Question 1 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 distribu-

tion function with a mean and standard deviation of P(X=x)=(6x)(436−x)(496)

x=0,1,2,3,4,5,6. E(X)=muX=3649=0.7347 SD(X)=sigmaX=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.

Billy's claim that "on average, he manages to have at least one matching number on his 6-49 ticket" in a year can be analyzed based on the given information. Let’s break this down:

### Key Information:
- The random variable \( X \) tracks the number of matches Billy experiences per week.
- The mean of the distribution \( \mu_X = 0.7347 \), which represents the expected number of matches in a single week.
- The standard deviation \( \sigma_X = 0.76 \), providing a measure of how spread out Billy’s results are around the mean.
- Billy buys one lottery ticket per week for 52 weeks in a year.

### Understanding Billy’s Claim:
Billy's claim implies that, **on average**, he expects to have at least one match per week over the course of a year. This can be interpreted probabilistically as Billy expecting that the number of weeks where he has at least one match will be significant.

#### Mean Number of Matches in a Year:
Over 52 weeks, the **expected total number of matches** Billy will experience can be found by multiplying the weekly mean by 52:
\[
E(\text{Total Matches in a Year}) = 52 \times \mu_X = 52 \times 0.7347 = 38.2044
\]
This means that, on average, Billy expects to have **around 38 matches** across 52 weeks.

#### Probability of No Matches in a Week:
We can now consider the probability of having **no matches** on a single ticket in a given week, which is:
\[
P(X = 0)
\]
While this probability is not given directly, it can be inferred from the distribution of \( X \). Since \( \mu_X = 0.7347 \), the probability of zero matches is likely quite high. Thus, Billy might have many weeks where he doesn't match any number.

#### Expected Number of Weeks with No Matches:
Given the weekly probability distribution, we expect some weeks with no matches, but we also expect many weeks where Billy has at least one match. To estimate this, we could use the **binomial distribution**:
\[
P(\text{at least one match in 52 weeks}) = 1 - P(\text{no matches in all 52 weeks})
\]
However, the critical part of Billy’s claim is about **average behavior** rather than individual weekly outcomes. The total number of matches suggests that he will have at least one match on several weeks.

### Conclusion on Billy’s Claim:
Billy’s claim that **on average** he has at least one matching number per week is **reasonable** based on the expected value of 0.7347 matches per week. However, this does not imply that every week will have a match, as the expected value is less than 1. In fact, there will likely be several weeks with no matches at all, but in some weeks, he may have multiple matches, making the claim reasonable from a long-term average perspective.

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Do you want more details about any step? Feel free to ask!

Here are five related questions to expand on this topic:
1. How can we calculate the exact probability of no matches in a given week?
2. What does the probability distribution \( P(X = x) \) look like for the number of matches in a week?
3. How does the binomial distribution apply to this lottery scenario?
4. What is the impact of the standard deviation \( \sigma_X = 0.76 \) on the number of matches per week?
5. How can we estimate the number of weeks where Billy has no matches at all?

**Tip:** Understanding expected value gives you a powerful tool in probability for predicting long-term outcomes based on average behaviors!

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 P(X=x)=(6x)(436−x)(496) x=0,1,2,3,4,5,6. E(X)=muX=3649=0.7347 SD(X)=sigmaX=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.

Explore the probability of getting matching numbers in a 6-49 lottery ticket over a year. Understand concepts such as expected value and binomial distribution to evaluate the likelihood of Billy's claim. Suitable for students in Grades 11-12.

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