@utkarsh_lath: Thanks It is more clear now
The sum of those two digits is exactly 9, and there was a carry from i-1th position. The sum of digits is 9 for the pairs (0, 9), (1, 8), (2, 7), … (9, 0). The probability of this happening is 10 / 100 * E[i-1].
can someone plz explain why we r multiplying E[i-1]*10/100;
The probability that we get a carry from i-1 th place is E[i-1]. Since the digits at ith place are independent of the digits at i-1 th place, we have
Prob[there was a carry from i-1 th place and sum of digits at i th place is 9] = Prob[there was a carry from i-1 th place] * Prob[sum of digits at i th place is 9] = E[i-1] * 10/100.
just use:
cout<<fixed <<arr[n]<<endl;
and it will get AC
plz tell me how the equation E(n)=45/100+10/100*E(i-1) came??
have we include the case when the first digit for a number(N>1) is 0…we keep on counting if the sum is 9 without testing that the first number cant be 0…
shouldn’t it be (10/100)+E[i-1]
where E[1]=(45/100)??
@ utkarsh lath why have we multiplied 45/100 with i…???