Question

Fill in the blanks in following table:
$$\begin{array}{ccccc}& P\left(A\right)& P\left(B\right)& P(A\cap B)& P(A\cup B)\\ \left(i\right)& \frac{1}{3}& \frac{1}{5}& \frac{1}{15}& \cdots \\ \left(ii\right)& 0.35& \cdots & 0.25& 0.6\\ \left(iii\right)& 0.5& 0.35& \cdots & 0.7\end{array}$$

Answer 1

$\text{(i) Here}P\left(A\right)=\frac{1}{3},P\left(B\right)=\frac{1}{5}\text{and}P(A\cap B)=\frac{1}{15}\text{We know that}P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)=\frac{1}{3}+\frac{1}{5}-\frac{1}{15}=\frac{5+3-1}{15}=\frac{7}{15}\left(ii\right)\text{Here}P\left(A\right)=0.35,P(A\cap B)=0.25\text{and}P(A\cup B)=0.6\text{We know that}P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)\therefore 0.6=0.35+P\left(B\right)-025\Rightarrow 0.6=0.1+P\left(B\right)\Rightarrow P\left(B\right)=0.6-0.1=0.5\text{(iii) Here}P\left(A\right)=0.5,P\left(B\right)=0.35\text{and}P(A\cup B)=0.7\text{We know that}P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)\therefore 0.7=0.5+0.35-P(A\cap B)\Rightarrow 0.7=0.85-P(A\cap B)\Rightarrow P(A\cap B)=0.85-0.7=0.15$