In class 10A, there are 20 boys and 20 girls. In 10B, there are 15 boys and 25 girls. One student is to be selected from each class.
(i) What is the probability of both being girls?
(ii) What is the probability of both being boys?
(iii) What is the probability of one boy and one girl?

Hint: In this problem, first we need to find the total no of students in the class.
In the first part, find the ratio of total no of girls to the total no of students.
In the second part, find the ratio of total no of boys to the total no of students.
In the third part, find the ratio of no of boys to the total no of students in first class, and add the ratio of no of girls to the total no of students in second class.

Complete step-by-step solution -
When the two events \[A\] and \[B\] are disjoint, then, the probability of happening either \[A\] or \[B\] is the sum of their individual probabilities.
Total no of students is obtained as follows:
  ${\text{Total no of students = 20 + 20 + 15 + 25}} \\ $
 $ {\text{Total no of students}} = 80 \\ $
(i) The probability of both students being girl is as follows:
${\text{The probability of both students being girl}} = \dfrac{{20 + 25}}{{80}} \\ $
${\text{The probability of both students being girl}} = \dfrac{{45}}{{80}} \\ $
${\text{The probability of both students being girl}} = \dfrac{9}{{16}} \\ $
The probability of both students being girls is \[\dfrac{9}{{16}}\].
(ii) The probability of both students being boy is as follows:
 ${\text{The probability of both students being boy}} = \dfrac{{20 + 15}}{{80}} \\ $
 ${\text{The probability of both students being boy}} = \dfrac{{35}}{{80}} \\ $
 ${\text{The probability of both students being boy}} = \dfrac{7}{{16}} \\ $
The probability of both students being boys is \[\dfrac{7}{{16}}\].
(iii) The probability of one boy and one girl is as follows:
  $\text{The probability of one boy and one girl}$ = $\left( {{\text{one boy }}\left( {10A} \right){\text{and one girl }}\left( {{\text{10B}}} \right)} \right) \\ + \left( {{\text{one boy }}\left( {{\text{10B}}} \right){\text{ and one girl }}\left( {{\text{10A}}} \right)} \right) \\ $
$\text{The probability of one boy and one girl}$ = $ \left[ {\dfrac{{20}}{{40}} \times \dfrac{{25}}{{40}} + \dfrac{{20}}{{40}} \times \dfrac{{15}}{{40}}} \right] \\ $
$\text{The probability of one boy and one girl} =\dfrac{{500}}{{1600}} + \dfrac{{300}}{{1600}} \\ $
 $\text{probability of one boy and one girl}$ =$ \dfrac{800}{1600} $
 $\text{probability of one boy and one girl}$ =$ \dfrac{1}{2} $
Thus, the probability of one boy and one girl is $\dfrac{1}{2}$.

Note: In third subpart, the probability of one boy and one girl is obtained by considering following two cases:
(a) Boy from 10 A and girl from 10B $\left( {{}^{20}{C_1} \times {}^{25}{C_1}} \right) $.
(b) Boy from 10 B and girl from 10A $ \left( {{}^{20}{C_1} \times {}^{15}{C_1}} \right) $.