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IAL 2022 Oct Q7

A Level / Edexcel / S1

IAL 2022 Oct Paper · Question 7

题目

Problem

Adana selects one number at random from the distribution of XX which has the following probability distribution.

x0510P(X=x)0.10.20.7\begin{array}{c|ccc} x&0&5&10\\ \hline P(X=x)&0.1&0.2&0.7 \end{array}

(a) Given that the number selected by Adana is not 55, write down the probability it is 00

(1)

(b) Show that E⁡(X2)=75\operatorname{E}(X^2)=75

(1)

(c) Find Var⁡(X)\operatorname{Var}(X)

(3)

(d) Find Var⁡(4−3X)\operatorname{Var}(4-3X)

(2)

Bruno and Charlie each independently select one number at random from the distribution of XX

(e) Find the probability that the number Bruno selects is greater than the number Charlie selects.

(3)

Devika multiplies Bruno’s number by Charlie’s number to obtain a product, DD

(f) Determine the probability distribution of DD

(4)

解答

(a)

解法一

思路

展开

已知不是 55，剩下可能是 00 或 1010。条件概率分母是 P(X≠5)P(X\ne5)。

答题过程

展开 P(X=0∣X≠5)=0.10.1+0.7=18.\begin{align*} P(X=0\mid X\ne5)=\frac{0.1}{0.1+0.7}=\frac{1}{8}. \end{align*}

(b)

解法一

思路

展开

用 E⁡(X2)=∑x2P(X=x)\operatorname{E}(X^2)=\sum x^2P(X=x)。

答题过程

展开 E⁡(X2)= 02(0.1)+52(0.2)+102(0.7)= 0+5+70= 75.\begin{align*} \operatorname{E}(X^2) =&\,0^2(0.1)+5^2(0.2)+10^2(0.7)\\[3mm] =&\,0+5+70\\[3mm] =&\,75. \end{align*}

(c)

解法一

思路

展开

先求 E⁡(X)\operatorname{E}(X)，再用 Var⁡(X)=E⁡(X2)−[E⁡(X)]2\operatorname{Var}(X)=\operatorname{E}(X^2)-[\operatorname{E}(X)]^2。

答题过程

展开 E⁡(X)= 0(0.1)+5(0.2)+10(0.7)= 8.\begin{align*} \operatorname{E}(X) =&\,0(0.1)+5(0.2)+10(0.7)\\[3mm] =&\,8. \end{align*}

Therefore

Var⁡(X)= 75−82= 11.\begin{align*} \operatorname{Var}(X) =&\,75-8^2\\[3mm] =&\,11. \end{align*}

(d)

解法一

思路

展开

加减常数不影响 variance；乘以 −3-3 会使 variance 乘以 (−3)2=9(-3)^2=9。

答题过程

展开 Var⁡(4−3X)= (−3)2Var⁡(X)= 9(11)= 99.\begin{align*} \operatorname{Var}(4-3X) =&\,(-3)^2\operatorname{Var}(X)\\[3mm] =&\,9(11)\\[3mm] =&\,99. \end{align*}

(e)

解法一

思路

展开

Bruno 的数比 Charlie 大，有 (5,0)(5,0)、(10,0)(10,0)、(10,5)(10,5) 三种组合。

答题过程

展开 P(B>C)= P(5,0)+P(10,0)+P(10,5)= 0.2(0.1)+0.7(0.1)+0.7(0.2)= 0.02+0.07+0.14= 0.23.\begin{align*} P(B>C) =&\,P(5,0)+P(10,0)+P(10,5)\\[3mm] =&\,0.2(0.1)+0.7(0.1)+0.7(0.2)\\[3mm] =&\,0.02+0.07+0.14\\[3mm] =&\,0.23. \end{align*}

(f)

解法一

思路

展开

列出两人数字相乘的可能结果：0,25,50,1000,25,50,100。因为两人独立，组合概率相乘；若有两种顺序，要把两种顺序相加。

答题过程

展开

The possible values of DD are

0, 25, 50, 100.\begin{align*} 0,\ 25,\ 50,\ 100. \end{align*}

Now

P(D=0)=1−P(B≠0, C≠0)=1−(0.9)(0.9)=0.19.\begin{align*} P(D=0)=1-P(B\ne0,\ C\ne0)=1-(0.9)(0.9)=0.19. \end{align*}

