Question

(i) Fill in the blanks in following table:
$\begin{array}{ccccc}& P\left(A\right)& P\left(B\right)& P(A\cap B)& P(A\cup B)\\ \left(i\right)& \frac{1}{3}& \frac{1}{5}& \frac{1}{15}& \dots \end{array}$

(ii) Fill in the blanks in following table:
$\begin{array}{ccccc}& P\left(A\right)& P\left(B\right)& P(A\cap B)& P(A\cup B)\\ \left(ii\right)& 0.35& ---& 0.25& 0.6\end{array}$

(iii) Fill in the blanks in following table:
$\begin{array}{ccccc}& P\left(A\right)& P\left(B\right)& P(A\cap B)& P(A\cup B)\\ \left(iii\right)& 0.5& 0.35& ---& 0.7\end{array}$

• A. 0.5
• B. 0.15
• C. \dfrac{7}{15}

Answer 1

Answer: (A), (B), (C)

(i) It is given that
$P\left(A\right)=\frac{1}{3},P\left(B\right)=\frac{1}{5},P(A\cap B)=\frac{1}{15}$ ...(i)
As we know that,
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$ ......(ii)
Substituting the values (i) in (ii)
$P(A\cup B)=\frac{1}{3}+\frac{1}{5}-\frac{1}{15}$

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

(ii) It is given that
$P\left(A\right)=0.35,P(A\cap B)=0.25,P(A\cup B)=0.6$ ...(i)
As we know that,
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$ ......(ii)
Substituting the values (i) in (ii)
$0.6=0.35+P\left(B\right)-0.25$
$0.6=0.10+P\left(B\right)$
$P\left(B\right)=0.6-0.10=0.5$
Hence, $P\left(B\right)=0.5$

(iii) It is given that
$P\left(A\right)=0.5,P\left(B\right)=0.35,P(A\cup B)=0.7$ ...(i)
As we know that,
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$ ......(ii)
Substituting the values (i) in (ii)
$0.7=0.5+0.35-P(A\cap B)$
$0.7=0.85-P(A\cap B)$
$P(A\cap B)=0.85-0.7=0.15$
Hence, $P(A\cap B)=0.15$