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IAL 2022 June Q5

A Level / Edexcel / S1

IAL 2022 June Paper · Question 5

题目

Problem

A red spinner is designed so that the score RR is given by the following probability distribution.

r23456P(R=r)0.250.30.150.10.2\begin{array}{c|ccccc} r&2&3&4&5&6\\ \hline P(R=r)&0.25&0.3&0.15&0.1&0.2 \end{array}

(a) Show that E⁡(R2)=15.8\operatorname{E}(R^2)=15.8

(1)

Given also that E⁡(R)=3.7\operatorname{E}(R)=3.7

(b) find the standard deviation of RR, giving your answer to 22 decimal places.

(2)

A yellow spinner is designed so that the score YY is given by the probability distribution in the table below. The cumulative distribution function F(y)F(y) is also given.

y23456P(Y=y)0.10.20.1abF(y)0.10.30.4cd\begin{array}{c|ccccc} y&2&3&4&5&6\\ \hline P(Y=y)&0.1&0.2&0.1&a&b\\ F(y)&0.1&0.3&0.4&c&d \end{array}

(c) Write down the value of dd

(1)

Given that E⁡(Y)=4.55\operatorname{E}(Y)=4.55

(d) find the value of cc

(5)

Pabel and Jessie play a game with these two spinners. Pabel uses the red spinner. Jessie uses the yellow spinner. They take turns to spin their spinner. The winner is the first person whose spinner lands on the number 22 and the game ends. Jessie spins her spinner first.

(e) Find the probability that Jessie wins on her second spin.

(2)

(f) Calculate the probability that, in a game, the score on Pabel’s first spin is the same as the score on Jessie’s first spin.

(3)

解答

(a)

解法一

思路

展开

用 E⁡(R2)=∑r2P(R=r)\operatorname{E}(R^2)=\sum r^2P(R=r)。

答题过程

展开 E⁡(R2)= 22(0.25)+32(0.3)+42(0.15) +52(0.1)+62(0.2)= 1+2.7+2.4+2.5+7.2= 15.8.\begin{align*} \operatorname{E}(R^2) =&\,2^2(0.25)+3^2(0.3)+4^2(0.15)\\[3mm] &\,\hspace{2pt}+5^2(0.1)+6^2(0.2)\\[3mm] =&\,1+2.7+2.4+2.5+7.2\\[3mm] =&\,15.8. \end{align*}

(b)

解法一

思路

展开

先求 variance，再开方。

答题过程

展开 Var⁡(R)= E⁡(R2)−[E⁡(R)]2= 15.8−3.72= 2.11.\begin{align*} \operatorname{Var}(R) =&\,\operatorname{E}(R^2)-[\operatorname{E}(R)]^2\\[3mm] =&\,15.8-3.7^2\\[3mm] =&\,2.11. \end{align*}

So

σR=2.11=1.4525…\begin{align*} \sigma_R=\sqrt{2.11}=1.4525\ldots \end{align*}

The standard deviation is

1.45.\begin{align*} 1.45. \end{align*}

(c)

解法一

思路

展开

d=F(6)d=F(6)，也就是所有概率累加到最后，必定为 11。

答题过程

展开 d=1.\begin{align*} d=1. \end{align*}

(d)

解法一

思路

展开

先由总概率得 a+b=0.6a+b=0.6，再由期望方程解出 aa，最后 c=F(5)=0.4+ac=F(5)=0.4+a。

答题过程

展开

Since the total probability is 11,

a+b=0.6.\begin{align*} a+b=0.6. \end{align*}

Using E⁡(Y)=4.55\operatorname{E}(Y)=4.55,

2(0.1)+3(0.2)+4(0.1)+5a+6b=4.55.\begin{align*} 2(0.1)+3(0.2)+4(0.1)+5a+6b=4.55. \end{align*}

So

5a+6b=3.35.\begin{align*} 5a+6b=3.35. \end{align*}

Using b=0.6−ab=0.6-a,

5a+6(0.6−a)= 3.355a+3.6−6a= 3.35a= 0.25.\begin{align*} 5a+6(0.6-a)=&\,3.35\\[3mm] 5a+3.6-6a=&\,3.35\\[3mm] a=&\,0.25. \end{align*}

Therefore

c=F(5)=0.1+0.2+0.1+0.25=0.65.\begin{align*} c=F(5)=0.1+0.2+0.1+0.25=0.65. \end{align*}

(e)

解法一

思路

展开

Jessie 第二次 spin 赢，表示 Jessie 第一次没出 22，Pabel 第一次也没出 22，然后 Jessie 第二次出 22。

答题过程

展开 P(Jessie wins on her second spin)= (1−0.1)(1−0.25)(0.1)= 0.9(0.75)(0.1)= 0.0675.\begin{align*} P(\text{Jessie wins on her second spin}) =&\,(1-0.1)(1-0.25)(0.1)\\[3mm] =&\,0.9(0.75)(0.1)\\[3mm] =&\,0.0675. \end{align*}

(f)

解法一

思路

展开

如果 Jessie 第一次出 22，游戏立即结束，Pabel 没有第一 spin。因此只能比较 3,4,5,63,4,5,6。

答题过程

展开

The score 22 is not included because if Jessie scores 22 first, Pabel does not spin.

Therefore

P(same first score)= 0.2(0.3)+0.1(0.15) +0.25(0.1)+0.35(0.2)= 0.06+0.015+0.025+0.07= 0.17.\begin{align*} P(\text{same first score}) =&\,0.2(0.3)+0.1(0.15)\\[3mm] &\,\hspace{2pt}+0.25(0.1)+0.35(0.2)\\[3mm] =&\,0.06+0.015+0.025+0.07\\[3mm] =&\,0.17. \end{align*}