Daily Lottery: Three Balls Numbered 0 To 9 ✓ Solved

A daily number lottery chooses three balls numbered 0 to 9.

A daily number lottery chooses three balls numbered 0 to 9. The probability of winning the lottery is 1/1000. Let x be the number of times you play the lottery before winning the first time. (a) Find the mean, variance, and standard deviation. Interpret the results. (b) how many times would you expect to have to play the lottery before winning? Assume that it costs $1 to play and winners are paid $500. Would you expect to make or lose money playing this lottery? Explain.

Paper For Above Instructions

The daily number lottery presents an interesting scenario in probability and expectation. In this case, the lottery allows players to choose three balls numbered from 0 to 9, creating a total of 1000 (10 x 10 x 10) different combinations, leading to a winning probability of 1/1000 for each play. This assignment requires us to delve into the mean, variance, and standard deviation of the number of plays required to win, along with the financial implications of participating in this lottery.

Mean, Variance, and Standard Deviation

In this context, the random variable x, which represents the number of times a player must play before winning, follows a geometric distribution. The properties of a geometric distribution are immensely useful for solving this part of the problem.

The mean (expected value) for a geometric distribution can be calculated using the formula:

E(X) = 1/p

In our scenario, p is the probability of winning, which is 1/1000. Therefore:

E(X) = 1/(1/1000) = 1000

This means that on average, a player can expect to play the lottery 1000 times before winning.

The variance of x in a geometric distribution is given by the formula:

Var(X) = (1-p)/p²

Applying our values:

Var(X) = (1-(1/1000))/(1/1000)²

Var(X) = (999/1000) / (1/1,000,000) = 999,000

This indicates a large spread of the number of lottery plays before a win occurs.

The standard deviation is the square root of the variance:

SD(X) = √Var(X) = √999,000 ≈ 999.5

Interpretation of Results

From these calculations, we can interpret that while the mean suggests a player will have to engage in 1000 attempts to win, the high variance and standard deviation imply significant variability in the number of plays required before achieving a victory. In simpler terms, while 1000 plays is the average, it could take considerably fewer or greater plays due to the nature of chance inherent in lotteries. The risk involved is substantial, given that most players will find themselves playing many times without hitting a win.

Financial Expectations from the Lottery

Next, we need to examine the financial aspects of playing the lottery. Each lottery ticket costs $1, and a win yields a payout of $500. However, to analyze whether a player would make or lose money, we can evaluate the expected value from one play of the lottery:

Let W be the winnings and L be the losses. Using the expected value formula, we have:

EV = (p × W) + ((1-p) × L)

Substituting the relevant values:

EV = (1/1000 × 500) + (999/1000 × (-1))

EV = (0.5) - (0.999) = -0.499

This calculation suggests that on average, a player loses $0.50 per ticket played. With an average expectation of 1000 plays before winning, the total expected cost before hitting a jackpot equates to:

Total spent = 1000 × $1 = $1000

Therefore, a player can expect to spend $1000 before winning $500, leading to a net loss of $500 on average.

Conclusion

In summary, the daily number lottery as described entails a probabilistic gamble with a very low chance of winning. The statistical outcomes indicate that, while a player can expect to need about 1000 plays to win, the associated variance and standard deviation suggest a range of potential outcomes. Financially, players should expect to incur losses due to the unfavorable expected value of participation, corroborated by the result that a net loss of $500 is the most probable outcome following the law of large numbers.

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