SOLUTION: The digits 0, 2, 3, 4, 5, 7, and 9 are to be formed with a three digit (without repetition) number. What is the probability that the number formed is an even number?

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Question 1118532: The digits 0, 2, 3, 4, 5, 7, and 9 are to be formed with a three digit (without repetition) number. What is the probability that the number formed is an even number?
Found 2 solutions by josmiceli, greenestamps:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The leading digit can't be a 0, so
The are +6+ possible digits
One digit is used up, but +0+ is
allowed, so there are +6+ possible
Two digits used up, so +5+ possible
for units digit
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+6%2A6%2A5+=+180+ possible, both odd
and even numbers
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As for all the possible even numbers,
The first 2 are +6+ possible still
The units digit can only be 9,2, or 4
+3+ possible
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+6%2A6%2A3+=+108+ possible even numbers
P( even 3-digit number ) = +108%2F180+=+.6+
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Feel free to get another opinion

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Total number of 3-digit numbers, since the first digit can't be 0: 6*6*5 = 180

Number of even 3-digit numbers with final digit 0: 1*6*5 = 30 (since 0 is the final digit, we don't have to worry about possibly choosing it for the first digit, which would not be allowed)
Number of even 3-digit numbers with final digit 2: 1*5*5 = 25 (only 5 choices for the first digit, since it can't be 0; then any 5 of the remaining digits for the second digit)
Number of even 3-digit numbers with final digit 4: 1*5*5 = 25 (same reason)

Total number of even 3-digit numbers: 30+25+25 = 80.

P(even) = 80/180 = 4/9