When you're thinking about probability, the common sense way to consider it is "what I want" / "total possible".

For this problem, the ultimate outcome that you want is to pick a sibling pair from among the juniors and seniors. So what has to happen for that outcome to occur?

First, from among the juniors you have to pick someone who has a sibling in the senior class. There are 60 of those, so the chances of that event are 60/1000, or 6/100

Next, you have to pick that peron's sibling from among the seniors. There is only going to be one person in the senior class who will be the sibling of whomever you picked from the junior class, so that will give you 1/800.

Multiply those together and you get 6/80,000. This reduces to 3/40,000.

Cybermusings wrote:A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of one junior and one senior student. if 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair.

1) 3/40,000

2) 1/3,600

3) 9/2000

4) 1/60

5) 1/15

\[\left( J \right)\,\,{\text{Junior}}\,\,:\,\,\,1000\,\,{\text{students}}\,\,,\,\,\,{J_1}\,,\,{J_2},\,\, \ldots \,\,,\,\,{J_{60}}\,\,{\text{among}}\,\,{\text{them}}\,\,\,\left( {{\text{the}}\,\,{\text{ones}}\,\,{\text{with}}\,\,{\text{senior}}\,\,{\text{siblings}}} \right)\]
\[\left( S \right)\,\,{\text{Senior}}\,\,:\,\,\,800\,\,{\text{students}}\,\,,\,\,\,{S_1}\,,\,{S_2},\,\, \ldots \,\,,\,\,{S_{60}}\,\,{\text{among}}\,\,{\text{them}}\,\,\,\left( {{\text{the}}\,\,{\text{ones}}\,\,{\text{with}}\,\,{\text{junior}}\,\,{\text{siblings}}} \right)\]
\[\left( {{J_1}\,,\,{S_1}} \right)\,\,;\,\,\left( {{J_2}\,,\,{S_2}} \right)\,\,;\,\, \ldots \,\,;\,\,\left( {{J_{60}}\,,\,{S_{60}}} \right)\,\,\,:\,\,\,{\text{pairs}}\,\,{\text{of}}\,\,{\text{siblings}}\,\]

\[? = P\left( {{\text{pair}}\,{\text{of}}\,{\text{siblings}}\,,\,\,{\text{in}}\,{\text{one}}\,J\,{\text{and}}\,{\text{one}}\,S\,{\text{extraction}}} \right)\]

\[{\text{Total}}\,\,:\,\,\,1000 \cdot 800\,\,\,{\text{equiprobables}}\,\,\,\left[ {\left( {{J_m},{S_n}} \right)\,\,,\,\,\,{\text{where}}\,\,1 \leqslant m \leqslant 1000\,\,{\text{and}}\,\,\,1 \leqslant n \leqslant 800} \right]\]
\[{\text{Favorable}}\,\,:\,\,60\,\,\,\,\,\left[ {\left( {{J_k},{S_k}} \right)\,\,,\,\,\,{\text{where}}\,\,1 \leqslant k \leqslant 60} \right]\,\,\,\,\]
\[? = \frac{{60}}{{1000 \cdot 800}} = \underleftrightarrow {\frac{{4 \cdot 15}}{{1000 \cdot 4 \cdot 200}}} = \frac{{4 \cdot 3 \cdot 5}}{{1000 \cdot 4 \cdot 5 \cdot 40}} = \frac{3}{{40 \cdot 1000}}\]


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.