Multiple choice

A coin whose faces are marked 3 and 5 is tossed 4 times; what are the odds against the sum of the numbers thrown being less than 15?

  1. The odds against are $5:11$
  2. The odds against are $11:5$
  3. The odds against are $16:5$
  4. The odds against are $5:16$
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B Correct answer
Explanation

Total outcomes for 4 tosses are 2^4 = 16. Sums less than 15: (3,3,3,3)=12, (3,3,3,5)=14, (3,3,5,3)=14, (3,5,3,3)=14, (5,3,3,3)=14. There are 5 outcomes with sum < 15 and 11 outcomes with sum >= 15. The odds against are 11:5.

AI explanation

The minimum possible sum is 12, so the sum is less than 15 only for the outcomes 13 and 14. There are 4 ways to get a sum of 13 and 6 ways to get a sum of 14, yielding 10 favorable outcomes out of 16 total possibilities. The odds against this event are calculated as the ratio of unfavorable to favorable outcomes, giving (16 minus 10) to 10, or 11 to 5.

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