can any one help me to solve this
out of 2000 employees in an office 48 % like cofee 54 % like tea and 64 % like smoke , out of total 28 % used c and t , 32 % used t and s 30 % preferred c and s only 6 % did none of these . the number having all the three is ?
a. referred to question no of employees using t and s but not c?
b . no of employees prefer only coffee ?
P(C)=48%
P(T)=54%
P(S)=64%
P(C&T)=28%
P(T&S)=32%
P(C&S)=30%
P(NONE)=6%
To find the number of people who do all:
100-(48+54+64-28-32-30+6)=18
Therefore 18% have all three.
a) Using T & S but not C:
32+28+30-48=42
Therefore 42% use T & S but not C.
b) only coffee:
48+28+32+30-54-64=20
Therefore 20% prefer only coffee.
If you understood and were helped, "thank user"
| Originally posted by : Rahul Agarwal | ||
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P(C)=48% P(T)=54% P(S)=64% P(C&T)=28% P(T&S)=32% P(C&S)=30% P(NONE)=6% To find the number of people who do all: 100-(48+54+64-28-32-30+6)=18 Therefore 18% have all three. a) Using T & S but not C: 32+28+30-48=42 Therefore 42% use T & S but not C. b) only coffee: 48+28+32+30-54-64=20 Therefore 20% prefer only coffee. If you understood and were helped, "thank user" |
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thanx a lot
In the answer a) is wrong answer it is not 42%
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