The exercise provided involves a game of chance where the player wins 1 Euro for each success and loses 1 Euro for each failure. The probability of success is $p = 0.4$.

The problem consists of two parts:

(a) $X$ is the number of successes in $n$ repetitions of the game. What is the distribution of $X$?

(b) $Y$ is the (net) number of Euros won in $n$ repetitions of the game. After determining a relation between $X$ and $Y$, determine the mean and variance of $Y$.

Solution:

(a) Distribution of $X$:

$X$ is the number of successes in $n$ repetitions of the game. Since each repetition is independent and has only two outcomes (success or failure), $X$ follows a Binomial distribution:

$X \sim \text{Binomial}(n, p)$

where $n$ is the number of repetitions (trials), and $p = 0.4$ is the probability of success in each trial.

(b) Distribution of $Y$:

Let $Y$ be the net number of Euros won after $n$ repetitions of the game. Since the player wins 1 Euro for each success and loses 1 Euro for each failure:

$Y = X - (n - X) = 2X - n$

Here, $X$ is the number of successes, so $n - X$ is the number of failures.

Mean of $Y$:

Using the expectation of $X$ (denoted as $\mathbb{E}[X]$):

[ \mathbb{E}[Y] = \mathbb{E}[2X - n] = 2\mathbb{E}[X] - n ]

Since $X$ follows a binomial distribution:

$\mathbb{E}[X] = np$

Thus:

$\mathbb{E}[Y] = 2(np) - n = 2n(0.4) - n = 0.8n - n = -0.2n$

Variance of $Y$:

The variance of $Y$ (denoted as $\text{Var}(Y)$) is:

$\text{Var}(Y) = \text{Var}(2X - n) = 4\text{Var}(X)$

Since $X$ follows a binomial distribution:

$\text{Var}(X) = np(1-p)$

Thus:

$\text{Var}(Y) = 4(np(1-p)) = 4n(0.4)(0.6) = 0.96n$

Summary:

• $X \sim \text{Binomial}(n, 0.4)$
• Mean of $Y$ is $\mathbb{E}[Y] = -0.2n$
• Variance of $Y$ is $\text{Var}(Y) = 0.96n$

Would you like further details or have any questions?

Here are five related questions to consider:

1. How would the mean and variance of $Y$ change if the probability of success $p$ were different?
2. How would you interpret the negative expected value of $Y$?
3. What is the significance of the variance in this context?
4. How would the analysis change if the game were repeated infinitely?
5. Could you simulate the game for a fixed $n$ to observe the distribution of $X$ and $Y$?

Tip: Understanding the relationship between the binomial distribution and its parameters is crucial for solving problems involving repeated independent trials, such as this one.