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Relevant wiki: Probability - By Outcomes

First we must determine the total number of possible $4$ digit number sequences. There are $10$ choices for the first digit. For each of these choices there are $10$ choices for the second digit. Therefore, there are $10 × 10 = 100$ choices for the first two digits. For each of these possibilities, there are $10$ choices for the third digit. Therefore, there are $100 × 10 = 1000$ possibilities for the first three digits. And finally, for each of these $1000$ choices for the first three digits there are $10$ choices for the fourth digit. Therefore there are $1000 × 10 = 10 000$ possibilities for license plates with four digits. Now we must determine the number of license plates with a digit sum of $34$ or higher. We will consider cases.

1. The license contains four nines. If the digits are all nines, the sum is $36$ which is acceptable. There is only one way to use all nines for digits.

2. The license contains three nines and one other digit. If three nines are used the sum is $27$ . In order to get $34$ or higher the fourth digit must be $7$ or $8$ . There are two choices for the fourth digit. Once the digit is chosen there are four places to put the digit. Once the digit is placed the remaining spots must be nines. Therefore there are $2 × 4 = 8$ license plates containing three nines. (It is possible to list them: $\text{7999, 8999, 9799, 9899, 9979, 9989, 9997, 9998}$ .)

3. The license contains two nines. If two nines are used the sum is $18$ . In order to get to $34$ or higher, we need a sum of $34 - 18 = 16$ or higher from the two remaining digits. The only way to do this, since we cannot use more nines, is to use $two$ eights. The two eights can be placed in six ways and then the nines must go in the remaining spots. Therefore there are six license plates containing two nines. (It is possible to list them: $\text{8899, 8989, 8998, 9889, 9898, 9988}$ .)

4. The license contains one nine. If one nine is used the sum is $9$ . In order to get to $34$ or higher, we need a sum of $34 - 9 = 25$ or higher from the remaining three digits. But this sum would be made from three digits chosen from the digits $0$ to $8$ . The maximum possible sum would be $24$ if three eights were used. We need $25$ or higher. Therefore no license plate containing only one $9$ will produce a sum of $34$ or higher. It should be noted that a license with no nines would not produce a sum of $34$ or higher either.

Therefore, the number of license plates with a digit sum of $34$ or higher is $1 + 8 + 6 = 15$ . To calculate the probability we divide the number of licenses with a digit sum of $34$ or more by the number of possible license plates. The probability of getting a license plate with a digit sum of $34$ or higher is $\frac{15}{10 000}$ = $\frac{3}{2000}$ . So $a+b=3+2000=2003$ .