Three distinct numbers a, b and c are chosen at random from the numbers 1, 2, …, 100. The probability that
`{:(" List I"," List II"),("a. a, b, c are in AP is","p. "(53)/(161700)),("b. a, b, c are in GP is","q. "(1)/(66)),("c. "(1)/(a). (1)/(b). (1)/(c) " are in GP is ","r. "(1)/(22)),("d. a + b + c is divisible by 2 is","s. "(1)/(2)):}`

Options

A. Option a.`{:("a","b","c","d"),("q","s","s","r"):}`
B. Option b.`{:("a","b","c","d"),("r","q","q","p"):}`
C. Option c.`{:("a","b","c","d"),("q","p","p","s"):}`
D. Option d.`{:("a","b","c","d"),("q","s","p","r"):}`

Answer

C

Explanation

`n(S) = .^(100)C_(3)`
a. 2b = a + c = even
This means that a and c are both even or both odd.
`n(E) = .^(50)C_(2) + .^(50)C_(2) = 50 xx 49`
b. Taking r = 2, 3, …, 10,
a, b, c can be in GP in 53 ways.
c. `(1)/(a), (1)/(b), (1)/(c)` are in GP = a, b, c are in GP
d. P(a + b + c is even) = `((.^(50)C_(3)+.^(50)C_(1) xx .^(50)C_(2))/(.^(100)C_(3))) = (1)/(2)`