Questions: Use the fact that the mean of a geometric distribution is μ=1/p and the variance is σ^2=q/p^2. A daily number lottery chooses two balls numbered 0 to 9. The probability of winning the lottery is 1/100. Let x be the number of times you play the lottery before winning the first time. (a) Find the mean, variance, and standard deviation. (b) How many times would you expect to have to play the lottery before winning? It costs 1 to play and winners are paid 900. Would you expect to make or lose money playing this lottery? Explain. (a) The mean is (Type an integer or a decimal.)

Solution Steps

The mean or expected number of plays before winning is given by \(\mu = \frac{1}{p}\), where \(p = 0.01\). So, \(\mu = \frac{1}{0.01} = 100\) plays.

The variance in the number of plays before winning is given by \(\sigma^2 = \frac{q}{p^2}\), where \(q = 1 - p = 0.99\). So, \(\sigma^2 = \frac{0.99}{0.01^2} = 9900\).

The standard deviation is the square root of the variance, \(\sigma = \sqrt{\sigma^2}\). So, \(\sigma = \sqrt{9900} = 99.5\) plays.

This is simply the mean, \(\mu = 100\) plays.

Calculate the expected net outcome by subtracting the expected total cost of playing from the expected total payout. The expected total cost is the mean number of plays before winning times the cost per play (\(\mu \times \$C\)), and the expected total payout is the payout for winning once (\$P\). The expected net outcome is \(\$P - (\mu \times \$C) = \$900 - (100 \times \$1) = \$800\).

Since the expected net outcome is positive, one would expect to make money by playing the lottery.

Final Answer: