一组100名学生被询问他们是否是学生领袖、是否参加了体育CCA,或者是否在理学院学习。学生领袖的人数为25人,参加体育CCA的人数为30人,在理学院学习的人数为40人。有15名学生领袖参加了体育CCA。既是学生领袖、参加了体育CCA又在理学院学习的学生人数为 xx。参加了体育CCA且在理学院学习但不是学生领袖的学生人数为 yy。 随机抽取其中一名学生。 AA 表示该学生是学生领袖的事件。
BB 表示该学生参加了体育CCA的事件。
CC 表示该学生在理学院学习的事件。
已知 AA 和 CC 是相互独立的。

  1. (a)

    完成下方的韦恩图以表示上述所有信息。允许用含有 xx 和 yy 的表达式表示。

  2. (b)

    另外已知 BB 和 CC 是相互独立的。

    求 yy 关于 xx 的表达式。由此求出 yy 的最大值和最小值。

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完整解法

6(a)

1. Determine the total number of students in $A \cap C$ using the independence of events $A$ and $C$.

Since choosing a student from AA and choosing a student from CC are independent events:
P(A∩C)=P(A)×P(C)\mathrm{P}(A \cap C) = \mathrm{P}(A) \times \mathrm{P}(C)
P(A∩C)=(25100)×(40100)=110\mathrm{P}(A \cap C) = \left(\frac{25}{100}\right) \times \left(\frac{40}{100}\right) = \frac{1}{10}
n(A∩C)=100×110=10n(A \cap C) = 100 \times \frac{1}{10} = 10

2. Find the expressions for the overlapping regions involving exactly two events.

We are given n(A∩B)=15n(A \cap B) = 15, n(A∩B∩C)=xn(A \cap B \cap C) = x, and n(B∩C only)=yn(B \cap C \text{ only}) = y.

The region for A∩BA \cap B only (excluding CC):
n(A∩B∩C′)=n(A∩B)−n(A∩B∩C)=15−xn(A \cap B \cap C') = n(A \cap B) - n(A \cap B \cap C) = 15 - x

The region for A∩CA \cap C only (excluding BB):
n(A∩C∩B′)=n(A∩C)−n(A∩B∩C)=10−xn(A \cap C \cap B') = n(A \cap C) - n(A \cap B \cap C) = 10 - x

3. Deduce the expressions for the regions representing exactly one event, and the region outside all three.

For AA only:
n(A only)=25−(15−x)−(10−x)−x=xn(A \text{ only}) = 25 - (15 - x) - (10 - x) - x = x

For BB only:
n(B only)=30−(15−x)−x−y=15−yn(B \text{ only}) = 30 - (15 - x) - x - y = 15 - y

For CC only:
n(C only)=40−(10−x)−x−y=30−yn(C \text{ only}) = 40 - (10 - x) - x - y = 30 - y

For the region outside A,B, and CA, B, \text{ and } C:
Total in A∪B∪C=25+(15−y)+y+(30−y)=70−y\text{Total in } A \cup B \cup C = 25 + (15 - y) + y + (30 - y) = 70 - y
n((A∪B∪C)′)=100−(70−y)=30+yn((A \cup B \cup C)') = 100 - (70 - y) = 30 + y

答案
  • AA only: xx
  • BB only: 15−y15 - y
  • CC only: 30−y30 - y
  • AcapBA cap B only (excluding CC): 15−x15 - x
  • AcapCA cap C only (excluding BB): 10−x10 - x
  • BcapCB cap C only (excluding AA): yy
  • AcapBcapCA cap B cap C: xx
  • Outside all three circles: 30+y30 + y

6(b)

1. Find the number of students in the intersection of B and C using independence.

Since BB and CC are independent events:
P(B∩C)=P(B)×P(C)P(B \cap C) = P(B) \times P(C)
P(B∩C)=30100×40100=0.12P(B \cap C) = \frac{30}{100} \times \frac{40}{100} = 0.12
The number of students in B∩CB \cap C is:
0.12×100=120.12 \times 100 = 12

2. Express $y$ in terms of $x$ using the Venn diagram.

From the Venn diagram, the region representing B∩CB \cap C consists of the overlapping parts xx and yy.
x+y=12x + y = 12
y=12−xy = 12 - x

3. Determine the constraints on $x$ to find the bounds for $y$.

The number of students in any region of the Venn diagram must be a non-negative integer (≥0\ge 0).
For the central intersection (A∩B∩CA \cap B \cap C):
x≥0x \ge 0
For the region in AA and CC but not BB:
10−x≥0  ⟹  x≤1010 - x \ge 0 \implies x \le 10
So, the domain for xx is 0≤x≤100 \le x \le 10.

4. Calculate the greatest and least possible values of $y$.

Since y=12−xy = 12 - x and 0≤x≤100 \le x \le 10:
The greatest value of yy occurs when xx is at its minimum (x=0x = 0):
ymax=12−0=12y_{\text{max}} = 12 - 0 = 12
The least value of yy occurs when xx is at its maximum (x=10x = 10):
ymin=12−10=2y_{\text{min}} = 12 - 10 = 2
(Checking the other regions ensures 15−y≥015 - y \ge 0 and 30−y≥030 - y \ge 0 are both safely satisfied for y∈[2,12]y \in [2, 12].)

答案

y=12−xy = 12 - x
Greatest possible value of yy is 1212.
Least possible value of yy is 22.

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