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IAL 2024 May Q6

A Level / Edexcel / S1

IAL 2024 May Paper · Question 6

题目

Problem

The Venn diagram shows the probabilities related to teenagers playing 33 particular board games.

Venn diagram for teenagers playing Chess, Scrabble and Go

CC is the event that a teenager plays Chess.

SS is the event that a teenager plays Scrabble.

GG is the event that a teenager plays Go.

where pp and qq are probabilities.

(a) Find the probability that a randomly selected teenager plays Chess but does not play Go.

(1)

Given that the events CC and SS are independent,

(b) find the value of pp.

(4)

(c) Hence find the value of qq.

(2)

(d) Find

(i) P((C∪S)∩G′)P((C\cup S)\cap G')

(ii) P(C∣(S∩G))P(C\mid(S\cap G))

(3)

A youth club consists of a large number of teenagers. In this youth club 7676 teenagers play Chess and Go.

(e) Use the information in the Venn diagram to estimate how many of the teenagers in the youth club do not play Scrabble.

(3)

解答

(a)

解法一

思路

展开

“plays Chess but does not play Go” 是 C∩G′C\cap G'。从图上对应 Chess 里但不在 Go 里的区域：0.040.04 和 0.120.12。

答题过程

展开 P(C∩G′)=0.04+0.12=0.16.\begin{align*} P(C\cap G')=0.04+0.12=0.16. \end{align*}

(b)

解法一

思路

展开

独立事件满足

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

从图上读出 P(C∩S)=0.15+0.12=0.27P(C\cap S)=0.15+0.12=0.27，P(S)=0.12+0.15+0.10+0.23=0.60P(S)=0.12+0.15+0.10+0.23=0.60，P(C)=0.04+0.12+0.15+p=0.31+pP(C)=0.04+0.12+0.15+p=0.31+p。

答题过程

展开

From the Venn diagram,

P(C∩S)=0.12+0.15=0.27.\begin{align*} P(C\cap S)=0.12+0.15=0.27. \end{align*}

Also,

P(S)=0.12+0.15+0.10+0.23=0.60.\begin{align*} P(S)=0.12+0.15+0.10+0.23=0.60. \end{align*}

And

P(C)=0.04+0.12+0.15+p=0.31+p.\begin{align*} P(C)=0.04+0.12+0.15+p=0.31+p. \end{align*}

Since CC and SS are independent,

P(C∩S)= P(C)P(S)0.27= (0.31+p)(0.60)0.45= 0.31+pp= 0.14.\begin{align*} P(C\cap S)=&\,P(C)P(S)\\[3mm] 0.27=&\,(0.31+p)(0.60)\\[3mm] 0.45=&\,0.31+p\\[3mm] p=&\,0.14. \end{align*}

解法二

思路

展开

条件概率独立性判定法。根据独立事件的性质，若事件 CC 和 SS 独立，则条件概率 P(C∣S)P(C \mid S) 必须等于无条件概率 P(C)P(C)：

P(C∣S)=P(C)\begin{align*} P(C \mid S) = P(C) \end{align*}

我们可以利用文氏图中的数据直接计算 P(C∣S)P(C \mid S)：

P(C∣S)=P(C∩S)P(S)=0.12+0.150.60=0.270.60=0.45\begin{align*} P(C \mid S) = \frac{P(C \cap S)}{P(S)} = \frac{0.12 + 0.15}{0.60} = \frac{0.27}{0.60} = 0.45 \end{align*}

然后由 P(C)=0.04+0.12+0.15+p=0.31+pP(C) = 0.04 + 0.12 + 0.15 + p = 0.31 + p 建立简单线性方程：

0.31+p=0.45⟹p=0.14\begin{align*} 0.31 + p = 0.45 \Longrightarrow p = 0.14 \end{align*}

这种方法不需要在乘法公式中展开括号或移动复杂的乘数，计算过程极为明晰。

答题过程

展开

Calculate the conditional probability P(C∣S)P(C \mid S):

