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}}), $ then the probability that $ {{x}_{3}}=30 $ is

• A. $ \frac{^{29}{{C}_{2}}{{\times }^{20}}{{C}_{2}}}{^{50}{{C}_{5}}} $
• B. $ \frac{^{30}{{C}_{1}}{{\times }^{29}}{{C}_{1}}}{^{50}{{C}_{5}}} $
• C. $ \frac{^{5}{{C}_{1}}{{\times }^{50}}{{C}_{2}}}{^{50}{{C}_{5}}} $
• D. $ \frac{^{50}{{C}_{2}}{{\times }^{29}}{{C}_{1}}}{^{50}{{C}_{5}}} $

Answer 1

Answer: (A)

Let E= Event of getting 5 tickets ascending order. Since, it is given that number 30 is on 3rd position. So we have to choose two numbers from 1 to 29 numbers and choose two numbers from 31 to 50
$ \therefore $ $ n(E){{=}^{29}}{{C}_{2}}{{\times }^{20}}{{C}_{2}} $
Total number of selection of 5 cards From 1 to 50 is,
$ n(S){{=}^{50}}{{C}_{5}} $
$ \therefore $ Required probability
$ =\frac{n(E)}{n(S)}=\frac{^{29}{{C}_{2}}{{\times }^{20}}{{C}_{2}}}{^{50}{{C}_{5}}} $