Statistics And Probability Question 147
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 $ (x_1<x_2<x_3<x_4<x_5). $ The probability that $ x_3=30 $ is
Options:
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
Show Answer
Answer:
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} $
BETA
