https://r2.erweima.ai/mathbot/image/1726516449023qovf6tex.jpg

The problem in the image revolves around a probabilistic model involving two types of games, A and B, with probabilities tied to whether the next game will be of the same type or different. I'll go step by step based on the questions.

### 1. a. Completing the probability tree

The tree diagram shows two branches from the state where the player finishes a game of type A or type B. Here’s how the tree should look:

- From \( A_n \), the probability of playing \( A_{n+1} \) is 0.8, and the probability of playing \( B_{n+1} \) is 0.2.
- From \( B_n \), the probability of playing \( A_{n+1} \) is 0.3, and the probability of playing \( B_{n+1} \) is 0.7.

Thus, the complete tree will have the following transitions:
- From \( A_n \) to \( A_{n+1} \) with probability 0.8, and from \( A_n \) to \( B_{n+1} \) with probability 0.2.
- From \( B_n \) to \( A_{n+1} \) with probability 0.3, and from \( B_n \) to \( B_{n+1} \) with probability 0.7.

### 1. b. Proving the recurrence relation for \( a_{n+1} \)

To prove that for any natural number \( n \geq 1 \), \( a_{n+1} = 0.5a_n + 0.3 \):

The probability \( a_{n+1} \) represents the chance that the \( (n+1) \)-th game is of type A. To find this probability, we account for the two cases:

1. **If the nth game is type A (with probability \( a_n \)):**
   - The next game is type A with probability 0.8.
   
2. **If the nth game is type B (with probability \( b_n = 1 - a_n \)):**
   - The next game is type A with probability 0.3.

Thus, the total probability of the \( (n+1) \)-th game being type A is the sum of these two cases:

\[
a_{n+1} = 0.8a_n + 0.3(1 - a_n)
\]

Simplifying:

\[
a_{n+1} = 0.8a_n + 0.3 - 0.3a_n = (0.8 - 0.3)a_n + 0.3 = 0.5a_n + 0.3
\]

This confirms the recurrence relation.

### 2. Proving monotonicity and boundedness of \( a_n \)

We want to show that the sequence \( a_n \) is increasing and bounded above by 0.6.

- **Monotonicity:**
  Since \( a_{n+1} = 0.5a_n + 0.3 \), we can see that if \( a_n < 0.6 \), then:

  \[
  a_{n+1} = 0.5a_n + 0.3 > a_n
  \]

  Hence, the sequence is increasing as long as \( a_n < 0.6 \).

- **Boundedness:**
  The upper bound comes from observing the recurrence formula. Suppose \( a_n = 0.6 \), then:

  \[
  a_{n+1} = 0.5(0.6) + 0.3 = 0.6
  \]

  This shows that 0.6 is a fixed point of the recurrence relation. Moreover, if \( a_n < 0.6 \), the sequence increases, but it never exceeds 0.6. Therefore, the sequence is bounded above by 0.6.

### 3. Determining the limit of \( a_n \)

From the recurrence relation \( a_{n+1} = 0.5a_n + 0.3 \), as \( n \to \infty \), the sequence converges to a limit \( L \). In the limit, \( a_{n+1} = a_n = L \), so:

\[
L = 0.5L + 0.3
\]

Solving for \( L \):

\[
L - 0.5L = 0.3 \quad \Rightarrow \quad 0.5L = 0.3 \quad \Rightarrow \quad L = \frac{0.3}{0.5} = 0.6
\]

Thus, the sequence converges to 0.6.

---

Would you like a deeper explanation on any step? Here are five related questions you might explore:

1. How does the concept of recurrence relations apply to stochastic processes?
2. What is the significance of monotonic sequences in probability theory?
3. Can we extend this type of analysis to a multi-type game system with more than two states?
4. How does the fixed point of the recurrence relation represent long-term behavior in probabilistic models?
5. How would the recurrence relation change if the transition probabilities were different?

**Tip:** Recurrence relations are often used to model systems that evolve step-by-step, where each state depends on previous states—these are common in both mathematics and computer science.

Stream est une plateforme informatique qui propose deux types de jeux vidéo : un jeu de type A et un jeu de type B. Lorsqu'un joueur achève une partie, il lui propose une nouvelle partie selon les probabilités suivantes :
- Si la partie était de type A, la prochaine partie sera de type A avec une probabilité de 0,8 et de type B avec une probabilité de 0,2.
- Si la partie était de type B, la prochaine partie sera de type A avec une probabilité de 0,3 et de type B avec une probabilité de 0,7.
Montrer que la suite a_n qui représente la probabilité que la n-ième partie soit de type A suit la relation de récurrence : a_{n+1} = 0.5a_n + 0.3. Ensuite, montrer que cette suite est croissante et bornée, et enfin, déterminer la limite de cette suite.

Explore the recurrence relation a_{n+1} = 0.5a_n + 0.3 representing the probability of selecting game type A in the n+1-th round. Learn how to prove the monotonicity and boundedness of this sequence and determine its limit as n approaches infinity.

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