IS_Math-2013-03

题目来源：Problem 3 日期：2024-08-09 题目主题: CS-概率论-古典概率

解题思路

题目涉及重复操作中颜色球的数量变化，考察概率计算。题目涉及对球的颜色变化的统计，需用到组合计数和递推公式来求解。考虑逐步增加颜色的可能性并结合递推关系式给出相应概率。

Solution

Question 1

Consider the case where Operation A is repeated 3 times. Find the probability that the number of colors of the balls in the bag (black is not counted) is 2 and 3, respectively.

1. Initial State: The bag contains 1 black ball (B).

2. After 1st operation:

   • The black ball (B) is drawn, and a new ball of color 1 (say R for red) is added.
   • Bag Content: {B, R}
   • Number of colors (not counting black): 1

   Illustration:

   [ B ] -> Draw B -> [ B, R ]
   

3. After 2nd operation:

   • There are two possibilities:
     • Draw R (Red): return R with another R. Bag Content: {B, R, R}
     • Draw B (Black): add a new color, say G (Green). Bag Content: {B, R, G}
   • Probabilities:
     • Probability of drawing R: $\frac{1}{2}$, results in 1 color.
     • Probability of drawing B: $\frac{1}{2}$, results in 2 colors.

   Illustration:

   [ B, R ] -> Draw R -> [ B, R, R ]  (1 color remains) 1/2
   [ B, R ] -> Draw B -> [ B, R, G ]  (2 colors now) 1/2
   

4. After 3rd operation:

   • If Bag Content = {B, R, R}, draw results:
     • Draw R: return R with another R. Bag Content: {B, R, R, R}
     • Draw B: add a new color, say G. Bag Content: {B, R, R, G}
   • If Bag Content = {B, R, G}, draw results:
     • Draw R or G: return with same color, stay with 2 colors.
     • Draw B: add a new color, say Y (Yellow). Bag Content: {B, R, G, Y}
   • Probabilities:
     • From 2 colors (after 2 operations):
       • Probability for 3 colors: $\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$
       • Probability for 2 colors: $\frac{1}{2}\times \frac{1}{3}+\frac{1}{2}\times \frac{2}{3}=\frac{1}{2}$
     • From 1 color (after 2 operations):
       • Probability remains 1 color: $\frac{1}{2}\times \frac{2}{3}=\frac{1}{3}$.

   Illustration:

   [ B, R, R ] -> Draw R -> [ B, R, R, R ] (1 color) 1/3
   [ B, R, R ] -> Draw B -> [ B, R, R, G ] (2 colors) 1/6
   
   [ B, R, G ] -> Draw R/G -> [ B, R, G, R/G ] (2 colors) 1/3
   [ B, R, G ] -> Draw B -> [ B, R, G, Y ] (3 colors) 1/6
   

Final result:

• The probability that the number of non-black colors is 2 is $\frac{1}{2}$
• The probability that it is 3 is $\frac{1}{6}$.

Question 2

Consider the case where Operation A is repeated 4 times. Find the probability that the number of colors of the balls in the bag (black is not counted) is 3.

Step 1: Analyze the outcome after 4 operations

Bag Content after 3 operations:

• It could be {B, R, G, R/G} (2 colors) or {B, R, G, Y} (3 colors).

Step 2: Probability computation for 4 operations

• If there are 2 colors after 3 operations:
  • Drawing B in the 4th operation adds a new color, achieving 3 colors.
• If there are already 3 colors after 3 operations:
  • Any non-colored ball drawn keeps it at 3 colors.

Calculation:

• After 3 operations, probability of having 2 colors: $\frac{1}{2}$.
• Probability of adding a 3rd color in the 4th operation: $\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}$.
• Probability of staying with 3 colors if already 3 colors: $\frac{1}{6}\times \frac{3}{4}=\frac{1}{8}$.

Final probability for 3 colors after 4 operations is $\frac{1}{8}+\frac{1}{8}=\frac{1}{4}$.

Question 3

To prove this by induction, we will proceed as follows:

Base Case

When $n=1$:

• Initially, there is only one black ball in the bag.
• After the first operation, we draw the black ball, and a new color (say color 1) is added to the bag.
• At this point, the bag contains two balls: one black ball and one non-black ball (color 1).
• The probability that there is only one non-black color in the bag is 1.

So, the base case is true: $\frac{1}{1}=1$.

Inductive Hypothesis

Assume that after $k$ operations, the probability that there is exactly 1 non-black color in the bag is $\frac{1}{k}$.

