In a room filled with 7 people, 2 people have exactly 1 sibling in the room and 5 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? Pick one:  A) 5/21  B) 3/7  C) 4/7  D) 5/7  E) 16/21
There are suppose A 3/7 C D E F G members in the room 4 people who have exactly one sibling....A 3/7 C D....(A is 3/7s∘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 EFG are 5 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 3/7)   (C and D);(EFG)   Now first selecting 2 people out of 7 is 7C2=21   first sibling pair----(A and 3/7)--selecting 2 people --2C2=1   second sibling pair (C and D)--selecting 2 people--2C2=1   third sibling pair (E F G)--selecting 2 out of 5 --5C2=10   total= 1+1+10= 
<gadget id="calculator">1+1+10</gadget>
<output>12</output>
12   but,a/c to formula P(success)-1-p(fail)   here,p(failure)is selecting 2 people who are siblings=12/21(21 is 7C2)   =1-12/21   =9/21   =3/7 ANS 3/7
<result>B</result>
In a room filled with 7 people, 2 people have exactly 1 sibling in the room and 5 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?