题目

Problem

In a school canteen, students can choose from a main course of meat ( $M$ ), fish ( $F$ ) or vegetarian ( $V$ ). They can then choose a drink of either water ( $W$ ) or juice ( $J$ ). The partially completed tree diagram, where $p$ and $q$ are probabilities, shows the probabilities of these choices for a randomly selected student.

(a) Complete the tree diagram, giving your answers in terms of $p$ and $q$ where appropriate.

(2)

(b) Find an expression, in terms of $p$ and $q$ , 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.

(2)

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

(d) Find the value of $p$ and the value of $q$ .

(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)

解答

(a)

解法一

思路

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第一层主菜概率相加为 1，所以 vegetarian 的概率是 $0.25$ 。每个主菜后的 water 和 juice 概率也相加为 1。

答题过程

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The probability of vegetarian is

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

The missing drink probabilities are

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

(b)

解法一

思路

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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

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\begin{aligned} 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{aligned}

(c)

解法一

思路

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

答题过程

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Since choosing vegetarian and choosing water are independent,

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

Therefore

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

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

Multiplying by 20 gives

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

(d)

解法一

思路

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

答题过程

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Since $P(W)=0.40$ ,

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

Given

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

we have

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

Thus

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

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

Hence

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

and

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

Using $8p+7q=6$ ,

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

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

and

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

(e)

解法一

思路

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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 $V$ 和 $W$ 独立， $P(V\mid W)=P(V)=0.25$ 。再算 fish 即可。

答题过程

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Since $V$ and $W$ are independent,

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

Also,

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

For fish,

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

Since

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

the manager’s claim is correct.

IAL 2020 Oct Q2

题目

解答

(a)

解法一

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答题过程

(b)

解法一

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(c)

解法一

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(d)

解法一

思路

答题过程

(e)

解法一

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