$$
\begin{tabular}{|l|l|l|l|l|}
\hline & $\mathrm{P}(\mathrm{A})$ & $\mathrm{P}(\mathrm{B})$ & $\mathrm{P}(\mathrm{A} \cap \mathrm{B})$ & $\mathrm{P}(\mathrm{AUB})$ \\
\hline $\text { (i) }$ & $\frac{1}{3}$ & $\frac{1}{5}$ & $\frac{1}{15}$ & \ldots \ldots \ldots \\
\hline $\text { (ii) }$ & $0.35$ & \ldots \ldots \ldots & 0.25 & $0.6$ \\
\hline $\text { (iii) }$ & $0.5$ & $0.35$ & \ldots \ldots \ldots & $0.7$ \\
\hline
\end{tabular}
$$

(i) By definition of P (A or B) under axiomatic approach we can write,

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

By using data from table, we get:

∴ P (A ∪ B) = 1/3 + 1/5 – 1/15

= 8/15 – 1/15

= 7/15

(ii) By definition of P (A or B) under axiomatic approach we write,

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

P (B) = P (A ∪ B) + P (A ∩ B) – P (A)

By using data from table, we get:

∴ P (B) = 0.6 + 0.25 – 0.35

= 0.5

(iii) By definition of P (A or B) under axiomatic approach we can write,

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

P (A ∩ B) = P (B) + P (A) – P (A ∪ B)

By using data from table, we get:

∴ P (A ∩ B) = 0.5 + 0.35 – 0.7

= 0.15

Hence the table is:

$$\begin{tabular}{|l|l|l|l|l|}
\hline & $\mathrm{P}(\mathrm{A})$ & $\mathrm{P}(\mathrm{B})$ & $\mathrm{P}(\mathrm{A} \cap \mathrm{B})$ & $\mathrm{P}(\mathrm{AUB})$ \\
\hline (i) & $\frac{1}{3}$ & $\frac{1}{5}$ & $\frac{1}{15}$ & $\frac{7}{15}$ \\
\hline (ii) & $0.35$ & $\mathbf{0 . 5}$ & $0.25$ & $0.6$ \\
\hline (iii) & $0.5$ & $0.35$ & $\mathbf{0 . 1 5}$ & $0.7$ \\
\hline
\end{tabular}$$