In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 3 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?	Choose the correct choice from the following:
A) 5/21
B) 3/7
C) 4/7
D) 5/7
E) 32/35
There are suppose A B C D 32/35 F G members in the room 4 people who have exactly one sibling....A B C D....(A is Bs∘SSsibl∈g∘SSand∘SSviceversa)∘SS(C∘SSis∘SSDs∘SSsibl∈g∘SSand∘SSviceversa)∘SS(C∘SSis∘SSDssibl∈gandviceversa)(CisDs sibling and viceversa) (C is Ds sibling and viceversa)...now remaning 32/35FG are 3 people who have exactly 3 siblings....(32/35 has F and G as his/her sibling and so on..)   there are now 3 different set of siblings (A and B)   (C and D);(32/35FG)   Now first selecting 3 people out of 7 is 7C3=35   first sibling pair----(A and B)--selecting 3 people --3C3=1   second sibling pair (C and D)--selecting 3 people--3C3=1   third sibling pair (32/35 F G)--selecting 3 out of 3 --3C3=1   total= 1+1+1= 
<gadget id="calculator">1+1+1</gadget>
<output>3</output>
3   but,a/c to formula P(success)-1-p(fail)   here,p(failure)is selecting 3 people who are siblings=3/35(35 is 7C3)   =1-3/35   =32/35   ANS 32/35
<result>E</result>
In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 3 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?