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If one three-digit number ( 0 cannot be a left digit) is chosen at random from all those that can be made from the following set all digits, find the probability that one chosen is not a multiple of 2. [0,1,2,3,4,5,6,7,8]

Answer :

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The probability of choosing a number that is not a multiple of 2 is P = 0.44

How to find the probability?

We need to count the number of options for each digit.

  • For the first digit, we have 8 options {1, 2, 3, 4, 5, 6, 7, 8}
  • For the second digit, we have 9 options {0 ,1, 2, 3, 4, 5, 6, 7, 8}
  • For the third digit, we have 9 options {0 ,1, 2, 3, 4, 5, 6, 7, 8}.

The total number of combinations is the product between the numbers of options:

C = 8*9*9 = 648

If we want our number to not be a multiple of 2 then it must end in a odd digit, the combinations that meet that condition are:

  • For the first digit, we have 8 options {1, 2, 3, 4, 5, 6, 7, 8}
  • For the second digit, we have 9 options {0 ,1, 2, 3, 4, 5, 6, 7, 8}
  • For the third digit, we have 4 options {1, 3,  5, 7}.

C = 8*9*4 = 288

Then the probability of selecting a 3 digit number that is not a multiple of 2 is:

P = 288/648 = 0.44

If you want to learn more about probability, you can read:

https://brainly.com/question/251701

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