Suppose n (> 6) people are asked successively in a random order and exactly 3 out of n people know the answer. Let p, denotes the probability that rth person asked is the first to know the answer, then 

	COLUMN - I		COLUMN - II
A)	Probability first four do not know the answer	p)	$\frac{3(\mathrm{n}-3)}{\mathrm{n}(\mathrm{n}-1)}$
B)	P2	q)	$\frac{(\mathrm{n}-4)(\mathrm{n}-5)(\mathrm{n}-6)}{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}$
C)	Pn–2	r)	$\frac{1}{{ }^{\mathrm{n}}{\mathrm{C}}_3}$
D)	Pr	s)	$\frac{3(\mathrm{n}-\mathrm{r})(\mathrm{n}-\mathrm{r}-1)}{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}$

	(A)	(B)	(C)	(D)
1)	q	p	r	s
2)	p	q	r	s
3)	s	p	r	q
4)	r	p	q	s

n people (n > 6) are asked a question one by one and out of n exactly 3 know the answer. P_r = Probability that r^th person is the 1st person to
know the answer. A) Probability that first four do not know the answer [n - 3 people do not know the answer]
$=\frac{\left({ }^{\mathrm{n}-3}{\mathrm{C}}_4\right)}{{ }^{\mathrm{n}}{\mathrm{C}}_4}[\because \mathrm{n}-3$people do not know the answer]
$=\frac{(\mathrm{n}-3)(\mathrm{n}-4)(\mathrm{n}-5)(\mathrm{n}-6)}{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}=\frac{(\mathrm{n}-4)(\mathrm{n}-5)(\mathrm{n}-6)}{\left(\mathrm{n}\right)(\mathrm{n}-1)(\mathrm{n}-2)}$
B) P₂ = P (First person do not know the answer & second person know the answer) 
$=\left[\frac{\left({ }^{\mathrm{n}-3}{\mathrm{C}}_1\right)}{\left({ }^{\mathrm{n}}{\mathrm{C}}_1\right)}\times \frac{{ }^3{\mathrm{C}}_1}{\left({ }^{\mathrm{n}-1}{\mathrm{C}}_1\right)}\right]=\frac{3(\mathrm{n}-3)}{\mathrm{n}(\mathrm{n}-1)}$
C) P_n–2 = P(First n-3 do not know the answer) ×
P((n–2)^th person know the answer which is sure event)
$=\left[\frac{\left({ }^{\mathrm{n}-3}{\mathrm{C}}_{\mathrm{n}-3}\right)}{\left({ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-3}\right)}\times \left(\frac{3}{3}\right)\right]=\frac{1}{\left({ }^{\mathrm{n}}{\mathrm{C}}_3\right)}$
D) P_r = P [First (r-1) persons do not know the answer and r^th person know the answer]
$=\left[\frac{\left({ }^{\mathrm{n}-3}{\mathrm{C}}_{\mathrm{r}-1}\right)}{\left({ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}-1}\right)}\times \frac{3}{(\mathrm{n}-\mathrm{r}+1)}\right]=\frac{3(\mathrm{n}-\mathrm{r})(\mathrm{n}-\mathrm{r}-1)}{\left(\mathrm{n}\right)(\mathrm{n}-1)(\mathrm{n}-2)}$