Statistics And Probability Question 123
Question: $ x_1,x_2,x_3,…x_{50} $ are fifty real numbers such that $ x_{r}<{x_{r+1}} $ for $ r=1,2,3….49. $ Five numbers out of these are picked up at random. The probability that the numbers have $ x_{20} $ as the middle numbers, is
Options:
A) $ \frac{^{20}C_2{{\times }^{30}}C_2}{^{50}C_5} $
B) $ \frac{^{30}C_2{{\times }^{19}}C_2}{^{50}C_5} $
C) $ \frac{^{19}C_2{{\times }^{31}}C_2}{^{50}C_5} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] $ n(S){{=}^{50}}C_5,n(E){{=}^{30}}C_2{{\times }^{19}}C_2 $
$ \therefore P(E)=\frac{^{30}C_2{{\times }^{19}}C_2}{^{50}C_5}. $
BETA
