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
Answers were Sorted based on User's Feedback
Answer / jamesk
p = prob(1st digit not 7)*prob(2nd digit not 7)*prob(3rd
digit not 7)
=(8/9)*(9/10)*(9/10)
=0.72
Sandhya had it right except that there are 900 numbers
between 100 and 999 inclusive (not 899).
Is This Answer Correct ? | 59 Yes | 6 No |
Answer / seiko
no 0f 7's in 100 to 200 is
107,117,127,137,147,157,167,170,171,172,173,174,175,176,177,
178,179,187,197.
so total of 19 numbers between 100 and 199
repeat that for 200-299, 300-399,...,600-699, 800-899, 900-
999, we have 19*8 = 152 occurences
for 700-799, all 100 of them contain a 7
so total numbers contain a 7 is 152 + 100 = 252
total numbers between 100 and 999 is 900
probability = (900-252)/900 = 0.72
Is This Answer Correct ? | 51 Yes | 8 No |
Answer / ajeet
100-199------19
similarly for all ranges till 999 except for:
700-799----100
total no.containing 7's= 19*8+100=252
Therefore,
probability=(900-252)/900
= 648/900
= 0.72
Is This Answer Correct ? | 10 Yes | 3 No |
Answer / mayank kumar
total numbers between 100 and 999 =900
number that have a digit 7 between 100 and 199 =19
similarly no. that have digit 7 between 200 and 699=5*19
number that have digit 7 between 700 and 799=100
number that have a digit 7 between 800 and 999 =2*19
so total number that have digit seven in them between 100 and 999 are=8*19+100 =252
so numbers that does not have 7 digit in them =900-252 =648
therefore required prob, p =648/900 = 0.72
Is This Answer Correct ? | 6 Yes | 3 No |
Answer / 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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