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 ≥ 1) > 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

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 ≥ 1) > 0.95

（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,

P(\text{even})=\frac15, \qquad P(\text{odd})=\frac45.

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}\left(4,\frac15\right).

If $R=r$ , then

X=5r+2(4-r)=8+3r.

Hence the possible scores are

8,\qquad11,\qquad14,\qquad17,\qquad20.

The corresponding probabilities are

P(X=8)=\left(\frac45\right)^4=\frac{256}{625},

\begin{align*} P(X=11)=&\,\binom41\left(\frac15\right) \left(\frac45\right)^3\\ =&\,\frac{256}{625}, \end{align*}

\begin{align*} P(X=14)=&\,\binom42\left(\frac15\right)^2 \left(\frac45\right)^2\\ =&\,\frac{96}{625}, \end{align*}

\begin{align*} P(X=17)=&\,\binom43\left(\frac15\right)^3 \left(\frac45\right)\\ =&\,\frac{16}{625}, \end{align*}

and

P(X=20)=\left(\frac15\right)^4=\frac1{625}.

Therefore, the sampling distribution of $X$ is

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

(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),

P(X=14)=\frac{96}{625}=0.1536.

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*} P(Y\geqslant1)\big|_{n=17}=&\,0.94128\ldots,\\ P(Y\geqslant1)\big|_{n=18}=&\,0.95030\ldots \end{align*}

Hence the minimum integer value is

\boxed{n=18}.

IAL 2024 Jan S2 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