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IAL 2025 Jan Q1

A Level / Edexcel / S1

IAL 2025 Jan Paper · Question 1

题目

Problem

Jen has one fair 44-sided red die and one fair 44-sided blue die.

  • The red die has sides numbered 11, 22, 33 and 44
  • The blue die has sides numbered 11, 33, 55 and 77

The discrete random variable RR represents the score from one roll of the red die. The discrete random variable BB represents the score from one roll of the blue die.

(a) Write down the name of the distribution of RR

(1)

(b) Find P(R<3)P(R<3)

(1)

(c) Write down the value of

(i) E(R)E(R)

(ii) E(B)E(B)

(2)

(d) Showing your working, find Var(B)\operatorname{Var}(B)

(3)

Jen rolls each die once.

(e) Find P(R+B5)P(R+B\leqslant5)

(2)

(f) Find P(R2<B)P(R^2<B)

(3)

The random variable DD is defined as the magnitude of the difference between the score on the red die and the score on the blue die. The table below shows the cumulative distribution function of DD

dd00112233445566
F(d)F(d)18\dfrac1838\dfrac38916\dfrac9{16}34\dfrac34pp1516\dfrac{15}{16}11

(g) Showing your working, find the value of pp

(3)

解答

(a)

解法一

思路

展开

红色骰子的四个面出现机会相同,所以 RR 是离散均匀分布。

答题过程

展开

RR has a discrete uniform distribution.

(b)

解法一

思路

展开

R<3R<3 表示红色骰子掷出 1122。四个面等可能,所以概率是 24\dfrac24

答题过程

展开 P(R<3)=P(R=1 or R=2)=24=12.\begin{align*} P(R<3) =&\,P(R=1\text{ or }R=2)\\[3mm] =&\,\frac24\\[3mm] =&\,\frac12. \end{align*}

(c)

解法一

思路

展开

公平离散均匀分布的期望就是所有可能取值的平均数。

答题过程

展开

For the red die,

E(R)=1+2+3+44=2.5.\begin{align*} E(R) =&\,\frac{1+2+3+4}{4}\\[3mm] =&\,2.5. \end{align*}

For the blue die,

E(B)=1+3+5+74=4.\begin{align*} E(B) =&\,\frac{1+3+5+7}{4}\\[3mm] =&\,4. \end{align*}

(d)

解法一

思路

展开

先算 E(B2)E(B^2),再用

Var(B)=E(B2)[E(B)]2.\begin{align*} \operatorname{Var}(B)=E(B^2)-[E(B)]^2. \end{align*}

答题过程

展开 E(B2)=12+32+52+724=1+9+25+494=21.\begin{align*} E(B^2) =&\,\frac{1^2+3^2+5^2+7^2}{4}\\[3mm] =&\,\frac{1+9+25+49}{4}\\[3mm] =&\,21. \end{align*}

Using E(B)=4E(B)=4,

Var(B)=E(B2)[E(B)]2=2142=5.\begin{align*} \operatorname{Var}(B) =&\,E(B^2)-[E(B)]^2\\[3mm] =&\,21-4^2\\[3mm] =&\,5. \end{align*}

(e)

解法一

思路

展开

两个骰子各有 44 种结果,一共有 1616 个等可能组合。列出满足 R+B5R+B\leqslant5 的组合即可。

答题过程

展开

The possible ordered pairs (R,B)(R,B) satisfying R+B5R+B\leqslant5 are

(1,1), (1,3), (2,1), (2,3), (3,1), (4,1).\begin{align*} (1,1),\ (1,3),\ (2,1),\ (2,3),\ (3,1),\ (4,1). \end{align*}

There are 66 favourable outcomes out of 1616 equally likely outcomes, so

P(R+B5)=616=38.\begin{align*} P(R+B\leqslant5)=\frac6{16}=\frac38. \end{align*}

(f)

解法一

思路

展开

仍然列等可能组合。注意 R2R^2 的值分别是 1,4,9,161,4,9,16,所以 R=3,4R=3,4 时已经不可能小于 BB

答题过程

展开

For R=1R=1, R2=1R^2=1, so BB can be 3,5,73,5,7.

For R=2R=2, R2=4R^2=4, so BB can be 5,75,7.

The favourable ordered pairs are

(1,3), (1,5), (1,7), (2,5), (2,7).\begin{align*} (1,3),\ (1,5),\ (1,7),\ (2,5),\ (2,7). \end{align*}

Therefore

P(R2<B)=516.\begin{align*} P(R^2<B)=\frac5{16}. \end{align*}

(g)

解法一

思路

展开

p=F(4)=P(D4)p=F(4)=P(D\leqslant4)。已知 F(3)=34F(3)=\dfrac34,所以只要补上 P(D=4)P(D=4)。也可以从 F(5)F(5) 减去 P(D=5)P(D=5)

答题过程

展开

The outcomes giving D=4D=4 are

(R,B)=(1,5), (3,7).\begin{align*} (R,B)=(1,5),\ (3,7). \end{align*}

So

P(D=4)=216=18.\begin{align*} P(D=4)=\frac2{16}=\frac18. \end{align*}

Since F(3)=34F(3)=\dfrac34,

p=F(4)=F(3)+P(D=4)=34+18=78.\begin{align*} p =&\,F(4)\\[3mm] =&\,F(3)+P(D=4)\\[3mm] =&\,\frac34+\frac18\\[3mm] =&\,\frac78. \end{align*}