anonymous
  • anonymous
What are the odds of rolling a 6 on a regular six sided die?
Algebra
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@Michele_Laino
anonymous
  • anonymous
i think its 1:6 @Michele_Laino
Michele_Laino
  • Michele_Laino
I think that they are 1, 3, and 5

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More answers

Michele_Laino
  • Michele_Laino
namely the odd numbers are 1, 3, and 5
Michele_Laino
  • Michele_Laino
the requested probability is: \[\frac{{{\text{favorable outcomes}}}}{{{\text{possible outcomes}}}} = \frac{1}{6}\]
anonymous
  • anonymous
IS ODDS DIFFERENT THE PROBABILITY @Michele_Laino
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
1/6
anonymous
  • anonymous
but isnt probability and odds different
Michele_Laino
  • Michele_Laino
for example, the odds is the probability when all the possible outcomes are different each from other
anonymous
  • anonymous
so 1:6 would be correct?
anonymous
  • anonymous
for odds?
Michele_Laino
  • Michele_Laino
I'm pondering
anonymous
  • anonymous
@Odds is defined as success : failures. There are two numbers greater than or equal to 5. Those numbers are 5 and 6. The failures would be the numbers that are not greater than or equal to 5. Those numbers are 1, 2, 3, and 4. This means that odds is 2 : 4 or 1 : 2.
anonymous
  • anonymous
im confused.
anonymous
  • anonymous
probabily would be 1:6 but for odds it would be 1:5 correct?
Michele_Laino
  • Michele_Laino
here is the right definition of odds: "odds is the ratio between the probability that an event occurs and the probability that the same event not occurs" so for question #1 probability to get "m" = 1/7 probability to get not an "m" = 1-(1/7)=6/7 \[\Large odds = \frac{{1/7}}{{6/7}} = \frac{1}{6}\]
Michele_Laino
  • Michele_Laino
another example: for a six sided dice, we have: probability to get a "6" = 1/6 probability to get not a "6" = 1-(1/6)= 5/6 so: \[\Large odds = \frac{{1/6}}{{5/6}} = \frac{1}{5}\]
Michele_Laino
  • Michele_Laino
finally: question #3 probability to get "g" = 2/7 probability to get not a "g" = 1-(2/7) = 5/7 so: \[\Large odds = \frac{{2/7}}{{5/7}} = \frac{2}{5}\]
anonymous
  • anonymous
Such a complicated process to find out something so simple! Would've never guessed. Thanks for broadening my horizons :) @Michele_Laino
Michele_Laino
  • Michele_Laino
:)
Jack1
  • Jack1
1 chance in 6 possibilities ;)
anonymous
  • anonymous
1/6

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