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IAL 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)

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

SB(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)

解法一

思路

展开

先求三次成功概率:p1=0.25,p2=0.40,p3=0.55p_1=0.25,p_2=0.40,p_3=0.55T=1T=1 时三种情况分别是一中二不中、二中其余不中、三中其余不中。

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*}

Also,

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

(e)

解法一

思路

展开

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.