Inductive Step

We now need to show that if the hypothesis holds for $n=k$, then it also holds for $n=k+1$.

After $k$ operations, the bag contains the following:

• A black ball.
• $k$ non-black balls of exactly one color (say color 1).

The probability that there is still only one non-black color after $k+1$ operations is calculated as follows:

• In the $(k+1)$-th operation, there are $k+1$ balls in the bag: 1 black ball and $k$ non-black balls of the same color.
• The probability of drawing the black ball (B) is $\frac{1}{k+1}$.
  • If the black ball is drawn, a new color is introduced, so we would no longer have exactly one non-black color.
• The probability of drawing one of the $k$ non-black balls (all of color 1) is $\frac{k}{k+1}$.
  • If a non-black ball is drawn, it is put back with another ball of the same color, and we still have only one non-black color.

Thus, the probability that after $k+1$ operations there is still only one non-black color in the bag is:

$$\text{Probability}=\text{(Probability there was 1 color after k operations)}\times \text{(Probability that k+1 operation keeps 1 color)}$$

Using the inductive hypothesis:

$$\text{Probability}=\frac{1}{k}\times \frac{k}{k+1}=\frac{1}{k+1}$$

This completes the inductive step.

Conclusion

Since we have shown that the hypothesis holds for $n=1$, and that if it holds for $n=k$, it also holds for $n=k+1$, by the principle of mathematical induction, the probability that there is exactly 1 non-black color in the bag after $n$ operations is $\frac{1}{n}$.

Question 4

To derive the probability $q_n(m)$ that after $n$ operations the number of non-black colors in the bag is exactly $m$, we can approach this problem purely from a combinatorial perspective.

Step 1: Understanding the Setup

1. Initial Condition: The bag starts with 1 black ball.
2. Operation Process:
   • If a black ball is drawn, a new color is introduced into the bag.
   • If a non-black ball is drawn, a ball of the same color is added to the bag.

Step 2: Counting the Configurations

To determine $q_n(m)$, we need to count the number of different ways to end up with exactly $m$ distinct non-black colors after $n$ operations and then divide this by the total number of possible outcomes.

1. Total Number of Outcomes: After $n$ operations, the total number of possible outcomes is $n!$ because each operation can occur in any sequence.

2. Favorable Configurations: We must count the number of ways to partition the sequence of $n$ operations such that exactly $m$ distinct non-black colors are introduced.

Step 3: Recursive Counting of Configurations

Consider the situation where we have performed $n-1$ operations and are performing the $n$-th operation.

• Case 1: The $n$-th operation introduces a new color (this occurs when a black ball is drawn).

  • For this to happen, there must have been exactly $m-1$ distinct colors after $n-1$ operations.
  • There is exactly 1 way to introduce a new color in the $n$-th operation.
  • The number of favorable configurations for this scenario is the number of ways to end up with $m-1$ colors after $n-1$ operations.

• Case 2: The $n$-th operation does not introduce a new color (this occurs when a non-black ball is drawn).

  • For this to happen, there must have been exactly $m$ distinct colors after $n-1$ operations.
  • There are $n-1$ ways to perform this operation since it can involve any of the $n-1$ existing balls (which already represent the $m$ colors).
  • The number of favorable configurations for this scenario is the number of ways to have $m$ colors after $n-1$ operations, multiplied by $n-1$ (because any of the previous operations could have been the one that added the ball of an existing color).

The recurrence relation for the number of favorable configurations $C(n,m)$ is therefore:

$$C(n,m)=C(n-1,m-1)+(n-1)C(n-1,m)$$

Step 4: Initial Conditions

We define the initial conditions as follows:

• $C(1,1)=1$: There is exactly one way to introduce one color after the first operation.
• $C(n,m)=0$ if $m>n$: You cannot have more colors than operations.
• $C(n,m)=0$ if $m=0$ and $n>0$: You must have at least one color after any positive number of operations.

Step 5: Calculating the Probability

Now, we can calculate the probability $q_n(m)$ by dividing the number of favorable configurations $S(n,m)$ by the total number of outcomes $n!$:

$$q_n(m)=\frac{S(n,m)}{n!}$$

Where $S(n,m)$ is computed using the recursive formula:

$$S(n,m)=S(n-1,m-1)+(n-1)S(n-1,m)$$

This recursive relation allows us to calculate the exact number of configurations that result in exactly $m$ non-black colors after $n$ operations.

