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From a class of 12 girls and 18 boys, two students are chosen randomly. What is the probability that both of them are girls?
  {\text{A}}{\text{. }}\dfrac{{22}}{{145}} \\
  {\text{B}}{\text{. }}\dfrac{{13}}{{15}} \\
  {\text{C}}{\text{. }}\dfrac{1}{{18}} \\
  {\text{D}}{\text{. }}\dfrac{1}{{15}} \\

Last updated date: 20th Jun 2024
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Hint- In this question we have to find the probability that when two students are chosen randomly both of them are girls. So, to solve this we will use combinations and formulas for probability to reach the answer.

Complete step-by-step answer:
We have been given that there are 12 girls and 18 boys in the class. So, there are 30 students in the class in total.
Let A be the event that both of the selected students are girls.
So, $n\left( A \right) = {}^{12}{C_2}$ as we have to select 2 girls out of 12 girls of the class.
Now, let S be the event of selecting two students from the class that is the sample space.
So, $n\left( S \right) = {}^{30}{C_2}$ as we have to select 2 students out of 30 students in the class.
Now, we have the number of favourable outcomes, that is n(A) and the sample space, that is n(S).
So, using the formula of probability we get,
${\text{Probability}} = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}$
${\text{Probability}} = \dfrac{{{}^{12}{C_2}}}{{{}^{30}{C_2}}} = \dfrac{{12!}}{{2! \times \left( {12 - 2} \right)!}} \times \dfrac{{2! \times \left( {30 - 2} \right)!}}{{30!}}$
$ \Rightarrow {\text{Probability}} = \dfrac{{12 \times 11}}{{30 \times 29}} = \dfrac{{22}}{{145}}$
Hence, the correct answer is A. $\dfrac{{22}}{{145}}$
Note- Whenever we face such types of problems the key point to remember is that we need to have a good grasp over probability and combinations. In these types of questions, we should always use combinations to compute the number of favourable outcomes and the sample space as it reduces the probability to make an error to solve the question.