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S1 2020 Oct Q2

A Level / Edexcel / S1

IAL 2020 Oct Paper · Question 2

题目

Problem

In a school canteen, students can choose from a main course of meat (MM), fish (FF) or vegetarian (VV). They can then choose a drink of either water (WW) or juice (JJ). The partially completed tree diagram, where pp and qq are probabilities, shows the probabilities of these choices for a randomly selected student.

(a) Complete the tree diagram, giving your answers in terms of pp and qq where appropriate.

(2)

(b) Find an expression, in terms of pp and qq, for the probability that a randomly selected student chooses water to drink.

(1)

The events “choosing a vegetarian main course” and “choosing water to drink” are independent.

(c) Find a linear equation in terms of pp and qq.

(2)

A student who has chosen juice to drink is selected at random. The probability that they chose fish for their main course is 730\frac7{30}

(d) Find the value of pp and the value of qq.

(5)

The canteen manager claims that students who choose water to drink are most likely to choose a fish main course.

(e) State, showing your working clearly, whether or not the manager’s claim is correct.

(3)
题目中文翻译

在一所学校的食堂里，学生的主菜可以选择肉类（MM）、鱼类（FF）或素食（VV），随后饮料可以选择水（WW）或果汁（JJ）。

题图是一棵尚未完成的树状图，其中 pp 和 qq 为概率；该图表示随机选取的一名学生作出这些选择的概率。

(a) 补全树状图；适当之处用 pp 和 qq 表示答案。

(b) 用 pp 和 qq 表示随机选取的一名学生选择喝水的概率。

“选择素食主菜”与“选择喝水”这两个事件相互独立。

(c) 求一个关于 pp 和 qq 的线性方程。

现从选择喝果汁的学生中随机选取一人。已知此人主菜选择鱼类的概率为 730\frac7{30}。

(d) 求 pp 与 qq 的值。

食堂经理声称：选择喝水的学生最有可能选择鱼类主菜。

(e) 清楚写出计算过程，说明经理的说法是否正确。

解答

(a)

解法一

思路

展开

第一层主菜概率相加为 1，所以 vegetarian 的概率是 0.250.25。每个主菜后的 water 和 juice 概率也相加为 1。

答题过程

展开

The probability of vegetarian is

1−0.40−0.35=0.25.\begin{align*} 1-0.40-0.35=0.25. \end{align*}

The missing drink probabilities are

P(J∣M)=1−p,P(J∣F)=1−q,P(J∣V)=0.60.P(J\mid M)=1-p,\qquad P(J\mid F)=1-q,\qquad P(J\mid V)=0.60.

The completed tree diagram is

(b)

解法一

思路

展开

选择 water 可以来自三个主菜分支，把三条路径的概率相加。

答题过程

展开 P(W)= P(M∩W)+P(F∩W)+P(V∩W)= 0.40p+0.35q+0.25(0.40)= 0.40p+0.35q+0.10.\begin{align*} P(W) =&\,P(M\cap W)+P(F\cap W)+P(V\cap W)\\ =&\,0.40p+0.35q+0.25(0.40)\\ =&\,0.40p+0.35q+0.10. \end{align*}

(c)

解法一

思路

展开

VV 和 WW 独立，所以 P(W∣V)=P(W)P(W\mid V)=P(W)。树图中 P(W∣V)=0.40P(W\mid V)=0.40。

答题过程

展开

Since choosing vegetarian and choosing water are independent,

P(W)=P(W∣V).\begin{align*} P(W)=P(W\mid V). \end{align*}

Therefore

0.40p+0.35q+0.10=0.40.\begin{align*} 0.40p+0.35q+0.10=0.40. \end{align*}

So

0.40p+0.35q=0.30.\begin{align*} 0.40p+0.35q=0.30. \end{align*}

Multiplying by 20 gives

8p+7q=6.\begin{align*} 8p+7q=6. \end{align*}

(d)

解法一

思路

展开

先利用独立性得到 P(J)=0.60P(J)=0.60，再用给出的条件概率 P(F∣J)=730P(F\mid J)=\frac7{30} 求 qq，最后代回线性方程求 pp。

答题过程

展开

Since P(W)=0.40P(W)=0.40,

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

Given

P(F∣J)=730,\begin{align*} P(F\mid J)=\frac7{30}, \end{align*}

we have

P(F∩J)P(J)=730.\begin{align*} \frac{P(F\cap J)}{P(J)}=\frac7{30}. \end{align*}

Thus

0.35(1−q)0.60=730.\begin{align*} \frac{0.35(1-q)}{0.60}=\frac7{30}. \end{align*}

So

0.35(1−q)=0.14.\begin{align*} 0.35(1-q)=0.14. \end{align*}

Hence

1−q=0.40,\begin{align*} 1-q=0.40, \end{align*}

and

q=0.60.\begin{align*} q=0.60. \end{align*}

Using 8p+7q=68p+7q=6,

8p+7(0.60)=6.\begin{align*} 8p+7(0.60)=6. \end{align*}

So

8p=1.8,\begin{align*} 8p=1.8, \end{align*}

and

p=0.225.\begin{align*} p=0.225. \end{align*}

(e)

解法一

思路

展开

要比较的是在已经选择 water 的条件下，三种主菜哪个概率最大。由于 VV 和 WW 独立，P(V∣W)=P(V)=0.25P(V\mid W)=P(V)=0.25。再算 fish 即可。

答题过程

展开

Since VV and WW are independent,

P(V∣W)=P(V)=0.25.\begin{align*} P(V\mid W)=P(V)=0.25. \end{align*}

Also,

P(M∣W)=P(M∩W)P(W)=0.40(0.225)0.40=0.225.P(M\mid W)=\frac{P(M\cap W)}{P(W)} =\frac{0.40(0.225)}{0.40}=0.225.

For fish,

P(F∣W)=P(F∩W)P(W)=0.35(0.60)0.40=0.525.P(F\mid W)=\frac{P(F\cap W)}{P(W)} =\frac{0.35(0.60)}{0.40} =0.525.

Since

0.525>0.25>0.225,\begin{align*} 0.525>0.25>0.225, \end{align*}

the manager’s claim is correct.