Also,

P(D=25)=0.2(0.2)=0.04.\begin{align*} P(D=25)=0.2(0.2)=0.04. \end{align*}

For D=50D=50, the outcomes are (5,10)(5,10) and (10,5)(10,5), so

P(D=50)=2(0.2)(0.7)=0.28.\begin{align*} P(D=50)=2(0.2)(0.7)=0.28. \end{align*}

Finally,

P(D=100)=0.7(0.7)=0.49.\begin{align*} P(D=100)=0.7(0.7)=0.49. \end{align*}

Thus the probability distribution of DD is

d02550100P(D=d)0.190.040.280.49\begin{array}{c|cccc} d&0&25&50&100\\ \hline P(D=d)&0.19&0.04&0.28&0.49 \end{array}

解法二

思路

展开

直接路徑加總法。對於乘積 D=0D = 0 的概率，除了使用補集（對立事件）計算之外，我們也可以直接列出所有相乘後等於 00 的組合。 當且僅當 Bruno (B)(B) 和 Charlie (C)(C) 中至少有一人抽到 00 時，乘積 DD 才為 00。 這些組合分別為：

  • (B=0,C=0)(B=0, C=0)：概率為 0.1×0.1=0.010.1 \times 0.1 = 0.01
  • (B=0,C=5)(B=0, C=5) 和 (B=5,C=0)(B=5, C=0)：概率為 2×(0.1×0.2)=0.042 \times (0.1 \times 0.2) = 0.04
  • (B=0,C=10)(B=0, C=10) 和 (B=10,C=0)(B=10, C=0)：概率為 2×(0.1×0.7)=0.142 \times (0.1 \times 0.7) = 0.14 我們直接將這幾條路徑的概率相加得到 0.190.19。這種方法不依賴對立事件，計算邏輯更加直接，不易出錯。

答题过程

展开

The possible values of DD are 0,25,50,1000, 25, 50, 100.

Calculate the probability P(D=0)P(D = 0) directly by listing all disjoint winning pairs (B,C)(B, C):

  • Both select 00: P(B=0∩C=0)=0.1×0.1=0.01P(B = 0 \cap C = 0) = 0.1 \times 0.1 = 0.01
  • One selects 00 and the other selects 55: P((0,5)∪(5,0))=2×(0.1×0.2)=0.04P((0, 5) \cup (5, 0)) = 2 \times (0.1 \times 0.2) = 0.04
  • One selects 00 and the other selects 1010: P((0,10)∪(10,0))=2×(0.1×0.7)=0.14P((0, 10) \cup (10, 0)) = 2 \times (0.1 \times 0.7) = 0.14

Summing these probabilities:

P(D=0)=  0.01+0.04+0.14=  0.19.\begin{align*} P(D = 0) =&\,\, 0.01 + 0.04 + 0.14\\[3mm] =&\,\, 0.19. \end{align*}

The other probabilities are calculated as:

P(D=25)=P(B=5∩C=5)=0.2×0.2=0.04\begin{align*} P(D = 25) = P(B = 5 \cap C = 5) = 0.2 \times 0.2 = 0.04 \end{align*} P(D=50)=P((5,10)∪(10,5))=2×(0.2×0.7)=0.28\begin{align*} P(D = 50) = P((5, 10) \cup (10, 5)) = 2 \times (0.2 \times 0.7) = 0.28 \end{align*} P(D=100)=P(B=10∩C=10)=0.7×0.7=0.49.\begin{align*} P(D = 100) = P(B = 10 \cap C = 10) = 0.7 \times 0.7 = 0.49. \end{align*}

Thus the probability distribution of DD is:

d02550100P(D=d)0.190.040.280.49\begin{array}{c|cccc} d&0&25&50&100\\ \hline P(D=d)&0.19&0.04&0.28&0.49 \end{array}