1. MAKE A MODEL AND CONDUCT A SIMULATION We design a model and simulate solutions to problems where an exact answer can only be estimated and probability is involved or when we do not have the math knowledge needed to solve the problem. Simulation is based on the premise that the more trials we conduct, the closer we will get to the actual solution. It involves carefully designing a model that represents the situation, conducting a series of trials using that model, and averaging the results.

2. SAMPLE PROBLEM: MRS. CHUNG’S CLASS HAS 30 STUDENTS IN IT, WITH 13 BOYS AND 17 GIRLS. IF CHARLES VOLUNTEERS TO SUPERVISE HIS SON AND FOUR OTHER RANDOMLY SELECTED STUDENTS ON A FIELD TRIP, HOW MANY OTHER BOYS CAN HE EXPECT TO SUPERVISE? SOLUTION: FIRST, WE WILL DESIGN OUR MODEL. WE NEED TO REPRESENT THE OTHER 29 STUDENTS IN THE CLASS (12 BOYS AND 17 GIRLS). TO DO THIS, WE WILL USE A DECK OF PLAYING CARDS WHERE WE WILL TAKE 12 BLACK CARDS (FOR BOYS) AND 17 RED CARDS (FOR GIRLS). FOR EACH TRIAL, WE WILL SHUFFLE THE CARDS AND THEN PICK OUT 4 RANDOM CARDS. WE WILL MARK ‘B’ FOR EACH BLACK CARD AND ‘R’ FOR EACH RED CARD. THEN WE WILL TOTAL THE NUMBER OF B’S WE HAVE. AFTER ALL THE TRIALS HAVE BEEN COMPLETED, WE WILL AVERAGE THE NUMBER OF ‘B’ CARDS. THIS WILL ESTIMATE HOW MANY OTHER BOYS CHARLES CAN EXPECT TO SUPERVISE. THE TRIALS AND ESTIMATE ARE DISPLAYED ON THE NEXT SLIDE.

3. EACH TIME A TRIAL WAS CONDUCTED, ALL THE CARDS WERE PLACED BACK IN THE DECK, AND IT WAS SHUFFLED AT LEAST 7 TIMES TO ENSURE A RANDOM SAMPLE. Trial #1BRBRB = 2 Trial #2RBBBB = 3 Trial #3BRRBB = 2 Trial #4RRRRB = 0 Trial #5RBBRB = 2 Trial #6RRBBB = 2 Trial #7RRRRB = 0 Trial #8RRRBB = 1 Trial #9BRRBB = 2 Trial #10 RRBRB = 1 Average number of B’s for the 10 trials (sum/10) 15/10 = 1.5 AFTER CONDUCTING 10 TRIALS, I OBTAINED AN AVERAGE OF 1.5. IF I ROUND THIS UP, I GET 2. THIS ALSO WORKS WITH THE TRIAL OUTCOMES, SINCE 5 OUT OF 10 OF THEM PRODUCED 2. SO, MY ESTIMATE FROM THE DATA WOULD BE THAT CHARLES CAN EXPECT TO TAKE TWO OTHER BOYS WITH HIS FIELD TRIP GROUP, GIVING A GROUP OF HIS SON, 2 OTHER BOYS, AND 2 GIRLS.

4. DESIGN A SIMULATION AND WORK THROUGH AT LEAST FIVE TRIALS TO ESTIMATE A REASONABLE ANSWER. NOTE: BECAUSE THESE ARE SIMULATIONS, ANSWERS MAY VARY. 1)YOUR BOSS CALLS A LUNCH MEETING BUT PROVIDES LUNCH BOXES FOR ALL THE EMPLOYEES. THESE BOXES CONTAIN EITHER A TURKEY SANDWICH, A ROAST BEEF SANDWICH, OR AN EGG SALAD SANDWICH; AND EITHER AN APPLE OR A BANANA. HOW MANY BOXES SHOULD YOU EXPECT TO EXAMINE UNTIL YOU FIND A BOX WITH A ROAST BEEF SANDWICH AND A BANANA? 2)CHOCO-PUFFS HAS SIX DIFFERENT PRIZES IN THEIR CEREAL BOXES THIS YEAR. HOW MANY BOXES WOULD YOU EXPECT TO BUY BEFORE YOU GET ALL SIX PRIZES? 3)RODNEY’S BASKETBALL TEAM WAS DOWN 1 POINT AS THE FINAL BUZZER SOUNDED. HOWEVER, HE WAS FOULED AT THE FINAL BUZZER AND IS ALLOWED TO SHOOT 2 FREE THROWS. IF HE NORMALLY MAKES 2 OUT OF 3 FREE THROWS, WILL HIS TEAM BE ABLE TO WIN THE GAME? WHAT ABOUT TIE THE GAME? 4)JOAN WORKS AS A SURGERY CENTER VALET. THE SPACES IN HER LOT ARE NUMBERED AND THE CAR KEYS ARE KEPT IN A BOX WITH HOOKS THAT ARE NUMBERED TO MATCH THE SPACE IN WHICH THE CAR IS PARKED. ONE DAY SHE HAS TWELVE CARS IN HER LOT WHEN SHE DROPS THE BOX. ALL THE KEYS FALL OFF THEIR HOOKS AND OUT OF THE BOX. IF SHE PICKS UP THE KEYS AND RANDOMLY PUTS THEM ON HOOKS, ABOUT HOW MANY WILL SHE GET IN THE RIGHT PLACE?

