What is the ration of winning if your sell 1,000 calendars and there are 356 prizes
asked Nov 5, 2016 in Algebra 1 Answers by JOLENE

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If each calendar is eligible to win a prize, the probability of one calendar to win a prize is 1/356.

The probability of 2 calendars c1 and c2 is summarised as:

c1 wins, c2 wins; c1 wins, c2 doesn't win; c1 doesn't win, c2 wins; neither wins.

The probabilities are calculated as follows.

  1. c1 has a 1 in 356 chance of winning=1/356; now given that c1 wins there will only be 355 prizes left so c2 has a chance of 1/355 of winning a prize=1/355. Combine these probabilities: 1/356*1/355=1/126380.
  2. c1's probability is 1/356; c2's probability of not winning is 354/355. Combined: 177/63190.
  3. c1's probability of not winning is 355/356; c2's probability of winning is 1/356. Combined: 355/126736.
  4. The probability of c1 or c2 not winning is 355/356 each; combined=126025/126736.

When added together these come to 1, which simply means we've got these probabilities right and we haven't missed any out.

To find out the probability of at least one winning, we only have to subtract the probability of neither winning from 1: 1-126025/126736=711/126736.

Now we consider 1000 calendars. We subtract the chance of none winning from 1. So we work out 1-(355/356)^1000=0.94 approximately or 94%. Expressed as a ratio this is 94/100=47:50.

 

 

answered Nov 5, 2016 by Rod Top Rated User (424,700 points)
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