Question 1082450: A spinner labeled 1 to 9 gives each of the numbers 3, 6, 7, and 8 a 22% chance of being landed upon. The chance of landing on each of the other five numbers is equal. If the spinner is spun 2,000 times, which choice is the most likely outcome for the 2,000 spins?
Number on Spinner 1 2 3 4 5 6 7 8 9
Number of Occurrences 521 520 501 518 519 502 499 504 520
Number on Spinner 1 2 3 4 5 6 7 8 9
Number of Occurrences 605 602 640 606 606 639 641 638 601
Number on Spinner 1 2 3 4 5 6 7 8 9
Number of Occurrences 350 349 301 346 350 303 299 301 351
Number on Spinner 1 2 3 4 5 6 7 8 9
Number of Occurrences 419 418 441 420 420 438 442 440 421
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
None of the answers are possible. The sum of all of the occurrences must, perforce, equal 2,000 and the smallest sum for any of the four given answers is, just by cursory examination, in the neighborhood of 3000.
Secondly, if the chance of 3, 6, 7, and 8 are EACH 22%, then the probability of 3 or 6 or 7 or 8 is 88%, and the probability of any of the other numbers is, in the aggregate, 12%, or individually 2.4%. That means 3, 6, 7, or 9 are each roughly 9 times more likely than any of the others individually.
I would expect a distribution that was closer to:
1 2 3 4 5 6 7 8 9
48 48 440 48 48 440 440 440 48
Which doesn't even come close to any of your given answers
John
My calculator said it, I believe it, that settles it
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