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IAL 2020 Jan Q6

A Level / Edexcel / S1

IAL 2020 Jan Paper · Question 6

题目

Problem

A tennis tournament has 5 rounds. After each round, winners go into the next round and losers are knocked out of the tournament. To enter the tournament players must pay an entry fee of 10butonlythepersonwhowinsall5roundsreceivestheprizeof10 but only the person who wins all 5 rounds receives the prize of 260

Serena enters this tennis tournament. The random variable SS represents the total number of rounds Serena wins. The probability distribution for SS is given in the following table.

ss012345
P(S=s)P(S=s)kkk2\frac{k}{2}k3\frac{k}{3}k4\frac{k}{4}k5\frac{k}{5}k6\frac{k}{6}

(a) Show that k=2049k=\frac{20}{49}

(2)

(b) Find E(S)E(S)

(3)

(c) Find Serena’s expected profit if she enters the tennis tournament.

(3)

Roger also enters this tennis tournament. Given that Roger is still in the tournament, the probability that he wins the next round is a constant pp.

The random variable RR represents the total number of rounds that Roger wins.

(d) Explain why P(R=2)=p2(1−p)P(R=2)=p^2(1-p)

(2)

(e) Find, in terms of pp, the probability distribution for RR.

(3)

(f) Find the smallest value of pp such that Roger’s expected profit is at least as great as Serena’s.

(4)

解答

(a)

解法一

思路

展开

概率总和为 1，把六项加起来即可。

答题过程

展开

Since the probabilities sum to 1,

k+k2+k3+k4+k5+k6=1.\begin{align*} k+\frac{k}{2}+\frac{k}{3}+\frac{k}{4}+\frac{k}{5}+\frac{k}{6}=1. \end{align*}

Using common denominator 6060,

k(60+30+20+15+12+1060)=1.\begin{align*} k\left(\frac{60+30+20+15+12+10}{60}\right)=1. \end{align*}

So

k⋅14760=1.\begin{align*} k\cdot\frac{147}{60}=1. \end{align*}

Therefore

k=60147=2049.\begin{align*} k=\frac{60}{147}=\frac{20}{49}. \end{align*}

(b)

解法一

思路

展开

用期望公式 ∑sP(S=s)\sum sP(S=s)，其中 s=0s=0 的项为 0。

答题过程

展开 E(S)= 0(k)+1(k2)+2(k3)+3(k4)+4(k5)+5(k6)= k(12+23+34+45+56).\begin{aligned} E(S) =&\,0(k)+1\left(\frac{k}{2}\right)+2\left(\frac{k}{3}\right) +3\left(\frac{k}{4}\right)+4\left(\frac{k}{5}\right)+5\left(\frac{k}{6}\right)\\ =&\,k\left(\frac12+\frac23+\frac34+\frac45+\frac56\right). \end{aligned}

Using k=2049k=\frac{20}{49},

E(S)=2049(7120)=7149.\begin{align*} E(S)=\frac{20}{49}\left(\frac{71}{20}\right)=\frac{71}{49}. \end{align*}

(c)

解法一

思路

展开

只有赢 5 轮才拿到 $260，但一定先付 $10 入场费。期望利润是期望奖金减入场费。

答题过程

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Serena receives the prize only when S=5S=5.

So

P(S=5)=k6=2049⋅16=10147.\begin{align*} P(S=5)=\frac{k}{6}=\frac{20}{49}\cdot\frac16=\frac{10}{147}. \end{align*}

Her expected profit is

260(10147)−10=7.687….260\left(\frac{10}{147}\right)-10 =7.687\ldots.

Therefore Serena’s expected profit is approximately

$7.69.\begin{align*} \$7.69. \end{align*}

(d)

解法一

思路

展开

Roger 总共赢 2 轮，表示先赢第 1、2 轮，然后第 3 轮输掉。

答题过程

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For R=2R=2, Roger must win the first two rounds and then lose the third round.

The probability is therefore

p⋅p⋅(1−p)=p2(1−p).\begin{align*} p\cdot p\cdot(1-p)=p^2(1-p). \end{align*}

(e)

解法一

思路

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如果 Roger 赢 r<5r<5 轮，他必须先赢 rr 轮，再输下一轮。若赢 5 轮，则没有后续输的那一轮。

答题过程

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The probability distribution is

r012345P(R=r)1−pp(1−p)p2(1−p)p3(1−p)p4(1−p)p5\begin{array}{c|cccccc} r&0&1&2&3&4&5\\ \hline P(R=r)&1-p&p(1-p)&p^2(1-p)&p^3(1-p)&p^4(1-p)&p^5 \end{array}

(f)

解法一

思路

展开

Roger 的期望利润只取决于赢 5 轮的概率 p5p^5。令他的期望利润至少等于 Serena 的期望利润。

答题过程

展开

Roger’s expected profit is

260p5−10.\begin{align*} 260p^5-10. \end{align*}

This must be at least Serena’s expected profit:

260p5−10≥7.687….\begin{align*} 260p^5-10\geq7.687\ldots. \end{align*}

So

260p5≥17.687….\begin{align*} 260p^5\geq17.687\ldots. \end{align*}

Hence

p5≥17.687…260.\begin{align*} p^5\geq\frac{17.687\ldots}{260}. \end{align*}

Therefore

p≥0.58418….\begin{align*} p\geq0.58418\ldots. \end{align*}

The smallest value of pp is approximately

0.584.\begin{align*} 0.584. \end{align*}