give me r code for

Billy purchases one 6-49 lottery ticket every week and

keeps track of the number of “matches”” he has on each of his

tickets.  To be clear, a “match” will occur when a number on his

ticket matches a number that appears in the winning combination.

A random variable X that keeps track of the number of matching

numbers Billy experiences per week has the probability distribution

function with a mean and standard deviation of

푃 (푋 = 푥) =

(

6

푥

)(

43

6−푥

)

(

49

6

)

x = 0, 1, 2, 3, 4, 5, 6

퐸[푋] = 휇

푥

=

36

49

= 0.7347

푆퐷(푋) = 휎

푥

= 0.75998 ≈ 0.76

Billy claims that in a year (52 weeks), on average, he manages

to have at least one matching number on his 6-49 ticket.  What

do you think about Billy’s claim?  Provide a brief commentary

about Billy’s claim using your current knowledge of statistics and

probability theory.

To analyze Billy's claim using R and the provided probability distribution, we need to calculate the expected number of weeks with at least one matching number over a year. We'll also simulate the lottery scenario to see how likely it is that Billy gets at least one match over a year (52 weeks).

The probability that Billy gets zero matches is crucial for this. We will:

1. Calculate \( P(X = 0) \), which represents the probability of getting zero matches.
2. Use this to determine the probability of getting at least one match in any given week.
3. Multiply that by the number of weeks in a year (52) to estimate the expected number of weeks Billy gets at least one match.
4. Simulate this scenario using R to verify Billy's claim.

Here’s the R code to do that:

```r
# Set parameters for the lottery distribution
x <- 0:6  # possible number of matches (0 to 6)
n <- 6    # number of numbers on the ticket
total_numbers <- 49  # total numbers in the lottery
remaining_numbers <- total_numbers - n  # remaining numbers after ticket

# Calculate the probability distribution P(X = x)
P_X <- choose(n, x) * choose(remaining_numbers, n - x) / choose(total_numbers, n)

# Display the probability distribution
data.frame(x = x, P_X = P_X)

# Mean (mu) and standard deviation (sigma) as given
mu_x <- 36/49
sigma_x <- 0.76

# Probability of getting at least one match
P_at_least_one_match <- 1 - P_X[1]  # P(X >= 1) = 1 - P(X = 0)

# Expected number of weeks with at least one match in 52 weeks
expected_weeks_with_match <- P_at_least_one_match * 52
cat("Expected number of weeks with at least one match in a year:", expected_weeks_with_match, "\n")

# Simulate the lottery for 52 weeks, repeating 10,000 times
set.seed(123)  # for reproducibility
simulations <- replicate(10000, sum(rbinom(52, size=1, prob=P_at_least_one_match)))

# Calculate the average number of weeks with at least one match
avg_simulated_weeks <- mean(simulations)
cat("Average number of weeks with at least one match from simulation:", avg_simulated_weeks, "\n")

# Compare Billy's claim
if (avg_simulated_weeks >= 52 * mu_x) {
  cat("Billy's claim is reasonable.\n")
} else {
  cat("Billy's claim might be overly optimistic.\n")
}
```

### Explanation:
1. **Probability Calculation**: We calculate the probabilities of getting 0 to 6 matches based on combinatorial probability.
2. **At Least One Match**: The probability of getting at least one match is \(1 - P(X = 0)\).
3. **Expected Value**: We compute the expected number of weeks Billy gets at least one match by multiplying by 52 weeks.
4. **Simulation**: We simulate 10,000 instances of Billy’s lottery experience over a year to check if his claim holds up in practice.

### Commentary:
Billy claims he manages to have at least one matching number on his ticket, on average, over 52 weeks. The probability of getting zero matches is non-negligible, but the expected number of weeks with at least one match can be approximated based on the probability distribution. If simulations and calculations show that this number is close to or greater than 1 per week, Billy’s claim may be reasonable. Otherwise, his claim could be overly optimistic.

---

Do you want more details on any step or have further questions? Here are 5 related questions:

1. How do you interpret the binomial distribution used in this case?
2. What are the assumptions behind this probability calculation?
3. How can we adjust this model for different lottery types (e.g., different number pools)?
4. What impact does increasing the number of weeks have on the results?
5. How could we use other methods, like a Poisson approximation, for large numbers?

**Tip**: Always check assumptions in probabilistic models to ensure they match real-world scenarios!

Billy purchases one 6-49 lottery ticket every week and keeps track of the number of “matches” he has on each of his tickets. A random variable X that keeps track of the number of matching numbers per week has a probability distribution with a mean and standard deviation. Billy claims that in a year (52 weeks), he manages to have at least one matching number. What do you think about Billy’s claim? Provide commentary using statistics and probability theory.

Analyze Billy's claim of having at least one match in a year with a 6-49 lottery ticket using probability theory. Learn how to calculate the probability distribution, expected values, and simulate scenarios with R code. Suitable for Grades 11-12 students.

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