Three children are selected at random from a group of 6 boys and 4 girls. It is known that in this group exactly one girl and one boy belong to same parents. The probability that the selected group of children have no blood relations, is equal to

$\frac{14}{15}$

Probability that the boy and his sister both are selected $=\frac{{ }^8 C_1}{{ }^{10} C_3}=\frac{1}{15}$

Thus, required probability = $1 - \frac{1}{15} = \frac{14}{15}$