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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.

In un gioco d'azzardo si vincono due Euro in caso di successo, e si perde un Euro in caso di insuccesso. La probabilità di successo è p = 0.4. (a) Sia X il numero di successi in n ripetizioni del gioco. Qual è la distribuzione di X? (b) Sia Y il numero (netto) di Euro vinti in n ripetizioni del gioco. Dopo aver determinato una relazione che leghi X e Y, si determinino media e varianza di Y.

This problem involves determining the distribution, mean, and variance of a number of successes in a gambling game using the binomial distribution. It covers key concepts in probability and statistics, including expectation and variance.

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