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S1 2019 Jan Q5

A Level / Edexcel / S1

IAL 2019 Jan Paper · Question 5

题目

Problem

Some children are playing a game involving throwing a ball into a bucket. Each child has 3 throws and the number of times the ball lands in the bucket, xx, is recorded. Their results are given in the table below.

xx0123
Frequency1636244

(a) Find xˉ\bar x

(1)

Sandra decides to model the game by assuming that on each throw, the probability of the ball landing in the bucket is 0.4 for every child on every throw and that the throws are all independent. The random variable SS represents the number of times the ball lands in the bucket for a randomly selected child.

(b) Find P(S=2)P(S=2)

(2)

(c) Complete the table below to show the probability distribution for SS.

ss0123
P(S=s)P(S=s)0.4320.064
(1)

Ting believes that the probability of the ball landing in the bucket is not the same for each throw. He suggests that the probability will increase with each throw and uses the model pi=0.15i+0.10p_i=0.15i+0.10, where i=1,2,3i=1,2,3 and pip_i is the probability that the iith throw of the ball, by any particular child, will land in the bucket.

The random variable TT represents the number of times the ball lands in the bucket for a randomly selected child using Ting’s model.

(d) Show that

(i) P(T=3)=0.055P(T=3)=0.055

(ii) P(T=1)=0.45P(T=1)=0.45

(5)

(e) Complete the table below to show the probability distribution for TT, stating the exact probabilities in each case.

tt0123
P(T=t)P(T=t)0.450.055
(3)

(f) State, giving your reasons, whether Sandra’s model or Ting’s model is the more appropriate for modelling this game.

(3)
题目中文翻译

一些儿童进行把球投进桶里的游戏。每人投 3 次,以 xx 记录球进入桶中的次数,结果如表所示。

(a) 求 xˉ\bar x。

Sandra 假设每名儿童每次投球入桶的概率均为 0.4,且各次投球相互独立。随机变量 SS 表示随机选取的一名儿童投球入桶的次数。

(b) 求 P(S=2)P(S=2)。

(c) 补全 SS 的概率分布表。

Ting 认为每次投球入桶的概率并不相同,且会逐次增加。他采用模型 pi=0.15i+0.10p_i=0.15i+0.10,其中 i=1,2,3i=1,2,3,pip_i 表示任一儿童第 ii 次投球入桶的概率。随机变量 TT 表示采用 Ting 模型时,随机选取的一名儿童投球入桶的次数。

(d) 证明:(i) P(T=3)=0.055P(T=3)=0.055;(ii) P(T=1)=0.45P(T=1)=0.45。

(e) 补全 TT 的概率分布表,并在每种情况下写出精确概率。

(f) 说明 Sandra 的模型还是 Ting 的模型更适合描述该游戏,并给出理由。

解答

(a)

解法一

思路

展开

用频数表求平均。

答题过程

展开 xˉ=0(16)+1(36)+2(24)+3(4)80=9680=1.2.\begin{align*} \bar x=\frac{0(16)+1(36)+2(24)+3(4)}{80}=\frac{96}{80}=1.2. \end{align*}

(b)

解法一

思路

展开

S∼B(3,0.4)S\sim B(3,0.4)。恰好 2 次成功。

答题过程

展开 P(S=2)=(32)(0.4)2(0.6)=0.288.\begin{align*} P(S=2)=\binom32(0.4)^2(0.6)=0.288. \end{align*}

(c)

解法一

思路

展开

用总概率为 1 求 P(S=0)P(S=0)。

答题过程

展开 P(S=0)=1−(0.432+0.288+0.064)=0.216.\begin{align*} P(S=0)=1-(0.432+0.288+0.064)=0.216. \end{align*}

The distribution is

s0123P(S=s)0.2160.4320.2880.064\begin{array}{c|cccc} s&0&1&2&3\\ \hline P(S=s)&0.216&0.432&0.288&0.064 \end{array}

(d)

(i)

思路

展开

先由模型求三次投球各自的成功概率,再把三次都成功的概率相乘。

答题过程

展开

The probabilities are

p1=0.25,p2=0.40,p3=0.55.\begin{align*} p_1=0.25,\qquad p_2=0.40,\qquad p_3=0.55. \end{align*}

Therefore

P(T=3)=0.25(0.40)(0.55)=0.055.\begin{align*} P(T=3)=0.25(0.40)(0.55)=0.055. \end{align*}

(ii)

思路

展开

T=1T=1 有三种互斥情形:只有第一次、只有第二次或只有第三次投中。

答题过程

展开 P(T=1)= 0.25(0.60)(0.45)+ 0.75(0.40)(0.45)+ 0.75(0.60)(0.55)= 0.45.\begin{align*} P(T=1) =&\,0.25(0.60)(0.45)\\ +&\,0.75(0.40)(0.45)\\ +&\,0.75(0.60)(0.55)\\ =&\,0.45. \end{align*}

(e)

解法一

思路

展开

求 T=0T=0,再用总概率为 1 求 T=2T=2。

答题过程

展开 P(T=0)=0.75(0.60)(0.45)=0.2025.\begin{align*} P(T=0)=0.75(0.60)(0.45)=0.2025. \end{align*}

Then

P(T=2)=1−(0.2025+0.45+0.055)=0.2925.\begin{align*} P(T=2)=1-(0.2025+0.45+0.055)=0.2925. \end{align*}

So

t0123P(T=t)0.20250.450.29250.055\begin{array}{c|cccc} t&0&1&2&3\\ \hline P(T=t)&0.2025&0.45&0.2925&0.055 \end{array}

(f)

解法一

思路

展开

把实际频率 16/80,36/80,24/80,4/8016/80,36/80,24/80,4/80 与两个模型比较。Ting 的每个概率都更接近实际频率。

答题过程

展开

The observed probabilities are

0.20,0.45,0.30,0.05.\begin{align*} 0.20,\quad0.45,\quad0.30,\quad0.05. \end{align*}

Sandra’s model gives

0.216,0.432,0.288,0.064.\begin{align*} 0.216,\quad0.432,\quad0.288,\quad0.064. \end{align*}

Ting’s model gives

0.2025,0.45,0.2925,0.055.\begin{align*} 0.2025,\quad0.45,\quad0.2925,\quad0.055. \end{align*}

Ting’s model is more appropriate, since its probabilities are closer to the observed probabilities.