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IAL 2024 Jan Q3

A Level / Edexcel / S1

IAL 2024 Jan Paper · Question 3

题目

Problem

In a sixth form college each student in Year 1212 and Year 1313 is either left-handed (L)(L) or right-handed (R)(R).

The partially completed tree diagram, where pp is a probability, gives information about these students.

Tree diagram for year group and handedness

(a) Complete the tree diagram, in terms of pp where necessary.

(1)

The probability that a student is left-handed is 0.110.11.

(b) Find the value of pp.

(3)

(c) Find the probability that a student selected at random is in Year 1212 and left-handed.

(2)

Given that a student is right-handed,

(d) find the probability that the student is in Year 1212.

(2)

解答

(a)

解法一

思路

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每一处分支概率相加为 11。第一层 Year 12 是 pp，所以 Year 13 是 1−p1-p。Year 12 下的 left-handed 是 18\frac18，所以 right-handed 是 78\frac78。Year 13 下的 left-handed 是 110\frac1{10}，所以 right-handed 是 910\frac9{10}。

答题过程

展开

The missing probabilities are

1−p,78,910.\begin{align*} 1-p,\qquad \frac78,\qquad \frac9{10}. \end{align*}

(b)

解法一

思路

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左撇子的总概率来自两条路径：Year 12 且 left-handed，或 Year 13 且 left-handed。

答题过程

展开

Since P(L)=0.11P(L)=0.11,

18p+110(1−p)=0.11.\begin{align*} \frac18p+\frac1{10}(1-p)=0.11. \end{align*}

Multiplying by 4040,

5p+4(1−p)= 4.4p+4= 4.4p= 0.4.\begin{align*} 5p+4(1-p)=&\,4.4\\[3mm] p+4=&\,4.4\\[3mm] p=&\,0.4. \end{align*}

Therefore

p=25.\begin{align*} p=\frac25. \end{align*}

解法二

思路

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纯小数计算方程法。因为在统计学中分率经常与小数互换，我们可以直接将分数转换为小数进行运算：

18=0.125,110=0.1\begin{align*} \frac{1}{8} = 0.125, \quad \frac{1}{10} = 0.1 \end{align*}

我们列出方程：

0.125p+0.1(1−p)=0.11\begin{align*} 0.125p + 0.1(1-p) = 0.11 \end{align*}

然后直接化简并计算出 pp。这种方法非常适合习惯在草稿纸上进行直接小手算、不倾向把方程转化成整数系数的同学。

答题过程

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Convert the fractions to decimals:

18=0.125,110=0.1.\begin{align*} \frac{1}{8} = 0.125, \quad \frac{1}{10} = 0.1. \end{align*}

We set up the equation:

0.125p+0.1(1−p)=0.11.\begin{align*} 0.125p + 0.1(1-p) = 0.11. \end{align*}

Expand and solve:

0.125p+0.1−0.1p=  0.110.025p=  0.01p=  0.010.025p=  0.4.\begin{align*} 0.125p + 0.1 - 0.1p =&\,\, 0.11\\[3mm] 0.025p =&\,\, 0.01\\[3mm] p =&\,\, \frac{0.01}{0.025}\\[3mm] p =&\,\, 0.4. \end{align*}

Therefore, p=25p = \frac{2}{5} (or 0.40.4).

(c)

解法一

思路

展开

沿树图路径相乘：Year 12 的概率乘以 Year 12 条件下 left-handed 的概率。

答题过程

展开 P(Year 12 and L)= p⋅18= 25⋅18= 120.\begin{align*} P(\text{Year 12 and }L) =&\,p\cdot\frac18\\[3mm] =&\,\frac25\cdot\frac18\\[3mm] =&\,\frac1{20}. \end{align*}

(d)

解法一

思路

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这是条件概率：

P(Year 12∣R)=P(Year 12 and R)P(R).\begin{align*} P(\text{Year 12}\mid R)=\frac{P(\text{Year 12 and }R)}{P(R)}. \end{align*}

已知 P(L)=0.11P(L)=0.11，所以 P(R)=0.89P(R)=0.89。

答题过程

展开 P(R)=1−0.11=0.89=89100.\begin{align*} P(R)=1-0.11=0.89=\frac{89}{100}. \end{align*}

Also,

P(Year 12 and R)= 25⋅78= 720.\begin{align*} P(\text{Year 12 and }R) =&\,\frac25\cdot\frac78\\[3mm] =&\,\frac7{20}. \end{align*}

Therefore

P(Year 12∣R)= P(Year 12 and R)P(R)= 72089100= 3589.\begin{align*} P(\text{Year 12}\mid R) =&\,\frac{P(\text{Year 12 and }R)}{P(R)}\\[3mm] =&\,\frac{\frac7{20}}{\frac{89}{100}}\\[3mm] =&\,\frac{35}{89}. \end{align*}

解法二

思路

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直接分支概率求和法。不通过 1−P(L)1 - P(L) 来求 P(R)P(R)，而是根据树状图的分支直接计算右撇子（right-handed）的概率。 右撇子可以来自：

  1. 处于 Year 12 且是右撇子（P(Year 12 and R)P(\text{Year 12 and } R)）
  2. 处于 Year 13 且是右撇子（P(Year 13 and R)P(\text{Year 13 and } R)） 我们将这两部分概率分别算出并求和得到 P(R)P(R)，然后再套入条件概率公式中。这种方法比较直接，适合喜欢从树图直接求出各项分支概率的同学。

答题过程

展开

Calculate the probability of right-handed students directly from the tree branches:

P(R)=  P(Year 12 and R)+P(Year 13 and R)=  p⋅78+(1−p)⋅910\begin{align*} P(R) =&\,\, P(\text{Year 12 and } R) + P(\text{Year 13 and } R)\\[3mm] =&\,\, p \cdot \frac{7}{8} + (1-p) \cdot \frac{9}{10} \end{align*}

Substitute p=0.4p = 0.4 (or 25\frac{2}{5}):

P(R)=  25⋅78+35⋅910=  720+2750=  0.35+0.54=  0.89=  89100.\begin{align*} P(R) =&\,\, \frac{2}{5} \cdot \frac{7}{8} + \frac{3}{5} \cdot \frac{9}{10}\\[3mm] =&\,\, \frac{7}{20} + \frac{27}{50}\\[3mm] =&\,\, 0.35 + 0.54\\[3mm] =&\,\, 0.89\\[3mm] =&\,\, \frac{89}{100}. \end{align*}

Now calculate the conditional probability:

P(Year 12∣R)=  P(Year 12 and R)P(R)=  7/2089/100=  3589.\begin{align*} P(\text{Year 12} \mid R) =&\,\, \frac{P(\text{Year 12 and } R)}{P(R)}\\[3mm] =&\,\, \frac{7/20}{89/100}\\[3mm] =&\,\, \frac{35}{89}. \end{align*}