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.