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
Correct Answer: C
Solution :
[c] Five tickets out of 50 can draw 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 n number of cases \[{{=}^{29}}{{C}_{2}}^{29}{{C}_{2}}\] Hence, required probability \[=\frac{^{20}{{C}_{2}}{{\times }^{29}}{{C}_{2}}}{^{52}{{C}_{5}}}\]
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