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S1 2025 Jan Q3

A Level / Edexcel / S1

IAL 2025 Jan Paper · Question 3

题目

Problem

A factory makes rugs and blankets from wool or cotton or polyester.

  • 30%30\% of the items made are rugs and 70%70\% of the items made are blankets
  • 60%60\% of the rugs are made from wool
  • 35%35\% of the blankets are made from wool
  • 10%10\% of the rugs are made from cotton
  • 20%20\% of the blankets are made from cotton
  • the rest of the rugs and blankets are made from polyester

An item made by the factory is selected at random.

(a) Complete the tree diagram below to illustrate this information.

(2)

(b) (i) Find the probability that the item selected is not made from wool.

(2)

(ii) Given that this item is not made from wool, find the probability that it is a blanket.

(2)
题目中文翻译

一家工厂使用羊毛、棉或聚酯纤维制作地毯和毯子。

  • 产品中 30%30\% 是地毯，70%70\% 是毯子。
  • 地毯中 60%60\% 由羊毛制成，10%10\% 由棉制成。
  • 毯子中 35%35\% 由羊毛制成，20%20\% 由棉制成。
  • 其余地毯和毯子由聚酯纤维制成。

从工厂生产的产品中随机选取一件。

(a) 完成上方树状图，表示这些信息。（2 分）

(b)(i) 求所选产品不是由羊毛制成的概率。（2 分）

(ii) 已知所选产品不是由羊毛制成，求它是毯子的概率。（2 分）

解答

(a)

解法一

思路

展开

第一层分 rug 和 blanket，概率分别是 0.30.3 和 0.70.7。第二层按材料分，polyester 是剩下的概率。

答题过程

展开

The first branches are

P(rug)=0.3,P(blanket)=0.7.\begin{align*} P(\text{rug})=0.3,\qquad P(\text{blanket})=0.7. \end{align*}

For rugs,

P(wool∣rug)= 0.6,P(cotton∣rug)= 0.1,P(polyester∣rug)= 1−0.6−0.1=0.3.\begin{align*} P(\text{wool}\mid\text{rug})=&\,0.6,\\ P(\text{cotton}\mid\text{rug})=&\,0.1,\\ P(\text{polyester}\mid\text{rug})=&\,1-0.6-0.1=0.3. \end{align*}

For blankets,

P(wool∣blanket)= 0.35,P(cotton∣blanket)= 0.20,P(polyester∣blanket)= 1−0.35−0.20=0.45.\begin{align*} P(\text{wool}\mid\text{blanket})=&\,0.35,\\ P(\text{cotton}\mid\text{blanket})=&\,0.20,\\ P(\text{polyester}\mid\text{blanket})=&\,1-0.35-0.20=0.45. \end{align*}

(b)(i)

解法一

思路

展开

“Not wool” 可以来自 rug 且不是 wool，也可以来自 blanket 且不是 wool。沿 tree diagram 两条大分支分别乘，再相加。

答题过程

展开 P(W′)= P(rug and not wool) +P(blanket and not wool)= 0.3(0.1+0.3)+0.7(0.2+0.45)= 0.3(0.4)+0.7(0.65)= 0.12+0.455= 0.575.\begin{align*} P(W') =&\,P(\text{rug and not wool})\\ &\,\hspace{2pt}+P(\text{blanket and not wool})\\ =&\,0.3(0.1+0.3)+0.7(0.2+0.45)\\ =&\,0.3(0.4)+0.7(0.65)\\ =&\,0.12+0.455\\ =&\,0.575. \end{align*}

解法二

思路

展开

先计算羊毛制品的概率 P(W)P(W)，再用 P(W′)=1−P(W)P(W')=1-P(W)。

答题过程

展开

First calculate the probability that the selected item is made from wool:

P(W)= P(rug and wool) +P(blanket and wool)= 0.3(0.6)+0.7(0.35)= 0.18+0.245= 0.425\begin{align*} P(W) =&\, P(\text{rug and wool})\\ &\,\hspace{2pt}+P(\text{blanket and wool})\\ =&\, 0.3(0.6) + 0.7(0.35)\\ =&\, 0.18 + 0.245\\ =&\, 0.425 \end{align*}

Then calculate the complementary probability:

P(W′)= 1−P(W)= 1−0.425= 0.575.\begin{align*} P(W') =&\, 1 - P(W)\\ =&\, 1 - 0.425\\ =&\, 0.575. \end{align*}

(b)(ii)

解法一

思路

展开

这是条件概率：

P(blanket∣W′)=P(blanket and not wool)P(W′).P(\text{blanket}\mid W') =\frac{P(\text{blanket and not wool})}{P(W')}.

答题过程

展开 P(blanket∣W′)= P(blanket and not wool)P(W′)= 0.7(0.2+0.45)0.575= 0.4550.575= 91115.\begin{align*} P(\text{blanket}\mid W') =&\,\frac{P(\text{blanket and not wool})}{P(W')}\\ =&\,\frac{0.7(0.2+0.45)}{0.575}\\ =&\,\frac{0.455}{0.575}\\ =&\,\frac{91}{115}. \end{align*}