dear , friends
1. out of 2000 employees in an office 48% preferred coffee , 54 % liked tea and 64 % used to smoke (s ) out of total 28% C& t 32 % used t & s .... and 30 % preferred c & s only .... 6 % did none of these
1 ... the number having all three ??
2. no of emoployees having t & s but not c
3.. no of preferring coffee only ??
answers
1) 420
2) 220
3) 220
De Morgan’s law
n(A u B u C) = n(A) + n(B) + n(C) - n(A n B ) - n(B n C) - n(A n C) + n (A n B n C)
n(C u T u S) = n(C) + n(T) + n(S) - n(C n T) - n(T n S) - n(C n S) + n (C n T n S) -------- (1)
n(C) = 960
n(T) = 1080
n(S) = 1280
n(C n T) = 560
n(T n S) = 640
n(C n S) = 600
n(C c n T c n S c) = 60
n(C c n T c n S c) = n(C u T u S) c
total number of employees = n(C u T u S) + n(C u T u S) c ------------------------------(2)
substitute values in eqn (1) & (2), we get
2000 = 960 + 1080 + 1280 – 560 - 640 - 600 + n (C n T n S) + 60
n (C n T n S) = 2000 – 1580
the number having all three, n (C n T n S) = 420
2) no of emoployees having t & s but not c =
n(T n S) - n (C n T n S) = 640- 420
= 220
3) no of preferring coffee only =
n(C) - n(C n T) - n(C n S) + n (C n T n S) = 960 – 560 – 600 + 420
= 220
venn diagram for more explanation
CAclubindia
