Illinois State University Mathematics Department

 MAT 312: Probability and Statistics for Middle School Teachers Dr. Roger Day (day@ilstu.edu)

### Possible Solutions: Assignment #4

Assignment due on Tuesday, 7/6/04

1. If Sanchez is scheduled to play Novatna in the next tournament, what is the probability Sanchez will win?

P(Sanchez wins in Sanchez-vs-Novatna match) = (# of wins by Sanchez vs Novatna) / (total matches Sanchez vs Novatna) = 4/12

2. Which player-to-player match-up indicates the least probable chance of winning? Who is the player with the least probable chance and what is the probability she will win?

Here are the possible winners of the six match-ups shown in the table, together with the probability for winning.

• P(Graf wins in Graf-vs-Sabatini match) = (# of wins by Graf vs Sabatini ) / (total matches Graf vs Sabatini ) = 27/38 ---> P(Sabatini wins here) = 11/38
• P(Sanchez wins in Sanchez-vs-Novatna match) = (# of wins by Sanchez vs Novatna) / (total matches Sanchez vs Novatna) = 4/12 ---> P(Novatna wins here) = 8/12
• P(Sabatini wins in Sabatini-vs-Sanchez match) = (# of wins by Sabatini vs Sanchez ) / (total matches Sabatini vs Sanchez ) = 10/22 ---> P(Sanchez wins here) = 12/22
• P(Novatna wins in Novatna-vs-Graf match) = (# of wins by Novatna vs Graf ) / (total matches Novatna vs Graf ) = 3/25 ---> P(Graf wins here) = 22/25
• P(Graf wins in Graf-vs-Sanchez match) = (# of wins by Graf vs Sanchez ) / (total matches Graf vs Sanchez ) = 24/32 ---> P(Sanchez wins here) = 8/32
• P(Sabatini wins in Sabatini-vs-Novatna match) = (# of wins by Sabatini vs Novatna) / (total matches Sabatini vs Novatna) = 10/13 ---> P(Novatna wins here) = 3/13

The smallest fraction here is 3/25 for the probability that Novatna will win in a match against Graf.

B. All automobile license plates in Liberia have six characters. The characters can be letters of the alphabet (A, B, . . . , Z) or single-digit numbers (0, 1, . . . , 9).

3. If a license plate has four letters followed by two digits, with repetition allowed, how many different license plates are possible?

26*26*26*26*10*10

Choose from among 26 letters for each of the first four positions. Choose from among 10 digits for each of the two remaining spots. We use mutiplication because we are matching up possibilities.

4. If a license plate again has four letters followed by two digits, this time with repetition not allowed, how many different license plates are possible?

26*25*24*23*10*9

Choose from among 26 letters for the first position, from among 25 letters for the second position, among 24 letters for the third position, and among 23 letters for the fourth position. Now, choose from among 10 digits for the fifth position and among 9 digits for the sixth position. We use mutiplication because we are matching up possibilities.

5. If a license plate has three letters followed by three digits, with no repetitions allowed and no use of either the letter O or the digit 0, how many different license plates are possible?

P(25,3)*P(9,3)

Throwing out the letter O, we have 25 letters from which to arrange three letters in the first three positions. We have 9 digits from which to arrange three digits in the next three positions, given that the digit 0 cannot be used.

C. From a class of 20 students, a committee of 3 is to be formed to organize a fundraiser.

6. If all 20 students are eligible for committee membership, how many different committees could be formed?

C(20,3)

Select three students from among the 20 available, with no need to arrange them.

7. If there are 12 women and 8 men in the class, and the committee requires at least one member of each gender, how many different committees could be formed?

C(12,1)*C(8,2)+C(12,2)*C(8,1)

Two cases to consider:

Case #1: One woman and two men on committee: Choose one woman from among 12 women and match up with two men chosen from among those 8 available men.

Case #2: Two women and one man on committee: Choose two women from among 12 women and match up with one man chosen from among those 8 available men.

Now add the result of each case to get desired result.

D. For each situation below, determine whether the description represents a discrete random variable (DRV), a continuous random variable (CRV), or neither. Explain your choice.

8. The number of tacks that land head flat when 20 tacks are tossed on to a flat surface. DRV: We can count the number ranging from 0 to 20 tacks. No partial tacks can be counted or can otherwise result.

9. The length of time the department office must wait until its copy machine is repaired. CRV: We can measure the time length. Partial units of time can be measured or stated.

10. The height of students in a suburban elementary school. CRV: We can measure the heights. Partial units of height can be measured or stated.

11. The number of times an employee is late during each monthly pay period. DRV: We can count the number ranging from 0 onward No partial times can be counted or ocan therwise occur.

12. The number of blue tokens in a bag that contains 40 blue tokens. Neither: There is nothing that varies.

E. Many states have their own lotteries, games of chance where people buy tickets and select numbers and check whether their selected numbers match the numbers selected at random by the state. One of the games in the Big Mountain lottery is a 3-digit game, where players spend \$1 to select a three-digit number (repeats allowed).

13. A wager of \$1 buys a single play, composed of one three-digit combination. How many different three-digit combinations are possible in this version of the lottery? Note that the selection 3 2 2 is different from the selection 2 3 2.

10*10*10

Assuming all digits 0 through 9 are available, with repetition.

14. For this version of the lottery, what is the probability of winning on a single \$1 play?

There is exactly one way to win on a \$1 play, so campared to 1000 (10*10*10) possible outcomes, the probability of winning on a single \$1 play is 1/1000 = 0.001.

15. Consider the possible lottery winnings as a random variable x, knowing that on a successful \$1 play, a winner gets \$500.

(i) Show the distribution of the random variable x.

 Possible Winnings (x) \$500 \$0 P(x) 1/1000 999/1000

(ii) Determine the expected value of x.

E(x) = (1/1000)*\$500 + (999/1000)*\$0 = \$1/2 = \$0.50

(iii) Is this version of the lottery a fair game? Explain.

No. A fair game was expected winnings equal to cost of playing. In this game, we play \$1 per turn, but, over the long haul, expect to win only \$0.50 per turn (expected value). Therefore, over the long haul, we can expect to lose \$0.50 per play.

F. The random variable v has the following discrete probability distribution:

 v 0 1 2 3 4 p(v) 0.2 0.1 0.4 0.1 0.2

16. Determine these values:

(i) P(v = 2) = 0.4

(ii) P(v is less than or equal to 3) = P(w = 0) + P(w = 1) + P(w = 2) + P(w = 3) = 0.2 + 0.1 + 0.4 + 0.1 = 0.8

(iii) P(w > 2) = P(w = 3) + P(w = 4) = 0.1 + 0.2 = 0.3

17. Determine the expected value of v.

E(x) = (0.2)*0 + (0.1)*1 + (0.4)*2 + (0.1)*3 + (0.2)*4 = 2.0

18. Graph p(v), the discrete probability distribution.

19. Identify a random variable and then create a table to show the discrete probability distribution for the outcomes of this situation:

Two fair dice are rolled and the sum of the values on the face-up sides is noted.

Let s be the random variable representing the desired sum. Here is a table showing possible values of s and the associated probabilities:

 s 2 3 4 5 6 7 8 9 10 11 12 p(s) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

BONUS! NOT A REQUIRED PROBLEM!

Suppose the four players listed with questions 1 and 2 are the only entries in a single-elimination championship tournament. Here are the pairings for the first round matches:

Match #1: Sanchez vs Graf

Match #2: Novatna vs Sabatini

The winners of these two matches play a third match for the championship. For each player, determine her probability of winning the tournament.