The question is:
There are 1000 junior and 800 senior students in a class.And there are 60 sibling pairs where each pair has 1 junior and 1 senior. One student is chosen from senior and 1 from junior randomly.What is the probability that the two selected students are from a sibling pair?
My attempt:
The total number of ways to select students will be $=^{1000}C_1*^{800}C_1$ .
Number of ways for a student to be selected from senior and one from junior
$=^{60}C_1*^{60}C_1$.
Answer: $\frac{3600}{800000}$.
But the answer given in the book and the website is: $\frac{7140}{800000}$.
What is the mistake in my calculation?
The wording of the question is not as clear as it could be, so I will rewrite the possible interpretations and then address why the answer on the website does not make sense.
This means that each element of our sample space has the form $(j, s)$, where $j$ is a junior student and $s$ is a senior student. We have $1000$ choices for the junior student and $800$ choices for the senior student. Hence, there are $$\binom{1000}{1}\binom{800}{1} = 1000 \cdot 800 = 800000$$ such pairs of junior students and senior students in our sample space, as you found.
There are $60$ pairs of siblings. Hence, the probability that the selected students are siblings is $$\frac{60}{800000}$$
This is the question that DavidP answered correctly.
There are $60$ junior students who have a sibling among the senior students and $60$ senior students who have a sibling among the junior students, so there are $60 \cdot 60 = 3600$ favorable cases. Hence, the probability that the randomly selected junior student has a sibling who is a senior student and the randomly selected senior student has a sibling who is a junior student is $$\frac{3600}{800000}$$ This is the question you answered correctly.
Why is the answer posted on the linked website incorrect?
The author(s) of that answer took their sample space to be the same one as above.
However, in their numerator, they answered a different question:
Since there are $60$ junior students who have a sibling among the senior students and $60$ senior students who have a sibling among the junior students, the author(s) of that post selected two students among the $60 + 60 = 120$ students who have a sibling in the class, giving $$\binom{120}{2} = 7140$$ favorable cases.
However, in doing so, they overlooked the requirement that the randomly selected students include one junior student and one senior student. Therefore, there count includes $\binom{60}{2}$ selections in which both students are junior students and $\binom{60}{2}$ selections in which both students are senior students in addition to the $\binom{60}{1}\binom{60}{1}$ selections in which one student is a junior student and one student is a senior student.
$$\color{red}{\binom{60}{2}} + \color{red}{\binom{60}{2}} + \binom{60}{1}\binom{60}{1} = \color{red}{1770} + \color{red}{1770} + 3600 = \color{red}{3540} + 3600 = 7140$$ Therefore, they included $3540$ pairs of students who violate the condition that one randomly selected student is a junior student and one randomly selected student is a senior student.
If they wanted to find the probability that two students selected randomly from the junior students and senior students each have a sibling among the junior students and senior students, they should have taken as their denominator the $$\binom{1000 + 800}{2} = \binom{1800}{2}$$ ways of selecting a pair of students from among the $1000$ junior students and $800$ senior students in the class.