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 the same parents. The probability that the selected group of children have no blood relations, is equal to \[\]
A.$\dfrac{11}{15}$ \[\]
B. $\dfrac{13}{15}$ \[\]
C.$\dfrac{14}{15}$ \[\]
D.$\dfrac{2}{15}$ \[\]

Answer 1

Hint: We find the number of ways we can get the boy the girl in our selected group of 3 children as $n\left( A \right)$by counting number of ways we fill third spot with 1 child out of 8 without blood relation. . We find the number of all possible outcomes by finding the number of ways we can select 3 children out of 10 as $n\left( S \right)$. The required probability is $P\left( {{A}^{'}} \right)=1-P\left( A \right)=1-\dfrac{n\left( A \right)}{n\left( S \right)}$ \[\]

Complete step by step answer:
We know from definition of probability that if there is $n\left( A \right)$ number of ways of event $A$ occurring (or number of favourable outcomes) and $n\left( S \right)$ is the size of the sample space (number of all outcomes) then the probability of the event $A$ occurring is given by
\[P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}\]
We are given in the question that 3 children are selected at random from a group of 6 boys and 4 girls. So the total number of children is $6+4=10$. We can select 3 children out of 10 children in ${}^{10}{{C}_{3}}$ way. So the number of all possible outcomes is
\[n\left( S \right)={}^{10}{{C}_{3}}=120\]
We are further given the question that in this group of 10 children exactly one girl and one boy belong to the same parents. So the number of children who do not have blood relation with each other is $10-1-1=8$. \[\]
We are asked in the question to find the probability that the selected group of children have no blood relations. \[\]
 Let us denote the event of getting both the boy and the girl who are siblings in the randomly selected 3 children as $A$. We have already filled 2 spots out of 3 empty spots for selection and we can fill the rest 1 spot with 8 children who do not have blood relations in ${}^{8}{{C}_{1}}$ . So the number of favourable outcomes is the number of ways an event $A$ can happen is ${}^{8}{{C}_{1}}=8$. So we have
\[n\left( A \right)=8\]
So the probability of event $A$ happening is ;
\[P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{8}{120}=\dfrac{1}{15}\]
The required probability is the probability of event $A$ not happening which means there should not be the sibling in our selection which is
\[P\left( {{A}^{'}} \right)=1-P\left( A \right)=1-\dfrac{1}{15}=\dfrac{14}{15}\]

So, the correct answer is “Option C”.

Note: We note that the method we used here is called the method of negation. We must be careful that we are finding $P\left( A \right)$ first and we should not choose options on the basis of $P\left( A \right)$. The event of getting and not getting the sibling in the selection are mutually exclusive events since they cannot occur together and exhaustively too since the sum of their probabilities is 1.