Question

Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is : 

(1) 15/101 

(2) 5/101 

(3) 5/33 

(4) 10/99

Answer 1

(3) 5/33 

Out of 11 consecutive natural numbers either 6 even and 5 odd numbers or 5 even and 6 odd numbers 

When 3 numbers are selected at random then total cases = ¹¹C₃ 

Since these 3 numbers are in A.P. Let no's are a,b,c 

2b ⇒ even number 

a + c 

So favourable cases = ⁶C₂ + ⁵C₂ 

= 15 + 10 = 25 

P(3 numbers are in A.P. = 25/¹¹C₃ = 25/165 = 2/33)