5. 1) YOUR BOSS CALLS A LUNCH MEETING BUT PROVIDES LUNCH BOXES FOR ALL THE EMPLOYEES. THESE BOXES CONTAIN EITHER A TURKEY SANDWICH, A ROAST BEEF SANDWICH, OR AN EGG SALAD SANDWICH; AND EITHER AN APPLE OR A BANANA. HOW MANY BOXES SHOULD YOU EXPECT TO EXAMINE UNTIL YOU FIND A BOX WITH A ROAST BEEF SANDWICH AND A BANANA? WE WILL BEGIN BY DESIGNING OUR MODEL. TO SIMULATE THE SANDWICH CHOICES, WE WILL USE A DIE (SINGULAR FOR DICE). IF WE GET A 1 OR 2, WE WILL SAY THAT IS TURKEY. IF WE GET A 3 OR 4, WE WILL SAY THAT IS ROAST BEEF. IF WE GET A 5 OR 6, WE WILL SAY THAT IS EGG SALAD. NEXT, WE WILL FLIP A COIN FOR THE FRUIT CHOICE. IF WE GET HEADS, WE WILL SAY WE HAVE AN APPLE. IF WE GET TAILS, WE WILL SAY WE HAVE A BANANA. IN ORDER TO GET THE BOX WE WANT, WE WILL NEED A 3 OR 4 AND ALSO A TAILS (3T OR 4T). FOR EACH TRIAL, WE WILL KEEP ROLLING THE DIE AND FLIPPING THE COIN UNTIL WE GET ONE OF THE TWO DESIRED RESULTS. WE WILL THEN RECORD HOW MANY TIMES WE HAD TO DO THIS TO GET THE DESIRED RESULT. THE DATA IS ON THE NEXT SLIDE.

6. REMEMBER, THIS IS ONLY SAMPLE DATA. YOUR DATA MAY BE DIFFERENT. Trial #12H, 4H, 6H, 4H, 3T5 BOXES Trial #2 6T, 1T, 5T, 6T, 6H, 5H, 6T, 3H, 6H, 2T, 1T, 2T, 6T, 6T, 2T, 5T, 4H, 6H, 3T 19 BOXES Trial #36T, 3H, 4T3 BOXES Trial #46T, 3H, 3T3 BOXES Trial #56H, 5H, 1H, 3T4 BOXES Trial #66H, 3H, 6H, 1H, 3T5 BOXES Trial #75T, 4T2 BOXES Trial #8 1T, 5T, 2H, 1H, 5H, 6T, 4H, 2H, 1T, 2H, 3H, 6H, 5T, 5H, 1H, 1H, 4H, 5T, 6H, 2H, 4T 21 BOXES Trial #96H, 4T2 BOXES Trial #104H, 6H, 2T, 6H, 4H, 6T, 3T7 BOXES AVERAGE NUMBER OF BOXES (SUM/10)71/10 = 7.1 ACCORDING TO MY DATA, IT WILL TAKE 7 BOXES BEFORE I FIND THE ONE I WANT. IN REALITY, IT SHOULD TAKE ABOUT 6 BOXES, WHICH MEANS MY AVERAGE IS SLIGHTLY HIGH, ALTHOUGH IT IS SUPPORTED BY MY DATA. THE MORE TRIALS DONE (FOR EXAMPLE, IF 15 GROUPS OF TWO DID 10 TRIALS EACH FOR A TOTAL OF 150 TRIALS), WE WOULD GET CLOSER TO THAT CORRECT NUMBER OF 6.

7. 2) CHOCO-PUFFS HAS SIX DIFFERENT PRIZES IN THEIR CEREAL BOXES THIS YEAR. HOW MANY BOXES WOULD YOU EXPECT TO BUY BEFORE YOU GET ALL SIX PRIZES? AGAIN, WE WILL BEGIN BY DESIGNING OUR MODEL. SINCE THERE ARE SIX PRIZES, WE CAN USE A DIE TO SIMULATE THIS SITUATION. FOR THIS ONE, WE WANT ALL SIX PRIZES, SO WE WILL NEED TO KEEP ROLLING THE DIE UNTIL WE GET ALL SIX OF THE NUMBERS TO APPEAR. WE WILL RECORD EACH ROLL OF THE DIE UNTIL WE GET ALL SIX AND THEN COUNT HOW MANY TIMES WE HAD TO ROLL THE DIE.

