(ii) 至少 6 个瑕疵。

每卷包含 100 米 Bhavna 的布。 她随机选取 15 卷。 (b) 求恰好有 10 卷各自瑕疵少于 13 个的概率。

Bhavna 决定把布切成小段出售。 每段布长 4 米。 每段制作成本为 £5.00。 若一段布没有瑕疵，售价为 £7.40。 若一段布有瑕疵，售价为 £2.00。 (c) 求 Bhavna 每出售一段布所能获得的期望利润。

（第 1 题共 11 分）

解答

(a)(i)

解法一

思路

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瑕疵随机、独立地以固定平均速率出现，因此 15 米布中的瑕疵数服从参数为 1.51.5 的 Poisson 分布。「少于 3 个」即瑕疵数为 0、1 或 2，可直接使用累积概率。

答题过程

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Let FF be the number of faults in 15 metres of cloth. Then

F∼Po⁡(1.5).F\sim\operatorname{Po}(1.5).

Therefore,

P(F<3)=P(F⩽2)= 0.808846…= 0.809.\begin{align*} P(F<3)=P(F\leqslant2) =&\,0.808846\ldots \\[4mm] =&\,0.809. \end{align*}

(a)(ii)

解法一

思路

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「至少 6 个」是右尾事件。使用补事件，把它写成 1 减去「至多 5 个」的累积概率。

答题过程

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Using F∼Po⁡(1.5)F\sim\operatorname{Po}(1.5),

P(F⩾6)= 1−P(F⩽5)= 1−0.995544…= 0.00445598…= 0.00446.\begin{align*} P(F\geqslant6)=&\,1-P(F\leqslant5) \\[4mm] =&\,1-0.995544\ldots \\[4mm] =&\,0.00445598\ldots \\[4mm] =&\,0.00446. \end{align*}

(b)

解法一

思路

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先按长度比例把每卷 100 米布的平均瑕疵数换算为 10，从而求出一卷少于 13 个瑕疵的概率。再把每一卷视为一次独立试验，在 15 卷中计算恰好 10 卷符合条件的二项概率。

答题过程

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Let RR be the number of faults in a 100-metre roll. Its mean is

1.5×10015=10,1.5\times\frac{100}{15}=10,

so

R∼Po⁡(10).R\sim\operatorname{Po}(10).

The probability that one roll has fewer than 13 faults is

p=P(R⩽12)=0.791556…p=P(R\leqslant12)=0.791556\ldots

Let XX be the number, out of 15 rolls, that each have fewer than 13 faults. Then

X∼B⁡(15,0.791556…).X\sim\operatorname{B}(15,0.791556\ldots).

Therefore,

P(X=10)= (1510)(0.791556…)10 ×(1−0.791556…)5= 0.114108…= 0.114.\begin{align*} P(X=10)=&\, \binom{15}{10}(0.791556\ldots)^{10} \\[4mm] &\,\hspace{2pt}\times(1-0.791556\ldots)^5 \\[4mm] =&\,0.114108\ldots \\[4mm] =&\,0.114. \end{align*}

(c)

解法一

思路

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4 米布中的平均瑕疵数为 0.40.4。无瑕疵时每段利润为 7.40−5.00=£2.407.40-5.00=£2.40；有瑕疵时每段利润为 2.00−5.00=−£3.002.00-5.00=-£3.00。先用 Poisson 分布求两种情况的概率，再计算利润的期望。

答题过程

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Let HH be the number of faults in a 4-metre piece. Its mean is

1.5×415=0.4,1.5\times\frac4{15}=0.4,

so

H∼Po⁡(0.4).H\sim\operatorname{Po}(0.4).

Hence,

P(H=0)=e−0.4=0.670320…P(H=0)=e^{-0.4}=0.670320\ldots

and

P(H>0)=1−e−0.4=0.329680…P(H>0)=1-e^{-0.4}=0.329680\ldots

The profit on a fault-free piece is £2.40, while the loss on a faulty piece is £3.00. Therefore,

E(profit)= 2.40e−0.4−3.00(1−e−0.4)= 0.619728…\begin{align*} E(\text{profit})=&\, 2.40e^{-0.4}-3.00(1-e^{-0.4}) \\[4mm] =&\,0.619728\ldots \end{align*}

Thus the expected profit is £0.620\boxed{\pounds0.620} per piece.

解法二

思路

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也可以先计算每段布的期望销售收入，再减去固定制作成本 £5.00。收入在无瑕疵时为 £7.40、有瑕疵时为 £2.00，所需概率与解法一相同。

答题过程

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Using P(H=0)=e−0.4P(H=0)=e^{-0.4}, the expected revenue is

7.40e−0.4+2.00(1−e−0.4).7.40e^{-0.4}+2.00(1-e^{-0.4}).

Therefore,

E(profit)= 7.40e−0.4+2.00(1−e−0.4)−5.00= 0.619728…\begin{align*} E(\text{profit})=&\, 7.40e^{-0.4}+2.00(1-e^{-0.4})-5.00 \\[4mm] =&\,0.619728\ldots \end{align*}

Hence the expected profit is £0.620\boxed{\pounds0.620} per piece.