Match List I with List II.
Let A and B be events with $$P(A)=\frac{2}{3}, P(B)=\frac{1}{2}$$ and $$P(A\cap B)=\frac{1}{3}$$

(Here c stands for complement)
Choose the correct answer from the options given below:

Let the sample size of the above events be 6 units.

Since $$P(A)=\frac{2}{3}, P(B)=\frac{1}{2}$$ and $$P(A\cap B)=\frac{1}{3}$$,

Then event A will have 4 units($$\frac{2}{3}\times\ 6$$), event B will have 3 units and event $$A\cap B$$ will have 2 units.

This can be expressed in terms of the Venn diagram as follows:

Now, it can be observed that $$A\cap B^c\ =2$$ units.

Thus, $$P\left(A\cap B^c\right)\ =\frac{2}{6}=\frac{1}{3}$$.

Similarly, $$A\ U\ B^c\ =5$$ units($$A=4$$ and $$B^c-A=1$$ units)

So, $$P\left(A\ U\ B^c\right)\ =\frac{5}{6}$$

Since only option A has $$P\left(A\cap B^c\right)=\frac{1}{3}$$ and $$P\left(A\ U\ B^c\right)\ =\frac{5}{6}$$,

The correct option is A.