Question 160562This question is from textbook Texas Algebra I
: Suppose you select a 3 digit number at random from the set of all positive 3 digit numbers. Find each probability or odds.
P(odd numbers)
P(number less than 900)
This question is from textbook Texas Algebra I
Answer by Fombitz(32388)   (Show Source):
You can put this solution on YOUR website! First off, I'm assuming that the set of all positive 3 digit numbers contains 1000 possible choices (000 to 999).
Half (500) are even, half are odd.
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P(odd)=500/1000=1/2.
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From 000 to 899, there are 900 numbers.
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P(<900)=900/1000=9/10.
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If that's not correct and no leading zeros are allowed, then
the set starts at 100 and goes to 999.
Then, the total number is diminished by 100, or 900 total.
Even and odds total 400 each.
Probability of an odd choice is the same.
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P(odd)=450/900=1/2.
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The total less than 900 is reduced to 800.
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P(<900)=800/900=8/9.
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