to be perfectly balanced here, it's not totally fair to say there are two errors, since, technically, the 900 isn't wrong. (the 900 is incompatible with the 271, but only the 271 is actually erroneous.)
i'd be interested in seeing the answer key for this problem, though.
see, ok, i can totally get how someone would have made this mistake -- in fact, i misread the problem myself, as i do much too often (see above) -- but what i'm wondering is, "wouldn't the author notice the mistake while writing the answer key?"
i.e., if there's actually a key to this problem, the key will have to explain how to enumerate the 271 [sic] integers in question.... and then that explanation should explicitly mention the values between 000 and 099, thus tipping off readers (and, i would have thought, the author!) to the inherent contradiction.
does anyone have the answer key to this thing? it would be interesting to see how, exactly, the answer here is justified, and/or how much explanation is given for the 271.
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lunarpower
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Ron has been teaching various standardized tests for 20 years.
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Matt@VeritasPrep
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I'm guessing the OA is a blend of miscounting and overcounting. If you do this:
7xx = 9*9 ways
x7x = 9*9 ways
xx7 = 9*9 ways
7x7 = 9 ways
x77 = 9 ways
77x = 9 ways
777 = 1 way
you get 81*3 + 27 + 1 = 271. There are 900 three digit numbers, "hence" 271/900.
7xx = 9*9 ways
x7x = 9*9 ways
xx7 = 9*9 ways
7x7 = 9 ways
x77 = 9 ways
77x = 9 ways
777 = 1 way
you get 81*3 + 27 + 1 = 271. There are 900 three digit numbers, "hence" 271/900.
Hello Ron - can you please explain how you got this? Like a step by step if possible please?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.
"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 "
Thanks
Subhakam
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Matt@VeritasPrep
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The opposite situation says:
The first digit of a three-digit number can't be 0, but it could be any of the other 9 digits. We want something that isn't 7, so any of the other 8 digits is fine. Our odds of getting a non-7 for the first digit are thus 8/9.
For the second and third digits, there are 10 possible digits for each space. We want a non-7, so we'll take any of the other 9. Our odds are thus 9/10 each time.
Since we need a non-7 for each of the three digits, we need all three scenarios to be non-7. Thus we multiply the probabilities together:
8/9 * 9/10 * 9/10 = 18/25
Since these are the odds of getting NO 7's, the odds of getting at least one 7 are (1 - odds of no 7's) or (1 - 18/25) or 7/25.
The first digit of a three-digit number can't be 0, but it could be any of the other 9 digits. We want something that isn't 7, so any of the other 8 digits is fine. Our odds of getting a non-7 for the first digit are thus 8/9.
For the second and third digits, there are 10 possible digits for each space. We want a non-7, so we'll take any of the other 9. Our odds are thus 9/10 each time.
Since we need a non-7 for each of the three digits, we need all three scenarios to be non-7. Thus we multiply the probabilities together:
8/9 * 9/10 * 9/10 = 18/25
Since these are the odds of getting NO 7's, the odds of getting at least one 7 are (1 - odds of no 7's) or (1 - 18/25) or 7/25.
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lunarpower
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^^^ yep. that.
Ron has been teaching various standardized tests for 20 years.
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Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
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Pueden hacerle preguntas a Ron en castellano
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Voit esittää kysymyksiä Ron:lle myös suomeksi
--
Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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Learn more about ron
