Question 120785: In excercises 34-36,use the spinner to find the probability. The spinner is divided into equal parts.The spinner numbers are 1-12 in a circle
35. What are the odds in favor of stopping on a multiple of 3?
Any help with this equation would be Greatly appreciated.
Found 2 solutions by checkley71, stanbon:
Answer by checkley71(8403)   (Show Source):
You can put this solution on YOUR website! SEEING AS EACH NUMBER HAS THE PROBABILITY OF 1/12 OR 8.333%.
AND YOU HAVE 3,6,9,&12 ARE THE NUMBERS OF INTEREST.
WE HAVE 4*.08333=.33333 OR 33.33% OR 1/3 CHANCE OF LANDING ON A NMUMBER THAT IS A MULTIPLE OF 3.
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! In excercises 34-36,use the spinner to find the probability. The spinner is divided into equal parts.The spinner numbers are 1-12 in a circle
35. What are the odds in favor of stopping on a multiple of 3?
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The multiples of 3 are 3,6,9,12
There are four multiples of three.
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Probability(landing on a multiple of 3) = 4/12 = 1/3
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Odds "in favor of" an event = P(event happens)/P(event does not happen)
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Your Problem:
Odds in favor of stopping on multiple of 3) = (1/3)/(2/3) = 1/2 = 1:2
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Cheers,
Stan H.
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