题目

Problem

A bag contains a large number of counters with an odd number or an even number written on each.

Odd and even numbered counters occur in the ratio $4:1$

In a game a player takes a random sample of 4 counters from the bag.

The player scores

• 5 points for each counter taken that has an even number written on it

• 2 points for each counter taken that has an odd number written on it

The random variable $X$ represents the total score, in points, from the 4 counters.

(a) Find the sampling distribution of $X$

(6)

A random sample of $n$ sets of 4 counters is taken. The random variable $Y$ represents the number of these $n$ sets that have a total score of exactly 14

(b) Calculate the minimum value of $n$ such that $P(Y\geqslant1)>0.95$

(3)

(Total for Question 6 is 9 marks)

题目中文翻译

一个袋子里有大量筹码，每个筹码上写有奇数或偶数。

奇数和偶数筹码的比例为 $4:1$ 。

在一场游戏中，玩家从袋中随机抽取 4 个筹码。

玩家得分

• 每个抽到的偶数筹码得 5 分

• 每个抽到的奇数筹码得 2 分

随机变量 $X$ 表示 4 个筹码的总得分。

(a) 求 $X$ 的抽样分布。

(6)

随机抽取 $n$ 组 4 个筹码。随机变量 $Y$ 表示这 $n$ 组中总得分恰好为 14 的组数。

(b) 计算最小的 $n$ 值，使得 $P(Y\geqslant1)>0.95$ 。

(3)

（第 6 题共 9 分）

解答

(a)

解法一

思路

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每次抽到偶数筹码的概率为 $1/5$ ，抽到奇数筹码的概率为 $4/5$ 。若 4 个筹码中有 $r$ 个偶数，则总分为 $5r+2(4-r)=8+3r$ ；让 $r=0,1,2,3,4$ ，并用二项分布计算各个总分的概率。

答题过程

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For each counter,

\begin{align*} P(\text{even})=&\,\frac15,\\[2mm] P(\text{odd})=&\,\frac45. \end{align*}

Let $R$ be the number of even counters among the four selected. Since the bag contains a large number of counters,

R\sim\operatorname{B}\Big(4,\frac15\Big).

If $R=r$ , then

\begin{align*} X=&\,5r+2(4-r)\\ =&\,5r+8-2r\\ =&\,8+3r. \end{align*}

Hence the possible scores are

8,\ 11,\ 14,\ 17,\ 20.

The corresponding probabilities are

\begin{align*} &\,P(X=8)\\[2mm] &\,=\Big(\frac45\Big)^4\\[2mm] &\,=\frac{256}{625}, \end{align*}

\begin{align*} &\,P(X=11)\\[2mm] &\,=\binom41\Big(\frac15\Big) \Big(\frac45\Big)^3\\ &\,=\frac{256}{625}, \end{align*}

\begin{align*} &\,P(X=14)\\[2mm] &\,=\binom42\Big(\frac15\Big)^2 \Big(\frac45\Big)^2\\ &\,=\frac{96}{625}, \end{align*}

\begin{align*} &\,P(X=17)\\[2mm] &\,=\binom43\Big(\frac15\Big)^3 \Big(\frac45\Big)\\ &\,=\frac{16}{625}, \end{align*}

and

\begin{align*} &\,P(X=20)\\[2mm] &\,=\Big(\frac15\Big)^4\\[2mm] &\,=\frac1{625}. \end{align*}

Therefore, the sampling distribution of $X$ is

$x$	$P(X=x)$
8	$\dfrac{256}{625}$
11	$\dfrac{256}{625}$
14	$\dfrac{96}{625}$
17	$\dfrac{16}{625}$
20	$\dfrac1{625}$

(b)

解法一

思路

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由 (a)，每组总分恰好为 14 的概率是 $96/625=0.1536$ 。把 $n$ 组看作独立二项试验；“至少一组”用补事件表示为 1 减去一组都没有的概率。解指数不等式后取满足严格不等式的最小整数。

答题过程

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From part (a),

\begin{align*} P(X=14)=&\,\frac{96}{625}\\[2mm] =&\,0.1536. \end{align*}

Therefore,

Y\sim\operatorname{B}(n,0.1536).

The required condition is

\begin{align*} P(Y\geqslant1)>&\,0.95,\\ 1-(1-0.1536)^n>&\,0.95,\\ (0.8464)^n<&\,0.05. \end{align*}

Taking logarithms, and noting that $\log(0.8464)<0$ , gives

\begin{align*} n>&\,\frac{\log(0.05)}{\log(0.8464)}\\ >&\,17.96\ldots \end{align*}

Also,

\begin{align*} &\,n=17:\quad P(Y\geqslant1)\\[2mm] &\,=1-(0.8464)^{17}\\ &\,=0.9412799887\ldots\\[3mm] &\,n=18:\quad P(Y\geqslant1)\\[2mm] &\,=1-(0.8464)^{18}\\ &\,=0.9502993825\ldots \end{align*}

Hence the minimum integer value is

\boxed{n=18}.

S2 2024 Jan Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程