P(C∣S)=  P(C∩S)P(S)=  0.12+0.150.12+0.15+0.10+0.23=  0.270.60=  0.45.\begin{align*} P(C \mid S) =&\,\, \frac{P(C \cap S)}{P(S)}\\[3mm] =&\,\, \frac{0.12 + 0.15}{0.12 + 0.15 + 0.10 + 0.23}\\[3mm] =&\,\, \frac{0.27}{0.60}\\[3mm] =&\,\, 0.45. \end{align*}

Since CC and SS are independent, we have:

P(C)=P(C∣S)=0.45.\begin{align*} P(C) = P(C \mid S) = 0.45. \end{align*}

From the Venn diagram, express P(C)P(C) in terms of pp:

P(C)=  0.04+0.12+0.15+p=  0.31+p.\begin{align*} P(C) =&\,\, 0.04 + 0.12 + 0.15 + p\\[3mm] =&\,\, 0.31 + p. \end{align*}

Equating the two expressions for P(C)P(C):

0.31+p=  0.45p=  0.14.\begin{align*} 0.31 + p =&\,\, 0.45\\[3mm] p =&\,\, 0.14. \end{align*}

(c)

解法一

思路

展开

Venn diagram 里所有区域概率加起来等于 11。把 p=0.14p=0.14 代入即可求 qq。

答题过程

展开

The probabilities in the Venn diagram sum to 11.

So

q= 1−(0.04+0.12+0.15+0.10+0.23+0.14)= 1−0.78= 0.22.\begin{align*} q =&\,1-(0.04+0.12+0.15+0.10+0.23+0.14)\\[3mm] =&\,1-0.78\\[3mm] =&\,0.22. \end{align*}

(d)(i)

解法一

思路

展开

(C∪S)∩G′(C\cup S)\cap G' 表示不在 Go 里面，但在 Chess 或 Scrabble 至少一个里面。对应区域是 0.04,0.12,0.230.04,0.12,0.23。

答题过程

展开 P((C∪S)∩G′)= 0.04+0.12+0.23= 0.39.\begin{align*} P((C\cup S)\cap G') =&\,0.04+0.12+0.23\\[3mm] =&\,0.39. \end{align*}

(d)(ii)

解法一

思路

展开

已知在 S∩GS\cap G 内，分母是 SS 和 GG 的交集总概率，即 0.15+0.100.15+0.10。其中也属于 CC 的部分是三者交集 0.150.15。

答题过程

展开 P(C∣S∩G)= P(C∩S∩G)P(S∩G)= 0.150.15+0.10= 0.6.\begin{align*} P(C\mid S\cap G) =&\,\frac{P(C\cap S\cap G)}{P(S\cap G)}\\[3mm] =&\,\frac{0.15}{0.15+0.10}\\[3mm] =&\,0.6. \end{align*}

(e)

解法一

思路

展开

图上 C∩GC\cap G 的概率是 p+0.15=0.29p+0.15=0.29。现实中这对应 7676 人，所以先估计总人数。然后“不玩 Scrabble”的概率是 P(S′)=0.4P(S')=0.4。

答题过程

展开

From the Venn diagram,

P(C∩G)=p+0.15=0.14+0.15=0.29.\begin{align*} P(C\cap G)=p+0.15=0.14+0.15=0.29. \end{align*}

This corresponds to 7676 teenagers, so the estimated total number of teenagers is

760.29.\begin{align*} \frac{76}{0.29}. \end{align*}

Also,

P(S′)=1−P(S)=1−0.60=0.40.\begin{align*} P(S')=1-P(S)=1-0.60=0.40. \end{align*}

Therefore the estimated number who do not play Scrabble is

760.29(0.40)= 104.827….\begin{align*} \frac{76}{0.29}(0.40) =&\,104.827\ldots. \end{align*}

So the estimate is

105.\begin{align*} 105. \end{align*}