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what is the probability that the pointers will stop at an odd or divisible by 3​


Sagot :

Answer:

8/12 or 2/3

  • 6/12 or 1/2 probability of stopping at an odd number
  • 4/12 or 1/3 probability of stopping at a number that is divisible by 3

Step-by-step Explanation:

  • Odd numbers from 1 to 12 is O = {1,3,5,7,9,11} = 6 elements
  • Divisible by 3 is D = {3,6,9,12} = 4 elements
  • Union of O and D is = {3,9} = 2 elements

My solution:

  • 6 + 4 = 10
  • 10 - 2 = 8

Set of not odd and divisible by 3

  • {2,4,8,10} = 4 elements