Sum of all M+P+C circles is given to be 0.75..(1)
Where each circle represents respective probability in passing.
The probability of passing in at least 2=0.5
The probability of passing in exactly 2=0.4
Therefore the probability of passing in exactly 3=0.5-0.4=0.1=mpc.
Now, Equation (1) can be restated as
$p+m+c-(mp+pc+mc)+mpc=0.75$............(2)
This is similar to
$n(A \cup B \cup C)=n(A)+n(B)+n(C)-(n(A \cap B)+n(B \cap C)+n(C \cap A))+n(A \cap B \cap C)$
Now to get blue area, which counts to 0.4, you would have taken $mp+pc+mc$ and to this subtracted the red part($mpc=0.1$), thrice,
hence to get mp+pc+mc, add $3 \times 0.1$ to 0.4
which gives
$mp+pc+mc=0.7$
plugging in values into (2) we get
$p+m+c=1.35=\frac{27}{20}$
And $mpc=0.1=p(m \cap p \cap c)=p \times m \times c$(A student passing or failing in subject is independent event).
So, III is true.
Ans- (D)