Question
A bag contains $$50$$ tickets numbered $$1,\,2,\,3,\,.....,\,50$$ of which five are drawn at random and arranged in ascending order of magnitude $$\left( {{x_1} < {x_2} < {x_3} < {x_4} < {x_5}} \right).$$ The probability that $${x_3} = 30$$ is :
A.
$$\frac{{{}^{20}{C_2}}}{{{}^{50}{C_5}}}$$
B.
$$\frac{{{}^2{C_2}}}{{{}^{50}{C_5}}}$$
C.
$$\frac{{{}^{20}{C_2} \times {}^{29}{C_2}}}{{{}^{50}{C_5}}}$$
D.
none of these
Answer :
$$\frac{{{}^{20}{C_2} \times {}^{29}{C_2}}}{{{}^{50}{C_5}}}$$
Solution :
Five tickets out of $$50$$ can drawn in $${{}^{50}{C_5}}$$ ways.
Since $${{x_1} < {x_2} < {x_3} < {x_4} < {x_5}}$$ and $${x_3} = 30,\,{x_1},\,{x_2} < 30,$$ i.e., $${x_1}$$ and $${x_2}$$ should come from tickets numbered $$1$$ and $$29$$ and this may happen in $${{}^{29}{C_2}}$$ ways.
Remaining ways, i.e., $${x_4},\,{x_5} > 30,$$ should come from $$20$$ tickets numbered $$31$$ to $$50$$ in $${{}^{20}{C_2}}$$ ways.
So, favourable number of cases $$ = {}^{29}{C_2}{}^{29}{C_2}$$
Hence, required probability $$ = \frac{{{}^{20}{C_2} \times {}^{29}{C_2}}}{{{}^{50}{C_5}}}$$