probablity

[jainrahul1985]

What is the probability that 4 appear 5 times in the list of numbers 100000 to 999999 inclusive ?

OA [spoiler]53/900000 but not sure [/spoiler]

[mj78ind]

Essentially there are 6 digits to be filled with numbers since we are looking at numbers between 1,00,000 and 9,99,999

Assume that we fill all the digits except the first one with 4s in a number _ _ _ _ _ _ then the first digit can be filed in 8 ways,

can not have a 0 or a 4. If we now fix 4s in 1st 3,4,5,6 th place the 2nd can be filled in 9 ways (0 to 9 except 4) this can be

repeated 5 times as we keep moving the place of the 'free' digit (which was 2nd in this case).

Hence numbers possible with 5 4's is = 8 + 9*5 = 53

Total numbers between 1,00,000 and 9,99,999 including 1,00,000 = 9,99,999 - 1,00,000 +1 = 9,00,000

Hence probability = 53/9,00,000

[kstv]

If the first digit is 4, of the remaining 5 digits, 4 places wiill have 4. The remaining places can be occupied by any of the 9 digits (0 to 9) but not 4 then the no will be 4,44,444.
This digit(other than 4) can be in any of the 5 places.
Total no of ways = 9*5 = 45
If the first digit is not 4, then it can be filled in 8 ways excluding 0, rest will be 4s.
Total no of ways = 45+8 = 53