Use random numbers to copy a chance situation
Grade 7 Math Worksheet - Challenge Me

Circle the best answer, or write your answer on the line.
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Digits 0 to 3 are star cards. Each trial stops at its first digit from 0 to 3. Trials: 5 7 2 | 1 | 6 9 4 8 0 | 9 3 | 4 5 6 7 8 3 | 0 | 8 4 2 | 7 6 1 | 3 | 9 8 8 9 5 1 | 6 2 | 4 7 9 0.
Ari draws a card at random until a star shows, putting each back (40% are stars). From the trials shown, what is the estimated chance it takes at least 4 draws, in simplest form?
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Digits 0 and 1 are a prize. Each trial stops at its first 0 or 1. Trials: 5 8 0 | 4 6 7 1 | 9 1 | 8 2 6 4 1 | 7 3 9 0 | 1 | 2 4 0 | 4 4 9 5 7 0 | 0 | 6 3 1 | 9 7 4 5 1 | 8 0.
One box in 5 holds a prize. Bo opens boxes at random until a prize shows. From the trials shown, what is the estimated chance it takes at least 3 boxes, in simplest form?
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Digits 0 to 2 are rain. Each trial stops at its first digit from 0 to 2. Trials: 5 8 7 1 | 3 0 | 9 2 | 4 7 9 8 5 0 | 8 1 | 7 7 2 | 6 4 9 1 | 2 | 9 5 8 3 8 0 | 5 4 0 | 1 | 8 9 6 7 4 2.
Each day has a 30% chance of rain. Esme counts days until the first rain. From the trials shown, what is the estimated chance it takes at least 3 days, in simplest form?
Use this
Digits 0 to 4 are made shots. Each trial stops at its first digit from 0 to 4. Trials: 6 4 | 9 1 | 8 7 0 | 5 2 | 7 9 8 3 | 6 4 | 9 7 6 0 | 2 | 8 3 | 9 9 7 4 | 7 6 8 6 1 | 5 1.
A player makes half of shots. Jun counts shots until the first made shot. From the trials shown, what is the estimated chance it takes at least 3 shots, in simplest form?
Kira copies a 30% event happening on both of two days. The plan: digits 0 to 2 mean yes, one digit per trial, count trials that show a yes. What is wrong?
- Nothing is wrong, because a 30% event takes 3 of the 10 digits, 0 to 2, which is what the plan uses.
- The trials should count the digits 3 to 9, which stand for the days when the event does not happen.
- Digits 0 to 2 are too few, because both days together are more likely than one day alone is.
- One digit cannot copy two days, so each trial needs two digits and both must be yes.
Luz draws 2 cards from a hat of 5 (2 red) without putting the first back. Digits 0 to 4 are the cards, 0 and 1 red; two digits per trial. What needs fixing?
- The hat has 5 cards, so each trial needs 5 digits instead of 2, one digit for each card in the hat.
- Digits 0 and 1 are too few for red, so 0 to 3 should be red to match the 2 red cards in the hat.
- Nothing needs fixing, because drawing without putting back works the same as drawing with putting back.
- A digit can repeat, but a card cannot be drawn twice, so skip any trial with a repeated digit.
Milo simulates two flips of a fair coin for 8 trials, and 3 show two heads. Milo says the chance is 3/8, so the coin is rigged. What is the best response?
- 3/8 is right, because the trials are real outcomes, and real outcomes cannot be wrong or misleading.
- Run 8 more trials and average the two results together to get exactly 1/4 for the chance of two heads.
- Eight trials is too few to settle anything; run many more trials and see where the share goes.
- The coin is rigged, because 3/8 is more than the 1/4 chance of two heads, so the model must be wrong.
Noah runs the same simulation twice. One run of 20 trials gives 40%. Another run of 400 trials gives 28%. Which estimate should Noah trust more, and why?
- The 20-trial run, because a short run is easy to check by hand, so it has fewer mistakes.
- The 28%, because a run with 400 trials usually gives a steadier share than 20 trials.
- The 40%, because it is a rounder number, so it is easier to check by hand when you are done.
- Neither, because a simulation cannot estimate a chance, however many trials it runs.
Pia simulated 60 turns of two free throws, and both shots were made in 18 of them. About how many of 500 real turns should end with both shots made?
Ren simulated 40 weekends, and it rained on both days in 12 of them. About how many of 300 weekends like these should have rain on both days?