CalcGospel 國際數學圖譜
返回

IAL 2023 Jan Q2

A Level / Edexcel / S1

IAL 2023 Jan Paper · Question 2

题目

Problem

Two bags, X and Y, each contain green marbles (G) and blue marbles (B) only.

  • Bag X contains 55 green marbles and 44 blue marbles
  • Bag Y contains 66 green marbles and 55 blue marbles

A marble is selected at random from bag X and placed in bag Y

A second marble is selected at random from bag X and placed in bag Y

A third marble is then selected, this time from bag Y

(a) Use this information to complete the tree diagram shown on page 77

Tree diagram
(3)

(b) Find the probability that the 22 marbles selected from bag X are of different colours.

(2)

(c) Find the probability that all 33 marbles selected are the same colour.

(2)

Given that all three marbles selected are the same colour,

(d) find the probability that they are all green.

(3)

解答

(a)

解法一

思路

展开

从 bag X 连续取两次且不放回，所以第二次概率会根据第一次颜色改变。第三次从 bag Y 取，要把前两次转移进 bag Y 的球也算进去。

答题过程

展开

The missing second-branch probabilities from bag X are

P(G∣G)=48,P(B∣G)=48,P(G\mid G)=\frac{4}{8},\qquad P(B\mid G)=\frac{4}{8},

and

P(G∣B)=58,P(B∣B)=38.P(G\mid B)=\frac{5}{8},\qquad P(B\mid B)=\frac{3}{8}.

For the third marble from bag Y:

GG:P(G)=813,P(B)=513,GB or BG:P(G)=713,P(B)=613,BB:P(G)=613,P(B)=713.\begin{gathered} GG:\quad P(G)=\frac{8}{13},\quad P(B)=\frac{5}{13},\\[3mm] GB\text{ or }BG:\quad P(G)=\frac{7}{13},\quad P(B)=\frac{6}{13},\\[3mm] BB:\quad P(G)=\frac{6}{13},\quad P(B)=\frac{7}{13}. \end{gathered}

(b)

解法一

思路

展开

前两次从 X 取出不同颜色，有 GBGB 和 BGBG 两种顺序。

答题过程

展开 P(different colours)= 59⋅48+49⋅58= 59.\begin{align*} P(\text{different colours}) =&\,\frac{5}{9}\cdot\frac{4}{8} +\frac{4}{9}\cdot\frac{5}{8}\\[3mm] =&\,\frac{5}{9}. \end{align*}

(c)

解法一

思路

展开

三颗同色只能是 GGGGGG 或 BBBBBB。

答题过程

展开 P(all same)= 59⋅48⋅813 +49⋅38⋅713= 61234.\begin{align*} P(\text{all same}) =&\,\frac{5}{9}\cdot\frac{4}{8}\cdot\frac{8}{13}\\[3mm] &\,\hspace{2pt} +\frac{4}{9}\cdot\frac{3}{8}\cdot\frac{7}{13}\\[3mm] =&\,\frac{61}{234}. \end{align*}

(d)

解法一

思路

展开

条件概率分母是“三颗同色”，分子是“三颗都是绿色”。

答题过程

展开 P(GGG)=59⋅48⋅813=20117.P(GGG)=\frac{5}{9}\cdot\frac{4}{8}\cdot\frac{8}{13} =\frac{20}{117}.

Therefore

P(GGG∣all same)= 2011761234= 4061.\begin{align*} P(GGG\mid \text{all same}) =&\,\frac{\frac{20}{117}}{\frac{61}{234}}\\[3mm] =&\,\frac{40}{61}. \end{align*}