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IAL 2021 Oct Q4

A Level / Edexcel / S1

IAL 2021 Oct Paper · Question 4

题目

Problem

Three bags A, B and C each contain coloured balls. Bag A contains 44 red balls and 22 yellow balls only. Bag B contains 44 red balls and 11 yellow ball only. Bag C contains 66 red balls only.

In a game Mike takes a ball at random from bag A, records the colour and places it in bag C. He then takes a ball at random from bag B, records the colour and places it in bag C. Finally, Mike takes a ball at random from bag C and records the colour.

(a) Complete the tree diagram on the page opposite, to illustrate the game by adding the remaining branches and all probabilities.

Tree diagram
(3)

(b) Show that the probability that Mike records a yellow ball exactly twice is 110\dfrac{1}{10}

(3)

Given that Mike records exactly 22 yellow balls,

(c) find the probability that the ball drawn from bag A is red.

(2)

Mike plays this game a large number of times, each time starting with the bags containing balls as described above. The random variable XX represents the number of yellow balls recorded in a single game.

(d) Find the probability distribution of XX

(3)

(e) Find E⁡(X)\operatorname{E}(X)

(2)

解答

(a)

解法一

思路

展开

Bag A 的黄球概率是 2/6=1/32/6=1/3，Bag B 的黄球概率是 1/51/5。第三次从 Bag C 抽时，要根据前两次放入 C 的颜色更新 C 中黄球数和总球数。

答题过程

展开

The probabilities from bag A are

P(R)=23,P(Y)=13.\begin{align*} P(R)=\frac23,\qquad P(Y)=\frac13. \end{align*}

The probabilities from bag B are

P(R)=45,P(Y)=15.\begin{align*} P(R)=\frac45,\qquad P(Y)=\frac15. \end{align*}

For the final draw from bag C:

first two coloursP(R from C)P(Y from C)RR10RY or YR7818YY6828\begin{array}{c|cc} \text{first two colours}&P(R\text{ from C})&P(Y\text{ from C})\\ \hline RR&1&0\\ RY\text{ or }YR&\frac78&\frac18\\ YY&\frac68&\frac28 \end{array}

(b)

解法一

思路

展开

恰好两次黄球有三种顺序：RYYRYY、YRYYRY、YYRYYR。

答题过程

展开 P(exactly 2 yellow)= P(RYY)+P(YRY)+P(YYR)= 23⋅15⋅18+13⋅45⋅18 +13⋅15⋅68= 160+130+120= 110.\begin{align*} P(\text{exactly 2 yellow}) =&\,P(RYY)+P(YRY)+P(YYR)\\[3mm] =&\,\frac23\cdot\frac15\cdot\frac18 +\frac13\cdot\frac45\cdot\frac18\\[3mm] &\,\hspace{2pt} +\frac13\cdot\frac15\cdot\frac68\\[3mm] =&\,\frac{1}{60}+\frac{1}{30}+\frac{1}{20}\\[3mm] =&\,\frac{1}{10}. \end{align*}

(c)

解法一

思路

展开

已知恰好两次黄球。若 Bag A 抽到红球，对应顺序只能是 RYYRYY。

答题过程

展开 P(R from A∣exactly 2 yellow)= P(RYY)P(exactly 2 yellow)= 23⋅15⋅18110= 16.\begin{align*} P(R\text{ from A}\mid \text{exactly 2 yellow}) =&\,\frac{P(RYY)}{P(\text{exactly 2 yellow})}\\[3mm] =&\,\frac{\frac23\cdot\frac15\cdot\frac18}{\frac{1}{10}}\\[3mm] =&\,\frac{1}{6}. \end{align*}

(d)

解法一

思路

展开

XX 可以是 0,1,2,30,1,2,3。把树图中对应 yellow 数量的路径概率相加。

答题过程

展开

The probability distribution of XX is

x0123P(X=x)6412042120121202120\begin{array}{c|cccc} x&0&1&2&3\\ \hline P(X=x)&\frac{64}{120}&\frac{42}{120}&\frac{12}{120}&\frac{2}{120} \end{array}

(e)

解法一

思路

展开

用上一小题的分布求 ∑xP(X=x)\sum xP(X=x)。

答题过程

展开 E⁡(X)= 0⋅64120+1⋅42120 +2⋅12120+3⋅2120= 72120= 0.6.\begin{align*} \operatorname{E}(X) =&\,0\cdot\frac{64}{120} +1\cdot\frac{42}{120}\\[3mm] &\,\hspace{2pt} +2\cdot\frac{12}{120} +3\cdot\frac{2}{120}\\[3mm] =&\,\frac{72}{120}\\[3mm] =&\,0.6. \end{align*}