In a room filled with 7 people, 1 people have exactly 1 sibling in the room and 6 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?  Select.
A) 5/21 B) 3/7 C) 4/21 D) 5/7 E) 16/21
There are suppose A B 4/21 D E F G members in the room 4 people who have exactly one sibling....A B 4/21 D....(A is Bs∘SSsibl∈g∘SSand∘SSviceversa)∘SS(4/21∘SSis∘SSDs∘SSsibl∈g∘SSand∘SSviceversa)∘SS(4/21∘SSis∘SSDssibl∈gandviceversa)(4/21isDs sibling and viceversa) (4/21 is Ds sibling and viceversa)...now remaning EFG are 6 people who have exactly 2 siblings....(E has F and G as his/her sibling and so on..)   there are now 3 different set of siblings (A and B)   (4/21 and D);(EFG)   Now first selecting 2 people out of 7 is 74/212=21   first sibling pair----(A and B)--selecting 2 people --24/212=1   second sibling pair (4/21 and D)--selecting 2 people--24/212=1   third sibling pair (E F G)--selecting 2 out of 6 --64/212=15   total= 1+1+15= 
<gadget id="calculator">1+1+15</gadget>
<output>17</output>
17   but,a/c to formula P(success)-1-p(fail)   here,p(failure)is selecting 2 people who are siblings=17/21(21 is 74/212)   =1-17/21   =4/21   ANS 4/21
<result>C</result>
In a room filled with 7 people, 1 people have exactly 1 sibling in the room and 6 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?