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IAL 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)

Tree diagram

(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}\\[2mm] &\,\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)

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

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The completed probabilities are:

P(first white)= 39,P(first yellow)= 69.\begin{align*} P(\text{first white})=&\,\frac39,\\[2mm] 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(second white∣first white)= 28,P(second yellow∣first white)= 68.\begin{align*} P(\text{second white}\mid\text{first white})=&\,\frac28,\\[2mm] P(\text{second yellow}\mid\text{first white})=&\,\frac68. \end{align*}

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

P(second white∣first yellow)= 38,P(second yellow∣first yellow)= 58.\begin{align*} P(\text{second white}\mid\text{first yellow})=&\,\frac38,\\[2mm] P(\text{second yellow}\mid\text{first yellow})=&\,\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

展开

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+\frac69\cdot\frac38\\[3mm] =&\,\frac{18}{72}+\frac{18}{72}\\[3mm] =&\,\frac{36}{72}\\[3mm] =&\,\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})\\[3mm] =&\,1-\frac69\cdot\frac58\\[3mm] =&\,1-\frac{30}{72}\\[3mm] =&\,\frac{42}{72}\\[3mm] =&\,\frac7{12}. \end{align*}

解法二

思路

展开

直接概率相加法。两次抽样中“至少抽到一个白球”可以细分为以下三个互斥的事件：

  1. 第一次是白球，第二次是黄球 (WYWY)
  2. 第一次是黄球，第二次是白球 (YWYW)
  3. 两次都是白球 (WWWW) 我们根据树状图分别计算这三个路径的概率，然后相加。这个方法不需要使用对立事件，思路更加直观。

答题过程

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The event “at least one white” is composed of three mutually exclusive outcomes: WYWY, YWYW, and WWWW.

From the tree diagram:

P(WY)=  39⋅68=1872P(YW)=  69⋅38=1872P(WW)=  39⋅28=672\begin{align*} P(WY) =&\,\, \frac39\cdot\frac68 = \frac{18}{72}\\[3mm] P(YW) =&\,\, \frac69\cdot\frac38 = \frac{18}{72}\\[3mm] P(WW) =&\,\, \frac39\cdot\frac28 = \frac6{72} \end{align*}

Summing these probabilities:

P(at least one white)=  P(WY)+P(YW)+P(WW)=  1872+1872+672=  4272=  712.\begin{align*} P(\text{at least one white}) =&\,\, P(WY) + P(YW) + P(WW)\\[3mm] =&\,\, \frac{18}{72} + \frac{18}{72} + \frac{6}{72}\\[3mm] =&\,\, \frac{42}{72}\\[3mm] =&\,\, \frac7{12}. \end{align*}

(d)

解法一

思路

展开

这是条件概率：

P(both white∣at least one white)=P(both white)P(at least one white).P(\text{both white}\mid\text{at least one white}) =\frac{P(\text{both white})}{P(\text{at least one white})}.

答题过程

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First,

P(both white)= 39⋅28= 672= 112.\begin{align*} P(\text{both white}) =&\,\frac39\cdot\frac28\\[3mm] =&\,\frac6{72}\\[3mm] =&\,\frac1{12}. \end{align*}

From part (c),

P(at least one white)=712.\begin{align*} P(\text{at least one white})=\frac7{12}. \end{align*}

Therefore

P(both white∣at least one white)= 112712= 17.\begin{align*} P(\text{both white}\mid\text{at least one white}) =&\,\frac{\frac1{12}}{\frac7{12}}\\[3mm] =&\,\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*}

The probability of getting red, red, yellow, white is

n2n⋅n2−1n−1⋅3n8n−2⋅n8n−3.\frac{\frac n2}{n} \cdot\frac{\frac n2-1}{n-1} \cdot\frac{\frac{3n}{8}}{n-2} \cdot\frac{\frac n8}{n-3}.

So

n2n⋅n2−1n−1⋅3n8n−2⋅n8n−3=3235.\begin{align*} \frac{\frac n2}{n} &\cdot\frac{\frac n2-1}{n-1} \cdot\frac{\frac{3n}{8}}{n-2} \cdot\frac{\frac n8}{n-3} =\frac3{235}. \end{align*}

Now simplify each factor:

n2n= 12,n2−1n−1= n−22(n−1).\begin{align*} \frac{\frac n2}{n}=&\,\frac12,\\[2mm] \frac{\frac n2-1}{n-1} =&\,\frac{n-2}{2(n-1)}. \end{align*}

Hence

12⋅n−22(n−1)⋅3n8(n−2)⋅n8(n−3)=3235.\begin{align*} \frac12 &\cdot\frac{n-2}{2(n-1)} \cdot\frac{3n}{8(n-2)} \cdot\frac{n}{8(n-3)} =\frac3{235}. \end{align*}

Cancel the common factor n−2n-2:

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)\\[3mm] 235n^2 =&\,256(n^2-4n+3)\\[3mm] 235n^2 =&\,256n^2-1024n+768\\[3mm] 21n^2-1024n+768=&\,0. \end{align*}

So

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