if a number is choosen between 100 and 999 includeing these
numberrs what is the provbabilty that the number selected
does not contain a 7 is
Answer Posted / shruti
here are 900 numbers between 100 and 999.
those that contain 7 are.....
107, 117, 127, ........ -- 10 of these
207, 217, 227, ........ -- 10 of these too
307, 317, 327, .......
....
9*10 numbers that end in 7.
170, 171, 172, 173, 174, 175, 176, (NOT 177 WE ALREADY COUNTED IT), 178, 179 -- 9 of these
270, 271, 272, 273, 274, 275, 276, (NOT 277 WE ALREADY COUNTED IT), 278, 279 -- 9 of these
...
and so on
so 9*9 numbers that have a middle digit of 7
and finally
700, 701, 702, 703, 704.................., but NOT
(707, 717, 727,....) or (770, 771, 772, 773, ....)
So that's (100 - 10 - 9) = 81 numbers that start with 7.
We had 900 numbers.
N = 900 - 90 - 81 - 81 = 648
So the probability is 648/900 = 0.72
Is This Answer Correct ? | 5 Yes | 3 No |
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