8. REMEMBER, THIS IS ONLY AN EXAMPLE OF HOW THE TRIALS COULD COME OUT. Trial #14 1 5 1 6 6 2 5 1 1 5 2 5 5 5 6 4 318 Boxes Trial #21 4 3 1 2 3 5 1 5 1 4 2 4 614 Boxes Trial #33 4 4 3 4 3 6 3 6 3 1 1 3 1 4 3 5 3 5 220 Boxes Trial #42 5 4 3 4 6 3 6 19 Boxes Trial #51 2 5 2 3 3 4 5 4 2 1 4 2 3 5 616 Boxes Average number of boxes (sum/5)77/5 = 15.4 IN THIS SAMPLE OF FIVE TRIALS, WE GOT AN AVERAGE OF 15.4, WHICH WOULD ROUND TO 15 BOXES. HOWEVER, BECAUSE THIS NUMBER IS NEAR THE MIDDLE, WE MAY WANT TO ROUND UP TO GET 16 BOXES. THIS IS PERSONAL PREFERENCE, THOUGH, SO THAT IS SOMETHING WE COULD DISCUSS WITH OUR CLASS TO GAIN THEIR PERSPECTIVE ON THE SITUATION.

9. 3) RODNEY’S BASKETBALL TEAM WAS DOWN 1 POINT AS THE FINAL BUZZER SOUNDED. HOWEVER, HE WAS FOULED AT THE FINAL BUZZER AND IS ALLOWED TO SHOOT 2 FREE THROWS. IF HE NORMALLY MAKES 2 OUT OF 3 FREE THROWS, WILL HIS TEAM BE ABLE TO WIN THE GAME? WHAT ABOUT TIE THE GAME? TO DESIGN THIS MODEL, WE CAN USE A DIE ONCE AGAIN. RODNEY MAKES 2 OUT OF 3 FREE THROWS, SO WE CAN LET A ROLL OF 1, 2, 3, OR 4 BE A SUCCESS AND A ROLL OF 5 OR 6 BE A FAILURE. WE ONLY NEED TO ROLL THE DIE TWICE FOR EACH TRIAL AND THEN RECORD HOW MANY POINTS HE SCORED. IF HE SCORED 0 POINTS, THE TEAM WOULD LOSE. IF HE SCORED 1 POINT, THE TEAMS WOULD TIE. IF HE SCORED 2 POINTS, THE TEAM WOULD WIN.

10. THIS IS THE SAMPLE DATA I COLLECTED FOR THIS DESIGN. YOURS MAY BE DIFFERENT. Trial #15 31 pointTie Trial #24 51 pointTie Trial #32 12 pointsWin Trial #43 42 pointsWin Trial #54 22 pointsWin Trial #64 32 pointsWin Trial #74 32 pointsWin Trial #85 60 pointsLoss Trial #93 51 pointTie Trial #104 12 pointsWin The team will win 6 times, tie 3 times, and lose 1 time. OUT OF THE TEN TRIALS, WE FOUND THAT RODNEY’S TEAM WILL WIN 6 OF THOSE TIMES. SO, THE TEAM WILL PROBABLY WIN. THEY WILL TIE 3 OF THOSE TIMES, SO THEY MAY TIE. THE TEAM WILL LOSE ONLY 1 OUT OF 10 TIMES, SO IT IS UNLIKELY THAT THEY WILL LOSE.

11. 4) JOAN WORKS AS A SURGERY CENTER VALET. THE SPACES IN HER LOT ARE NUMBERED AND THE CAR KEYS ARE KEPT IN A BOX WITH HOOKS THAT ARE NUMBERED TO MATCH THE SPACE IN WHICH THE CAR IS PARKED. ONE DAY SHE HAS TWELVE CARS IN HER LOT WHEN SHE DROPS THE BOX. ALL THE KEYS FALL OFF THEIR HOOKS AND OUT OF THE BOX. IF SHE PICKS UP THE KEYS AND RANDOMLY PUTS THEM ON HOOKS, ABOUT HOW MANY WILL SHE GET IN THE RIGHT PLACE? TO MAKE A MODEL FOR THIS PROBLEM, WE NEED SOMETHING THAT HAS TWELVE CHOICES. SINCE WE DON’T HAVE A 12-SIDED DIE, WE CAN USE A DECK OF CARDS FOR THIS SIMULATION. WE WILL PULL OUT AN ACE, 2, 3, 4, 5, 6, 7, 8, 9, 10, JACK, AND QUEEN OF ONE SUIT TO SIMULATE THE 12 KEYS. THE ACE WILL STAND FOR KEY #1, THE JACK FOR KEY #11, AND THE QUEEN FOR KEY #12. WE WILL SHUFFLE THE DECK EACH TIME AND THEN LIST THE CARDS AS THEY COME OFF THE PILE. WE WILL COMPARE THIS SOLUTION WITH WHAT WE WOULD SEE IN THE KEY BOX, WHERE THEY WOULD BE IN ORDER FROM 1 TO 12. MY DATA AND SOLUTION FOR THIS PROBLEM ARE ON THE NEXT PAGE.

12. THE DATA I COLLECTED FOR THIS SIMULATION IS LISTED HERE. 123456789101112 Number Right Trial #1 8171112310659420 Trial #2 9141271031156820 Trial #3 1142791283101650 Trial #4 4511109712368120 Trial #5 8391012457116120 In each trial, none of the keys matched up. So, the data shows that no key will be placed correctly.

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