SOLUTION: A three-digit number is chosen among all three-digit numbers. What is the probability that it will have distinct even digits?
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Question 1120137: A three-digit number is chosen among all three-digit numbers. What is the probability that it will have distinct even digits? Found 2 solutions by Fombitz, ikleyn:Answer by Fombitz(32388) (Show Source):
You can put this solution on YOUR website! The first digit can take on 5 values (0,2,4,6,8).
The second digit could take on 4 values (one less than 5 to be unique).
The third digit could take on 3 values (two less than 5 to be unique).
So that makes,
So there are 60 unique three digit even numbers.
There are 1000 possible outcomes.
The total number of all 3-digit numbers from 100 to 999 is 900.
The even digits are 0, 2, 4, 6 and 8 (five, in all).
The first (most-left) digit can be any of four 2, 4, 6 or 8 (it can not be 0, zero).
The second digit can be any of remaining four digits (including zero).
The third digit can be any of remaining three digits (including zero).
The total number of 3-digit numbers with even non-repeating digits is 4*4*3 = 48.
The probability under the question is = = = 0.0533 = 5.33% (approximately).