Question 5: Find the Minimum Number of Trials for Probability Exceeding 35% for at Least 3 Colors

We are tasked with finding the smallest number of trials $n$ such that the probability of having at least 3 non-black colors in the bag exceeds 35%.

Step 1: Understand the Probability Condition

The goal is to find the smallest $n$ such that:

$$P(\text{at least 3 colors after }n\text{ operations})>0.35$$

This can be expressed as:

$$q_n(3)+q_n(4)+\cdots +q_n(n)>0.35$$

However, since the total probability sums to 1, we can equivalently solve:

$$q_n(0)+q_n(1)+q_n(2)<0.65$$

Where:

• $q_n(0)$ is the probability of having 0 non-black colors (impossible for $n\ge 1$).
• $q_n(1)$ is the probability of having exactly 1 non-black color.
• $q_n(2)$ is the probability of having exactly 2 non-black colors.

Step 2: Calculate $q_n(0)$, $q_n(1)$, and $q_n(2)$ Recursively

The probability $q_n(m)$ is computed using the combinatorial approach discussed earlier:

$$q_n(m)=\frac{S(n,m)}{n!}$$

Where $C(n,m)$ is the number of favorable configurations, and it satisfies the recurrence relation:

$$S(n,m)=S(n-1,m-1)+(n-1)S(n-1,m)$$

With the initial conditions:

• $S(1,1)=1$ and $S(1,m)=0$ for $m>1$.
• $S(n,0)=0$ for all $n\ge 1$.

Step 3: Determine the Minimum $n$

We need to find the smallest $n$ such that:

$$q_n(0)+q_n(1)+q_n(2)<0.65$$

Step 4: Calculating $n$

We need to find the smallest $n$ such that:

$$q_n(3)+q_n(4)+\cdots +q_n(n)>0.35$$

Given the total probability must sum to 1, we can equivalently calculate:

$$q_n(0)+q_n(1)+q_n(2)<0.65$$

This means we need to compute the probabilities $q_n(0)$, $q_n(1)$, and $q_n(2)$ for increasing values of $n$ until their sum is less than 0.65. Let’s examine the results:

1. For $n=3$:

   • The sum $q_3(0)+q_3(1)+q_3(2)$ is approximately 0.833.
   • This is greater than 0.65, so the probability of having at least 3 colors is less than 0.35.

2. For $n=4$:

   • The sum $q_4(0)+q_4(1)+q_4(2)$ is approximately 0.708.
   • This is still greater than 0.65, so the probability of having at least 3 colors is still less than 0.35.
   • However, $q_4(3)+q_4(4)$ (the probability of having at least 3 colors) is approximately 0.292, which is less than 0.35.

3. For $n=5$:

   • The sum $q_4(0)+q_4(1)+q_4(2)$ is approximately 0.617.

Since this probability exceeds 0.35, $n=5$ is indeed the smallest value that satisfies the condition.

Question 6: Proof Using Mathematical Induction and the Recurrence Relation

We want to prove the following equality for all natural numbers $n$:

$$\sum _{m=1}^{n}m\cdot q_n(m)=1+\frac{1}{2}+\cdots +\frac{1}{n}$$

where $q_n(m)$ satisfies the recurrence relation derived from the problem:

$$q_n(m)=\frac{1}{n}{q}_{n-1}(m-1)+\frac{n-1}{n}{q}_{n-1}(m)$$

Step 1: Base Case $n=1$

For $n=1$:

• $q_1(1)=1$ since after the first operation, there is exactly one non-black color.
• There are no other possible outcomes, so $q_1(m)=0$ for all $m>1$.

Thus:

$$\sum _{m=1}^{1}m\cdot q_1(m)=1\cdot q_1(1)=1$$

And the right-hand side is also:

$$1=1$$

The base case holds true.

Step 2: Inductive Hypothesis

Assume that the statement holds for some $n=k$:

$$\sum _{m=1}^{k}m\cdot q_k(m)=1+\frac{1}{2}+\cdots +\frac{1}{k}$$

We need to prove that the statement holds for $n=k+1$:

$$\sum _{m=1}^{k+1}m\cdot q_{k+1}(m)=1+\frac{1}{2}+\cdots +\frac{1}{k+1}$$

Step 3: Prove for $n=k+1$

Using the recurrence relation for $q_{k+1}(m)$:

$$q_{k+1}(m)=\frac{1}{k+1}{q}_k(m-1)+\frac{k}{k+1}{q}_k(m)$$

Now, compute the sum:

$$\sum _{m=1}^{k+1}m\cdot q_{k+1}(m)=\sum _{m=1}^{k+1}m\left(\frac{1}{k+1}{q}_k(m-1)+\frac{k}{k+1}{q}_k(m)\right)$$

