Out of 8 doctors, 6 arespecialists and 2 are MBBS. 3 doctors are selected in random. Find prob of distribution of the of the number of specialist doctors ?

							There are three possibilities to select 3 people : 
(1) 1 is a specialist, 2 are non-specialists (X=1).
(2) 2 are specialists, 1 is non-specialist (X=2).
(3) all 3 are specialists (X=3).

Also, not that X=0 is not a possibility as there can't be a case that all 3 selected are non-specialists, as there are only 2 non-specialists. So, the probability distribution of specialists is as follows:-

$P$(X=0)=0
$P$(X=1)=68×12×11=38
$P$(X=2)=68×57×12=1556
$P$(X=3)=68×57×46=514
Thanks
Bharat Bajaj
IIT Delhi
						

							
There are three possibilities to select 3 people :
(1) 1 is a specialist, 2 are non-specialists (X=1).
(2) 2 are specialists, 1 is non-specialist (X=2).
(3) all 3 are specialists (X=3).

Also, not that X=0 is not a possibility as there can't be a case that all 3 selected are non-specialists, as there are only 2 non-specialists. So, the probability distribution of specialists is as follows:-

$P$(X=0)=0
$P$(X=1)=68×12×11=38
$P$(X=2)=68×57×12=1556
$P$(X=3)=68×57×46=514
Thanks
Bharat Bajaj
IIT Delhi