Answer to Question #107790 in Statistics and Probability for This website so good

Question

Which of the game would work with the Binomial Distribution?
Marble Draw : 15 marbles are in the bag , 7 of them are red. Draw 4 marbles without replacement , at a cost of 10 points. Rewards are 0 red : 30 points 1 red : 10 points 2 red : 0 points 3 red : 20 points 4 red : 50 points
Word Scramble : Costs 10 points. Total 4 letters . Depending on how many letters are in the correct spot , a prize is awarded
Only 1st letter in the correct position : Win 0
Only 2nd letter : win 5
Only 3rd letter : win 10
Only 4th letter : win 15
2 letters including the first : win 20
2 letters not including the first : win 25
All 4 letters : win 40
a) Create the probability distribution table
b) Calculate the expected value and compare to the 10 points it costs to play the game

Answer 1

Marble Draw.

Let random variable $X$ is the number of points earned by the player.

$p=\frac{7}{15}\text{ --- probability that the player will draw a red marble}.\\ q=\frac{8}{15}\text{ --- probability that the player will not draw a red marble}.$

The player draws 4 marbles without replacement at a cost of 10 points. So he can earn -10, 0, 10, 20 or 40 points.

$a) P\{X=-10\}=C_{4}^{2}(\frac{7}{15})^2(\frac{8}{15})^2\approx 0.372.\\ P\{X=0\}=C_{4}^{1}(\frac{8}{15})^3(\frac{7}{15})\approx 0.283.\\ P\{X=10\}=C_{4}^{3}(\frac{7}{15})^3(\frac{8}{15})\approx 0.217.\\ P\{X=20\}=(\frac{8}{15})^4\approx 0.081.\\ P\{X=40\}=(\frac{7}{15})^4\approx 0.047.\\ b) MX\approx 4.798 \text{ (sum of products of probabilities and points)}.$

We see that on the average the player will lose $\approx 10-4.798=5.202$ points.

Word Scramble.

Let random variable $Y$ is the number of points earned by the player.

The player can earn -10, -5, 0, 5, 10, 15 or 30 points.

$P\{Y=-10\}=\frac{2}{4!}=\frac{1}{12}\\ P\{Y=-5\}=\frac{2}{4!}=\frac{1}{12}\\ P\{Y=0\}=\frac{2}{4!}=\frac{1}{12}\\ P\{Y=5\}=\frac{2}{4!}=\frac{1}{12}\\ P\{Y=10\}=\frac{3}{4!}=\frac{1}{8}\\ P\{Y=15\}=\frac{12}{4!}=\frac{1}{2}\\ P\{Y=30\}=\frac{1}{4!}=\frac{1}{24}\\ b)MY\approx 9.166 \text{ (sum of products of probabilities and points)}.$

We see that on the average the player will lose $10-9.166=0.834$ points.

The first game works with the binomial distribution.