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S1 2025 May Q7

A Level / Edexcel / S1

IAL 2025 May Paper · Question 7

题目

Problem

Riley has a bag containing 33 white beads and 66 yellow beads only. Two beads are selected at random from the bag, one at a time without replacement.

(a) Complete the tree diagram on page 2121.

(3)

(b) Find the probability that exactly one of the beads selected is yellow. Show your working.

(2)

(c) Find the probability that at least one of the beads selected is white.

(2)

Given that at least one of the beads selected is white,

(d) find the probability that both of the beads selected are white.

(2)

Riley adds some beads to the bag so it contains a total of nn beads. The beads are either red or yellow or white such that

number of red beads:number of yellow beads :number of white beads=4:3:1.\begin{align*} \text{number of red beads} &:\text{number of yellow beads}\\ &\,\hspace{2pt}:\text{number of white beads}=4:3:1. \end{align*}

Four beads are selected at random from the bag, one at a time without replacement. The probability of getting 22 red beads, followed by a yellow bead, followed by a white bead is

3235.\frac{3}{235}.

(e) Show that nn satisfies the equation

21n2+bn+c=021n^2+bn+c=0

where bb and cc are integers to be found.

(4)
题目中文翻译

Riley 的袋中只有 33 颗白色珠子和 66 颗黄色珠子。随机逐颗抽取两颗，不放回。

(a) 完成官方试卷第 2121 页的树状图（本页上方已给出）。（3 分）

(b) 求所选珠子中恰有一颗黄色的概率，并写出计算过程。（2 分）

(c) 求所选珠子中至少有一颗白色的概率。（2 分）

已知至少选到一颗白色珠子，

(d) 求两颗珠子均为白色的概率。（2 分）

Riley 向袋中添加珠子，使总数为 nn。珠子为红色、黄色或白色，数量之比为 4:3:14:3:1。

随机逐颗抽取四颗，不放回。先抽到两颗红色，再抽到一颗黄色，最后抽到一颗白色的概率为 3235\dfrac3{235}。

(e) 证明 nn 满足 21n2+bn+c=021n^2+bn+c=0，并求整数 bb 和 cc。（4 分）

解答

(a)

解法一

思路

展开

第一次抽之前一共有 99 个 beads，其中白色 33 个、黄色 66 个。因为是不放回抽样，第二次的分母会变成 88，而分子要根据第一次抽到什么来调整。

答题过程

展开

The completed probabilities are:

Here WiW_i and YiY_i mean that bead ii is white or yellow, respectively.

P(first white)= 39,P(first yellow)= 69.\begin{align*} P(\text{first white})=&\,\frac39,\\ P(\text{first yellow})=&\,\frac69. \end{align*}

If the first bead is white, then 22 white and 66 yellow beads remain, out of 88 beads:

P(W2∣W1)= 28,P(Y2∣W1)= 68.\begin{align*} P(W_2\mid W_1)=&\,\frac28,\\ P(Y_2\mid W_1)=&\,\frac68. \end{align*}

If the first bead is yellow, then 33 white and 55 yellow beads remain, out of 88 beads:

P(W2∣Y1)= 38,P(Y2∣Y1)= 58.\begin{align*} P(W_2\mid Y_1)=&\,\frac38,\\ P(Y_2\mid Y_1)=&\,\frac58. \end{align*}

(b)

解法一

思路

展开

“Exactly one yellow” 有两种顺序：先白后黄，或先黄后白。分别乘路径概率，再相加。

答题过程

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Exactly one yellow means either white then yellow, or yellow then white.

P(exactly one yellow)= 39⋅68 +69⋅38= 1872+1872= 3672= 12.\begin{align*} P(\text{exactly one yellow}) =&\,\frac39\cdot\frac68\\ &\,\hspace{2pt}+\frac69\cdot\frac38\\ =&\,\frac{18}{72}+\frac{18}{72}\\ =&\,\frac{36}{72}\\ =&\,\frac12. \end{align*}

(c)

解法一

思路

展开

“At least one white” 的反面是“两次都是 yellow”。所以用 1−P(YY)1-P(YY) 最快。

答题过程

展开 P(at least one white)= 1−P(both yellow)= 1−69⋅58= 1−3072= 4272= 712.\begin{align*} P(\text{at least one white}) =&\,1-P(\text{both yellow})\\ =&\,1-\frac69\cdot\frac58\\ =&\,1-\frac{30}{72}\\ =&\,\frac{42}{72}\\ =&\,\frac7{12}. \end{align*}

解法二

思路

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至少一个白色可分成两个互斥事件：第一颗白色（第二颗任意），或第一颗黄色、第二颗白色。

答题过程

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Let W1W_1 mean the first bead is white and Y1W2Y_1W_2 mean yellow then white.

P(at least one white)= P(W1)+P(Y1W2)= 39+69×38= 2472+1872= 712.\begin{align*} P(\text{at least one white})=&\,P(W_1)+P(Y_1W_2)\\ =&\,\frac39+\frac69\times\frac38\\ =&\,\frac{24}{72}+\frac{18}{72}\\ =&\,\frac7{12}. \end{align*}

(d)

解法一

思路

展开

令 AA 表示两颗均为白色，BB 表示至少一颗白色。因为 A⊂BA\subset B，条件概率为 P(A∣B)=P(A)/P(B)P(A\mid B)=P(A)/P(B)。

答题过程

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Let AA mean both beads are white and BB mean at least one is white. Then

P(A)= 39⋅28= 672= 112.\begin{align*} P(A) =&\,\frac39\cdot\frac28\\ =&\,\frac6{72}\\ =&\,\frac1{12}. \end{align*}

From part (c),

P(B)=712.\begin{align*} P(B)=\frac7{12}. \end{align*}

Therefore

P(A∣B)= 112712= 17.\begin{align*} P(A\mid B) =&\,\frac{\frac1{12}}{\frac7{12}}\\ =&\,\frac17. \end{align*}

(e)

解法一

思路

展开

比例是 4:3:14:3:1，总份数是 88，所以 red、yellow、white 的数量分别是

n2,3n8,n8.\begin{align*} \frac n2,\quad \frac{3n}{8},\quad \frac n8. \end{align*}

不放回抽样时，分母依次是 n,n−1,n−2,n−3n,n-1,n-2,n-3。按题目指定顺序 R,R,Y,WR,R,Y,W 写出概率，然后令它等于 3235\dfrac3{235}。

答题过程

展开

The ratio is 4:3:14:3:1, so the numbers of red, yellow and white beads are

n2,3n8,n8.\begin{align*} \frac n2,\qquad \frac{3n}{8},\qquad \frac n8. \end{align*}

Let RRYWRRYW denote the required ordered outcome. The first probability is n/2n=12\frac{n/2}{n}=\frac12.

Since the bag contains at least the original 99 beads, n>3n>3 and n−2≠0n-2\neq0. Then

P(RRYW)= 12⋅n/2−1n−1 ⋅3n/8n−2⋅n/8n−3= 12⋅n−22(n−1) ⋅3n8(n−2)⋅n8(n−3)= 3n2256(n−1)(n−3).\begin{align*} P(RRYW)=&\,\frac12\cdot\frac{n/2-1}{n-1}\\ &\,\hspace{2pt}\cdot\frac{3n/8}{n-2}\cdot\frac{n/8}{n-3}\\ =&\,\frac12\cdot\frac{n-2}{2(n-1)}\\ &\,\hspace{2pt}\cdot\frac{3n}{8(n-2)}\cdot\frac{n}{8(n-3)}\\ =&\,\frac{3n^2}{256(n-1)(n-3)}. \end{align*}

Equating this to the given probability,

3n2256(n−1)(n−3)= 3235.\begin{align*} \frac{3n^2}{256(n-1)(n-3)}=&\,\frac3{235}. \end{align*}

Divide both sides by 33:

n2256(n−1)(n−3)= 1235.\begin{align*} \frac{n^2}{256(n-1)(n-3)} =&\,\frac1{235}. \end{align*}

Therefore

235n2= 256(n−1)(n−3)235n2= 256(n2−4n+3)235n2= 256n2−1024n+76821n2−1024n+768= 0.\begin{align*} 235n^2 =&\,256(n-1)(n-3)\\ 235n^2 =&\,256(n^2-4n+3)\\ 235n^2 =&\,256n^2-1024n+768\\ 21n^2-1024n+768=&\,0. \end{align*}

So

b=−1024,c=768.\begin{align*} b=-1024,\qquad c=768. \end{align*}