In a room filled with 7 people, 3 people have exactly 1 sibling in the room and 4 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 the correct option
A) 5/21	B) 3/7	C) 4/7	D) 5/7	E) 13/21
There are suppose A B C D 13/21 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 13/21FG are 4 people who have exactly 2 siblings....(13/21 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);(13/21FG)   Now first selecting 2 people out of 7 is 7C2=21   first sibling pair----(A and B)--selecting 2 people --2C2=1   second sibling pair (C and D)--selecting 2 people--2C2=1   third sibling pair (13/21 F G)--selecting 2 out of 4 --4C2=6   total= 1+1+6= 
<gadget id="calculator">1+1+6</gadget>
<output>8</output>
8   but,a/c to formula P(success)-1-p(fail)   here,p(failure)is selecting 2 people who are siblings=8/21(21 is 7C2)   =1-8/21   =13/21   ANS 13/21
<result>E</result>
In a room filled with 7 people, 3 people have exactly 1 sibling in the room and 4 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?