Willy is a Dumby

Let $E_{n}$ denote the number of n-digit strings composed of 0's and 1's such that there is at least one string of two consecutive 1's.

If the last digit of the string is a 0, then there must be at least one string of two consecutive 1's in the previous $n-1$ digits, and hence the number of possibilities in this case is $E_{n-1}$

If the last digit of the string is a 1, the ( $n-1$ )th digit can either be a 1 or a 0.

If the ( $n-1$ )th digit is a 0, we can see that there must be at least one string of two consecutive 1's in the previous $n-2$ digits, and hence the number of possibilities in this case is $E_{n-2}$

If the ( $n-1$ )th digit is a 1, we can see that we already have a string of two consecutive 1's in the nth and ( $n-1$ )th digit, and so it does not matter what the preceding $n-2$ digits are. The number of possibilities in this case is $2^{n-2}$

In this way, we can obtain a recursive formula: $E_{n}=E_{n-1}+E_{n-2}+2^{n-2}$

We can work out that $E_{2}=1$ as the only possibility is 11. The next term to calculate is $E_{3}=3$ as there are three possibilities: 011, 110, and 111.

Using the recursive formula, we can work out $E_{4}=E_{3}+E_{2}+2^{2}=3+1+4=8$

Similarly, $E_{5}=E_{4}+E_{3}+2^{3}=8+3+8=19$

Finally, $E_{6}=E_{5}+E_{4}+2^{4}=19+8+16=43$

The total number of six-digit strings of 0's and 1's such that there is at least one string of two consecutive 1's is $2^{6}=64$ , since there is a choice between 1 or 0 for each digit.

The probability that someone is unable to do the task is $\large\frac{64-43}{64}=\frac{21}{64}$

The probability that both of them are unable do the task is $\large(\frac{21}{64})^{2}=\frac{441}{4096}$

Therefore, $\large a+b=441+4096=4537$