We draw a Venn diagram with 3 sets ( the 3 circles) and a universal set (the big rectangle).  
We let the letters d through k indicate the probabilities of the 8 regions:

   

Since the probability of something in the Universal set occurring 
is 1, referring to the above Venn diagram:

p(U) = 1.0 = d+e+f+g+h+i+j+k
p(A) = 0.5 = d+e+g+h 
p(B) = 0.6 = e+f+h+i 
p(C) = 0.4 = g+h+i+j
p(AnB) = 0.3 = e+h 
p(AnC) = 0.1 = g+h
p(BnC) = 0.2 = f+i
p(AnBnC) = 0.05 = h

So we have this system of equations:
(1)     d+e+f+g+h+i+j+k = 1.0
(2)     d+e  +g+h       = 0.5 
(3)       e+f  +h+i     = 0.6 
(4)           g+h+i+j   = 0.4
(5)       e    +h       = 0.3 
(6)           g+h       = 0.1
(7)             h+i     = 0.2
(8)             h       = 0.05 

Use (8) to substitute 0.05 for h in (7) to get i = 0.15,
and in (6) to get g = 0.05, and in (5) to get e = 0.25.

Substitute for g,h and i in (4) to get j = 0.15
Substitute for e,h and i in (3) to get f = 0.15
Substitute for e,g and h in (2) to get d = 0.15
Substitute for d,e,f,g,h,i,j in (1) to get k = 0.05

So we have:

    

(1) p(AnBnC') = e = 0.25
(2) p(AuBuC) = h = 0.05  <--that was given!!!
(3) p(A'nBnC') = f = 0.15

Edwin

To calculate  P(A n B n C'), notice that  (A n B n C')  are those elements  (events)  of (A n B) that do not belong to C;

in other words,  (A n B n C') = (A n B) \ (A n B n C),      (1)

where the sign " \ " denotes subtraction of sets (of subsets).


Then equality (1) implies that

    P(A n B n C') = P(A n B) - P(A n B n C) = (substitute the given data) = 0.3 - 0.05 = 0.25.        ANSWER

Use the formula

    P(A U B U C) = P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + P(A n B n C)     (2)

which is valid for any subsets  A, B and C  of the universal set U.


    This formula is very well known in elementary set theory.

    If you do not know its proof, here it is in few lines:

    
        Without my explanations, you understand well why I added P(A), P(B) and P(C) in the right side of the formula (2).

        But when I added these terms, I counted the intersections P(AnB), P(AnC)  and P(BnC) twice - so I need to subtract 
        these terms in the right side of the formula (2).

        But again,  when I added the terms P(A), P(B) and P(C) in the right side of the formula (2),
        I counted the intersection P(A n B n C) thrice. When I later subtracted the terms P(AnB), P(AnC)  and P(BnC), I fully compensate it,
        but now this intersection is not covered in the formula at all.
      
        Therefore I must add the term P(A n B n C), and this last step makes everything in equilibrium / (in balance).  

        The proof is completed.



    Now, when the formula (2) is proved, simply substitute the given input data in it. You will get

        P(A U B U C) = 0.5 + 0.6 + 0.4 - 0.3 - 0.1 - 0.2 + 0.05 = 0.95      ANSWER

To calculate  P(A' n B n C'), notice that  (A' n B n C')  are those elements  (events)  of B that do belong NEITHER  A  NOR  C;

in other words,  (A' n B n C') = (B \ (A n B) \ (C n B))    (3)



Then equality (2) implies that

    P(A' n B n C') = P(B \ (A n B) \ (C n B)) = (P(B) - P(AnB)) + (P(B) - P(CnB)) + P(A n B n C)     (4)


Notice that in the right side of the formula (4), I subtracted the probability P(AnB) from P(B), then subtracted P(CnB) from P(B), 
so I subtracted the probability P(AnC) of the intersection (AnC) twice; therefore, I should add P(A n B n C) to restore the equilibrium.


Substitute the given data into the formula (4), you get  the ANSWER

    P(A' n B n C') =   (0.6  - 0.3   - 0.2)   +   0.05       = 0.15.    (5)