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? Pick one.
A) 5 / 21
B) 3 / 7
C) 4 / 7
D) 5 / 7
E) 13 / 21

Rationale: there are suppose a b c d e 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 efg are 4 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 ) ( c and d ) ; ( efg ) now first selecting 2 people out of 7 is 7 c 2 = 21 first sibling pair - - - - ( a and b ) - - selecting 2 people - - 2 c 2 = 1 second sibling pair ( c and d ) - - selecting 2 people - - 2 c 2 = 1 third sibling pair ( e f g ) - - selecting 2 out of 4 - - 4 c 2 = 6 total = 1 + 1 + 6 = 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 7 c 2 ) = 1 - 8 / 21 = 13 / 21 ans e
Answer: E