CalcGospel 國際數學圖譜
返回

IAL 2025 Jan Q4

A Level / Edexcel / S1

IAL 2025 Jan Paper · Question 4

题目

Problem

A manufacturer of pet food gives samples of three new flavours of cat food, A, B and C, to each of 100100 customers who each own one cat. These customers are surveyed to find out whether or not their cat likes each flavour with the following results

4040 cats like A

4545 cats like B

5050 cats like C

1919 cats like both A and B

2020 cats like both A and C

2222 cats like both B and C

1212 cats like all three of these flavours.

(a) Using these results complete the Venn diagram below.

Venn diagram
(4)

For these cats,

(b) determine whether or not a cat liking flavour A is independent of it liking flavour C

(2)

(c) One of these 100100 cats is chosen at random.

(i) Find the probability that it likes only one of the three flavours.

(1)

(ii) Given that it likes flavour C, find the probability that it also likes flavour B

(2)

解答

(a)

解法一

思路

展开

先填中心 A∩B∩C=12A\cap B\cap C=12。题目中的 “both” 包含三个都喜欢，所以每个两两交集区域都要减去 1212。最后再用每个圆的总数求只属于单个圆的区域。

答题过程

展开

The centre region is

A∩B∩C=12.\begin{align*} A\cap B\cap C=12. \end{align*}

The pairwise-only regions are

A∩B only= 19−12=7,A∩C only= 20−12=8,B∩C only= 22−12=10.\begin{align*} A\cap B\text{ only}=&\,19-12=7,\\[2mm] A\cap C\text{ only}=&\,20-12=8,\\[2mm] B\cap C\text{ only}=&\,22-12=10. \end{align*}

Now find the single-flavour-only regions:

A only= 40−7−8−12=13,B only= 45−7−10−12=16,C only= 50−8−10−12=20.\begin{align*} A\text{ only} =&\,40-7-8-12=13,\\[2mm] B\text{ only} =&\,45-7-10-12=16,\\[2mm] C\text{ only} =&\,50-8-10-12=20. \end{align*}

The number outside the three circles is

100−(13+16+20+7+8+10+12)= 14.\begin{align*} 100&-(13+16+20+7+8+10+12)\\[3mm] =&\,14. \end{align*}

(b)

解法一

思路

展开

判断 A 和 C 是否独立，可以检查

P(A∩C)=P(A)P(C).\begin{align*} P(A\cap C)=P(A)P(C). \end{align*}

答题过程

展开 P(A)=40100,P(C)=50100.\begin{align*} P(A)=\frac{40}{100},\qquad P(C)=\frac{50}{100}. \end{align*}

Also,

P(A∩C)=20100.\begin{align*} P(A\cap C)=\frac{20}{100}. \end{align*}

Now

P(A)P(C)= 40100⋅50100= 20100= P(A∩C).\begin{align*} P(A)P(C) =&\,\frac{40}{100}\cdot\frac{50}{100}\\[3mm] =&\,\frac{20}{100}\\[3mm] =&\,P(A\cap C). \end{align*}

Therefore liking flavour A and liking flavour C are independent.

解法二

思路

展开

条件概率独立性判别法。根据独立性的定义，若两个事件独立，则其中一个事件发生的概率不随另一个事件的发生与否而改变。即验证：

P(A∣C)=P(A)(或 P(C∣A)=P(C))\begin{align*} P(A \mid C) = P(A) \quad (\text{或 } P(C \mid A) = P(C)) \end{align*}

我们分别算出条件概率与无条件概率并进行对比，如果不等则不独立，如果相等则独立。该方法特别适用于已知部分条件概率或能直接从文氏图中读出条件占比的情况。

答题过程

展开

First calculate the unconditional probability of a cat liking flavour A:

P(A)=40100=0.4.\begin{align*} P(A) = \frac{40}{100} = 0.4. \end{align*}

Next, calculate the conditional probability of liking flavour A given liking flavour C (from the Venn diagram, out of 5050 cats liking C, 2020 also like A):

P(A∣C)=  P(A∩C)P(C)=  20/10050/100=  2050=  0.4.\begin{align*} P(A \mid C) =&\,\, \frac{P(A \cap C)}{P(C)}\\[3mm] =&\,\, \frac{20/100}{50/100}\\[3mm] =&\,\, \frac{20}{50}\\[3mm] =&\,\, 0.4. \end{align*}

Since P(A∣C)=P(A)P(A \mid C) = P(A), liking flavour A is independent of liking flavour C.

(c)(i)

解法一

思路

展开

“Only one flavour” 指只在 A、只在 B、只在 C 三个区域。把这三个区域相加，再除以 100100。

答题过程

展开

The number of cats that like only one flavour is

13+16+20=49.\begin{align*} 13+16+20=49. \end{align*}

Therefore the probability is

49100.\begin{align*} \frac{49}{100}. \end{align*}

(c)(ii)

解法一

思路

展开

已知喜欢 C，所以分母是 C 圆内总数 5050。同时喜欢 B 和 C 的有 2222 只，包括三者都喜欢的那部分。

答题过程

展开 P(B∣C)= P(B∩C)P(C)= 22/10050/100= 2250= 1125.\begin{align*} P(B\mid C) =&\,\frac{P(B\cap C)}{P(C)}\\[3mm] =&\,\frac{22/100}{50/100}\\[3mm] =&\,\frac{22}{50}\\[3mm] =&\,\frac{11}{25}. \end{align*}