lunarpower wrote: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.

--

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.