Question

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

• A. $\frac{11}{15}$
• B. $\frac{13}{15}$
• C. $\frac{14}{15}$
• D. $\frac{2}{15}$

Answer 1

The correct option is C $\frac{14}{15}$

Total number of ways of selecting $3$ children from $10$ children$=$ ${}_3^{10}C=120$ ways.
If we want to include the siblings and If the siblings are already selected, then we need to select another child from the remaining $8$. This can be done in ${}_1^{8}C=8$ ways.
Thus, the probability that there is always a blood relation among the selected children$=$ $\frac{8}{120}=\frac{1}{15}$.

Thus, the probability that there is no blood relation among the selected children $=$ $1-\frac{1}{15}=\frac{14}{15}$.