Quiz 9 Solution Sheet Math 1680 Ms. Stonebraker

1. A die is rolled some number of times, and you win $1 if it shows a number greater than 4 less than 15% of the time. Which is better? (2 pts)

20 rolls is better. The law of averages say the more rolls you do the closer your percentage is to the probability of that event occurring. In our example, the event is that a die’s face shows a number greater than 4. 5 and 6 are the 2 1 1 only faces greater than 4, thus there is a probability of or or 33 %. We 6 3 3 win a dollar if the percent it occurs is less than 15%, thus it is undesirable to have a percentage close to the probability of that event occurring which we 1 found to be 33 %. Thus we want fewer rolls, so that we are less likely to see 3 the probability that our event occurs.

2. As in (1), but you win the dollar if the percentage of faces showing a number greater than 4 is more than 20% of the time. Which is better? (2 pts)

200 rolls is better. We win a dollar if the percentage of faces showing a number greater than 4 is more than 20% of the time. The more rolls we do the 1 closer our percentage will be to 33 %, which is greater than 20%, and is thus 3 a desirable result.

3. As in (1), but you win the dollar if the percentage of faces showing a number 1 greater than 4 is exactly 33 % of the time. Which is better? (2 pts) 3

20 rolls is better. We win a dollar if the percentage of faces showing a number 1 greater than 4 is EXACTLY 33 % of the time. Despite the fact that the more 3 1 rolls we get the closer our percentage will be to 33 %, our actual percentage 3 observed is still likely to be a bit off from the exact amount. The more rolls you have the less likely you are to have an exact percentage. You can see this more clearly when you compare these using the binomial formula. See the following: 1 P(exactly 33 % of 20 rolls showing more than 4) 3 7 13 20 1   2  = P(exactly 7 rolls showing more than 4) =       .18  7  3  3 

1 P(exactly 33 % of 200 rolls showing more than 4) 3 67 133 200 1   2  = P(exactly 67 rolls showing more than 4) =       .06  67  3  3 

4. You are pumping gas at a local 7-11 gas station. You and your friend decide to play a game. You bet your friend a dollar that the last digit of the dollar amount is divisible by 3. If you are right, she owes you a dollar. If you are wrong, she gets to keep your dollar. Assume that each digit is equally likely to occur. Construct a box model for such a scenario. (2 pts)

Our sample space is all possible digits, which is 0 through 9. Thus there are 10 total tickets in our box. We win a dollar if the digit is divisible by 3, thus we win if we have a 0, 3, 6, or a 9. Thus of our ten there are four winning tickets that have a “ 1” on them and there are six tickets that are not winning tickets that have a “ -1” on them. Thus our box would look like below.

1 1 1 1 -1 -1 -1 -1 -1 -1

5. What is the expected value of the sum if you play the game in (4) ten different times? (3 pts)

The number of draws we make from this box is equal to the number of times we play the game. Thus N = 10.

4(1)  6(1) 4  6  2 1 EV =   or BOX 10 10 10 5 1 10 EV = EV  N  10   2 SUM BOX 5 5

Thus the expected value of the sum is for you to lose 4 dollars or gain –4 dollars. (This would not be a bright game on your part.)