macattack -- yeah, you're right here. nice work.
as usual, i misread the problem (yay! dyslexia is fun) as "three-digit code" instead of "three-digit number". so, in other words, i was thinking about 000-999 instead of 100-999.
so, yes, that works.
using the "opposite situation" method:
the chance of getting NO sevens is
(8/9)(9/10)(9/10)
= (8/10)(9/10)
= 72/100
= 18/25
so, the chance of getting at least one seven is 1 - 18/25 = 7/25.
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you can also do this:
* 90 numbers with "7" in the units place (107, 117, ..., 997); 90 numbers with "7" in the tens place (170, 171, ..., 979), and 100 numbers with "7" in the hundreds place (700-799) ... so, 280.
BUT
* we've counted all the _77 numbers except 777 (eight of them) twice. there are eight of those, so subtract 8.
* we've counted all the 7_7 numbers except 777 (nine of them) twice. there are nine of those, so subtract 9.
* we've counted all the 77_ numbers except 777 (nine of them) twice. there are nine of those, so subtract 9.
so, right now, 254.
FINALLY
* we've counted 777 three times. so, subtract two of them (so that one still counts), leaving 252.
... but your way of counting seems to be more straightforward.
Ron has been teaching various standardized tests for 20 years.
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