In a survey of 1000 person it was found that  (1) 600 drink,(2) 720 smoke,(3) 560 chew,
    (4) 380 drink and smoke, (5)270 drink and chew,(6)350 smoke and chew,(7)80 drink, smoke  and chew.
    what is the probability that a person  is (a) do not drink or smoke (2) drink, smoke, but do not chew?

You have these subsets

    (1)  D,   n(D)  = 600;
    (2)  S,   n(S)  = 720;
    (3)  C,   n(C)  = 560;
    (4)  DS,  n(DS) = 380;
    (5)  DC,  n(DC) = 270;
    (6)  SC,  n(SC) = 350;
    (7)  DSC, n(DSC) = 80.       // All abbreviations are OBVIOUS and SELF-EXPLANATORY.


Then you can find

     n(D U S U C) = n(D) + n(S) + n(C) - n(DS) - n(DC) - n(SC) + n(DSC) = 

                  = 600 + 720 + 560 - 380 - 270 - 350 + 80 = 960.


It is the number of those who is ether D, or S, or C.

Hence, the number of those who is NEITHER D, NOR S, NOR C is 1000-960 = 40.



Now I am in position to answer the questions.



(a)  Now, the number of those who do not drink or smoke is 

     n(D U S U C) - n(D) - n(S) + n(DS) + 40 = 960 - 600 - 720 + 380 + 40 = 60.

     Hence, the probability (a) is   =  =  = 0.06 = 6%.    ANSWER



(b)  The number of those who D, S but not C is  n(DS) - n(DSC) = 380 - 40 = 340,

     and the probability (b) is   =  =  = 0.34 = 34%.      ANSWER