# Fill in the blanks in following table:

Question:

Fill in the blanks in following table:

Solution:

(i) Here, $P(A)=\frac{1}{3}, P(B)=\frac{1}{5}, P(A \cap B)=\frac{1}{15}$

We know that $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$

$\therefore \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{1}{3}+\frac{1}{5}-\frac{1}{15}=\frac{5+3-1}{15}=\frac{7}{15}$

(ii) Here, $P(A)=0.35, P(A \cap B)=0.25, P(A \cup B)=0.6$

We know that $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$\therefore 0.6=0.35+\mathrm{P}(\mathrm{B})-0.25$

$\Rightarrow P(B)=0.6-0.35+0.25$

$\Rightarrow P(B)=0.5$

(iii) Here, $P(A)=0.5, P(B)=0.35, P(A \cup B)=0.7$

We know that $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$\therefore 0.7=0.5+0.35-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$

$\Rightarrow P(A \cap B)=0.5+0.35-0.7$

$\Rightarrow P(A \cap B)=0.15$