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IAL 2022 Jan Q7

A Level / Edexcel / S1

IAL 2022 Jan Paper · Question 7

题目

Problem

A bag contains nn marbles of which 77 are green. From the bag, 33 marbles are selected at random. The random variable XX represents the number of green marbles selected. The cumulative distribution function of XX is given by

x0123F(x)ab37381\begin{array}{c|cccc} x&0&1&2&3\\ \hline F(x)&a&b&\frac{37}{38}&1 \end{array}

(a) Show that n(n−1)(n−2)=7980n(n-1)(n-2)=7980

(4)

(b) Verify that n=21n=21 satisfies the equation in part (a).

(1)

Given that n=21n=21

(c) find the exact value of aa and the exact value of bb

(6)

解答

(a)

解法一

思路

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F(3)−F(2)=P(X=3)F(3)-F(2)=P(X=3)。由表可得 P(X=3)=1−3738=138P(X=3)=1-\dfrac{37}{38}=\dfrac{1}{38}。另一方面，三颗全是绿色的概率可用不放回抽样相乘。

答题过程

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From the cumulative distribution function,

P(X=3)=F(3)−F(2)=1−3738=138.\begin{align*} P(X=3)=F(3)-F(2)=1-\frac{37}{38}=\frac{1}{38}. \end{align*}

Also,

P(X=3)=7n⋅6n−1⋅5n−2.\begin{align*} P(X=3)=\frac{7}{n}\cdot\frac{6}{n-1}\cdot\frac{5}{n-2}. \end{align*}

Therefore

7n⋅6n−1⋅5n−2=138.\frac{7}{n}\cdot\frac{6}{n-1}\cdot\frac{5}{n-2} =\frac{1}{38}.

So

210n(n−1)(n−2)=138.\begin{align*} \frac{210}{n(n-1)(n-2)}=\frac{1}{38}. \end{align*}

Hence

n(n−1)(n−2)=210(38)=7980.\begin{align*} n(n-1)(n-2)=210(38)=7980. \end{align*}

(b)

解法一

思路

展开

把 n=21n=21 代入左边验证即可。

答题过程

展开 21(21−1)(21−2)=21(20)(19)=7980.\begin{align*} 21(21-1)(21-2)=21(20)(19)=7980. \end{align*}

Therefore n=21n=21 satisfies the equation.

(c)

解法一

思路

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a=F(0)=P(X=0)a=F(0)=P(X=0)，即三颗都不是绿色。b=F(1)=P(X=0)+P(X=1)b=F(1)=P(X=0)+P(X=1)。

答题过程

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Since n=21n=21, there are 1414 non-green marbles.

For aa,

a= P(X=0)= 1421⋅1320⋅1219= 2695.\begin{align*} a =&\,P(X=0)\\[3mm] =&\,\frac{14}{21}\cdot\frac{13}{20}\cdot\frac{12}{19}\\[3mm] =&\,\frac{26}{95}. \end{align*}

For exactly one green marble,

P(X=1)= 3⋅721⋅1420⋅1319= 91190.\begin{align*} P(X=1) =&\,3\cdot\frac{7}{21}\cdot\frac{14}{20}\cdot\frac{13}{19}\\[3mm] =&\,\frac{91}{190}. \end{align*}

Therefore

b= F(1)= P(X=0)+P(X=1)= 2695+91190= 143190.\begin{align*} b =&\,F(1)\\[3mm] =&\,P(X=0)+P(X=1)\\[3mm] =&\,\frac{26}{95}+\frac{91}{190}\\[3mm] =&\,\frac{143}{190}. \end{align*}