Question

From a set of all eight-digit natural numbers, where only the digits from the set {0,1, 3, 5, 7, 9} are present in decimal notation. we draw one. Calculate the probability of the event that the sum of digits of the drawn number is equal to 3.

Answer 1

Since we are considering 8digit numbers so we can not put 0 at the beginning because then it will be a 7 digit number. for example is not an 8 digit number.

So, the total number of such 8 digit numbers is= because th place can be filled with any of 1,3,5,7,9 so in 5 ways and for the other places each place can be filled with one of 0,1,3,5,7,9 so in 6 ways.

Now for an 8 digit number sum of digits =3 is possible only when we make the number with only 0,1,3

now 3=0+3, 1+1+1

For 0+3 case

place can not get 0, so it will get 3 and rest places will be 0 only, there is only 1 such number

i.e. 30000000

For 1+1+1 case

place can not get 0, so it will get 1 (it can be done in only 1 way) and other two 1s can be placed in ways and rest empty places will get 0s only.

so total number of such numbers=1+42=43

Then the probability is =