How could it be so probable?

$P(\text{both correct}) = \frac18\cdot\frac1{12} = \frac1{96};$ $P(\text{both incorrect}) = \frac78\cdot\frac{11}{12} = \frac{77}{96};$ $P(\text{both incorrect but same answer}) = \frac{77}{96}\times\frac{1}{1+1000} = \frac{77}{96\cdot1001};$ $P(\text{same answer}) = P(\text{both correct} + \text{both incorrect but same answer}) = \frac{1001 + 77}{96\cdot 1001} = \frac{1078}{96\cdot 1001};$ $P(\text{both correct}|\text{same answer}) = \frac{P(\text{both correct})}{P(\text{same answer})} = \frac{1001/96\cdot 1001}{1078/96\cdot 1001} = \frac{1001}{1078}.$ Noting that $77|1001$ , we find $\dots = \frac{77\cdot 13}{77\cdot (13+1)} = \frac{13}{14}.$ Thus the answer is $13 + 14 = 27$ .