Distribute the sum:

$$\sum _{m=1}^{k+1}m\cdot q_{k+1}(m)=\frac{1}{k+1}\sum _{m=1}^{k+1}m\cdot q_k(m-1)+\frac{k}{k+1}\sum _{m=1}^{k+1}m\cdot q_k(m)$$

Step 4: Simplify the Sum

Let’s analyze each term separately:

1. First Term $\frac{1}{k+1}{\sum }_{m=1}^{k+1}m\cdot q_k(m-1)$:

   Notice that $q_k(m-1)$ shifts the index by 1. This can be rewritten as:

   $$\frac{1}{k+1}\sum _{m=1}^{k+1}m\cdot q_k(m-1)=\frac{1}{k+1}\sum _{m=0}^k(m+1)\cdot q_k(m)$$

   Simplifying this:

   $$=\frac{1}{k+1}\left(\sum _{m=1}^{k}m\cdot q_k(m)+\sum _{m=0}^k{q}_k(m)\right)$$

   Since ${\sum }_{m=0}^k{q}_k(m)=1$ (the total probability), this becomes:

   $$=\frac{1}{k+1}\left(\sum _{m=1}^{k}m\cdot q_k(m)+1\right)$$

2. Second Term $\frac{k}{k+1}{\sum }_{m=1}^{k+1}m\cdot q_k(m)$:

   This term simplifies directly as:

   $$\frac{k}{k+1}\sum _{m=1}^{k+1}m\cdot q_k(m)=\frac{k}{k+1}\sum _{m=1}^{k}m\cdot q_k(m)+(k+1)\cdot q_k(k+1)=\frac{k}{k+1}\sum _{m=1}^{k}m\cdot q_k(m)$$

   Since $q_k(k+1)=0$.

Step 5: Combine the Results

Combining the two terms:

$$\sum _{m=1}^{k+1}m\cdot q_{k+1}(m)=\frac{1}{k+1}\left(\sum _{m=1}^{k}m\cdot q_k(m)+1\right)+\frac{k}{k+1}\sum _{m=1}^{k}m\cdot q_k(m)$$

Factor out the common sum ${\sum }_{m=1}^{k}m\cdot q_k(m)$:

$$=\sum _{m=1}^{k}m\cdot q_k(m)\cdot \frac{1}{k+1}+\frac{k}{k+1}+\frac{1}{k+1}$$

Simplify:

$$=\sum _{m=1}^{k}m\cdot q_k(m)+\frac{1}{k+1}$$

By the inductive hypothesis:

$$\sum _{m=1}^{k}m\cdot q_k(m)=1+\frac{1}{2}+\cdots +\frac{1}{k}$$

Thus:

$$\sum _{m=1}^{k+1}m\cdot q_{k+1}(m)=1+\frac{1}{2}+\cdots +\frac{1}{k}+\frac{1}{k+1}$$

This completes the inductive step, proving that:

$$\sum _{m=1}^{n}m\cdot q_n(m)=1+\frac{1}{2}+\cdots +\frac{1}{n}$$

知识点

组合计数递推公式概率论Stirling数数学归纳法

难点思路

这道题目中最具挑战性的部分是第 4 问和第 6 问。第 4 问需要认识到问题与 Stirling 数的关系,这需要对组合数学有较深入的了解。第 6 问的证明需要巧妙地运用数学归纳法和概率的性质。

解题技巧和信息

1. 对于小规模的问题,穷举所有可能性是一个好方法。
2. 寻找递归关系可以帮助解决大规模的问题。
3. 熟悉常见的组合数学工具 (如 Stirling 数) 对解决此类问题很有帮助。
4. 在处理概率问题时,注意概率的加法规则和条件概率。
5. 数学归纳法是证明涉及自然数的命题的强大工具。

重点词汇

• probability 概率
• conditional probability 条件概率
• recursion relation 递归关系
• Stirling numbers of the second kind 第二类 Stirling 数
• mathematical induction 数学归纳法
• enumeration 枚举

参考资料

1. Feller, W. (1968). An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd Edition. Wiley. Chapter 2: Sample Spaces.
2. Graham, R. L., Knuth, D. E., & Patashnik, O. (1994). Concrete Mathematics: A Foundation for Computer Science. Addison-Wesley. Chapter 6: Special Numbers.