An exciting computer game is released. Fifty-five percent of players complete all the levels. Thirty percent of them will then buy an advanced version of the game. Among 20 users, what is the expected number of people who will buy the advanced version? What is the probability that at least two people will buy it?

We are given the following probabilities:

- 55% of players complete all the levels.
- 30% of those who complete the game will buy the advanced version.

Let's break the problem into two steps:

### Step 1: Expected number of people who will buy the advanced version.

We are interested in the expected number of people who will buy the advanced version among 20 users.

- The probability that a player buys the advanced version is the product of two probabilities:  
  \( P(\text{buy advanced version}) = 0.55 \times 0.30 = 0.165 \)

This means that the probability of any given player buying the advanced version is 16.5%.

Using the **Binomial Expectation Formula**:

\[
E(X) = n \times p
\]

Where:
- \( n = 20 \) (the total number of users),
- \( p = 0.165 \) (the probability that a player buys the advanced version).

Thus, the expected number \( E(X) \) is:

\[
E(X) = 20 \times 0.165 = 3.3
\]

So, the expected number of people who will buy the advanced version is **3.3**.

### Step 2: Probability that at least two people will buy the advanced version.

This is a **binomial probability problem**, where:
- \( n = 20 \),
- \( p = 0.165 \),
- We need to find \( P(X \geq 2) \), the probability that at least 2 people buy the advanced version.

The complement is \( P(X < 2) \), meaning that 0 or 1 person buys the advanced version. We calculate \( P(X = 0) \) and \( P(X = 1) \), and subtract this from 1.

The binomial probability formula is:

\[
P(X = k) = \binom{n}{k} \times p^k \times (1 - p)^{n - k}
\]

#### Probability for \( X = 0 \):

\[
P(X = 0) = \binom{20}{0} \times 0.165^0 \times 0.835^{20} = 1 \times 1 \times 0.835^{20}
\]

Let me calculate this value.The probabilities are as follows:

- \( P(X = 0) = 0.0271 \)
- \( P(X = 1) = 0.1073 \)
- \( P(X < 2) = P(X = 0) + P(X = 1) = 0.1344 \)

Thus, the probability that **at least 2** people will buy the advanced version is:

\[
P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1344 = 0.8656
\]

So, the probability that at least two people will buy the advanced version is approximately **0.866** or **86.6%**.

### Summary:
1. The expected number of people who will buy the advanced version is **3.3**.
2. The probability that at least two people will buy the advanced version is **86.6%**.

Would you like more details, or do you have any questions? Here are some related questions:

1. What is the probability that exactly three people will buy the advanced version?
2. How does the probability change if 30 users are considered instead of 20?
3. What is the standard deviation of the number of people who buy the advanced version?
4. How do we calculate the probability for more than five users buying the advanced version?
5. What if the completion rate of the game changes to 60%? How would the results vary?

**Tip**: Understanding binomial distributions can help analyze many real-world scenarios involving probability and success/failure outcomes.

Discover how to calculate the expected number of users who will buy an advanced game version and the probability that at least two users will do so. This problem covers probability concepts, expected value calculations, and binomial distributions, suitable for students in Grades 10-12.

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