Question

A determinant of order 2 is of the form A = . It is equal to ad-bc. The total number of ways of choosing a, b, c and d is 2x2x2x2=16. Now ∆≠0 if and only if either ad = 1, bc = 0 or ad = 0, bc = 1. But ad = 1, bc = 0 iff a = d = 1 and one of b, c is zero. Therefore ad = 1, bc = 0 in three cases, similarly ad = 0, bc = 1 in three cases. Therefore the required probability = 6/16 = 3/8.

• A. k/999
• B. k/1000
• C. k-1/1000
• D. None

Answer 1

Answer: (B)

k/1000

Answer 2

Let A denote the event that the stranger succeeds at the k^th trial. Then

P(A’) =

⇒ P(A’) =