Question 360786: a spinner has regions numbered 1 through 21. what is the probability that the spinner will stop on an even number or a mulitiple of 3? reduce fraction if necessary.
Found 2 solutions by ewatrrr, stanbon:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi,
21 numbers from 1 to 21
10 even numbers: 2,4,6,8,10,12,14,16,18,20
7 numbers which are multiple of 3: 3,6,9,12,15,18,21
P(even 'or' multiple of three) = P(even) + P(multiple of three)
P(even 'or' multiple of three) = 10/21 + 7/21 = 17/21
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! a spinner has regions numbered 1 through 21. what is the probability that the spinner will stop on an even number or a mulitiple of 3? reduce fraction if necessary.
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Even: 2,4,6,8,10,12,14,16,18,20: 10 ways
Even and multiple of 3: 6,12,18: 3 ways
Multiples of 3: 3,6,9,12,15,18,21: 7 ways
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Possible results: 21
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P(even or multiple of 3) = P(even) + P(multiple) - P(even and multiple of 3)
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= (10 + 7 - 3)/21 = 14/21 = 2/3
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Cheers,
Stan H.
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