1   Given that E and F are events such that $P ( E )= 0.6, P ( F )= 0.3$ and, find $P ( E | F )$ and $P ( F | E ) .$

Solution :

It is given that $P ( E )= 0.6, P ( F )= 0.3 $, and $P ( E \cap F)0.2\\ \implies P(E|F) =\dfrac{P(E\cap F)}{P(F)} =\dfrac{0.2}{0.3}=\dfrac{2}{3}\\ \implies P(F|E) =\dfrac{P(E \cap F)}{P(F)} =\dfrac{0.2}{0.6} =\dfrac{1}{3}$

2   Compute $P ( A | B ) ,$$\\$ if $P ( B )= 0.5$ an d $P ( A \cap B )= 0.32$

Solution :

It is given that $P(B)=0.5 $ and $ P(A \cap B)=0.32\\ \implies P(\dfrac{A}{B}) =\dfrac{P(A \cap B)}{P(B)} =\dfrac{0.32}{0.5}=\dfrac{16}{25}$

3   If $ P ( A )= 0.8, P ( B )= 0.5$ and $P ( B | A )= 0 . 4 ,$ find$\\$ $(i) P ( A \cap B )\\ (ii) P ( A | B )\\ (iii) P ( A \cup B )$

Solution :

It is given that $P ( A )= 0.8, P ( B )= 0.5 $and $P (B | A )= 0 . 4$ $\\$ $ (i) P(A \cap B) = 0.4 \\ \therefore \dfrac{P(A \cap B)}{P(A)} =0.4\\ \implies \dfrac{P(A\cap B)}{0.8} =0.4\\ \implies P(A\cap B) =0.32$$\\$ $(ii) P(A|B) =\dfrac{P(A \cap B)}{P(B)}\\ \implies P(A|B)=\dfrac{0.32 }{0.5} =0.64$ $\\$ $ (iii) P(A \cup B) =P(A)+ P(B )- P(A \cap B)\\ \implies P(A \cup B) = 0.8+0.5-0.32=0.98$

4   Evaluate $ P(A \cup B),$ if $2P (A) = P(B) =\dfrac{5}{13}$ and $ P(A|B)=\dfrac{2}{5}$

Solution :

It is given that,$2P(A)=P(B)=\dfrac{5}{13}\\ \implies P(A) =P(B)=\dfrac{5}{13}\\ P(A|B)=\dfrac{2}{5}\\ \implies \dfrac{P(A\cap B)}{P(B)} =\dfrac{2}{5}\\ \implies P( A\cap B)=\dfrac{2}{5}* P(B)=\dfrac{2}{5}*\dfrac{5}{13} =\dfrac{2}{13}$$\\$ It is known that,$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$\\$ $\implies P(A\cup B) =\dfrac{5}{26} +\dfrac{5}{13}-\dfrac{2}{13}\\ \implies P(A\cup B) =\dfrac{5+10-4}{26}\\ P(A\cup B)=\dfrac{11}{26}$

5   If $ P(A)=\dfrac{6}{11},P(B)=\dfrac{5}{11}$ and $ P(A \cup B)=\dfrac{7}{11},$ find $\\$ $(i) P(A\cap B)\\ (ii) P(A|B)\\ (iii) P(B|A)$

Solution :

It is given that $P( A) =\dfrac{6}{11} ,P(B)=\dfrac{5}{11}$ and $ P(A\cup B) =\dfrac{7}{11}$$\\$ $(i) P(A \cup B)=\dfrac{7}{11}\\ \therefore A(A)+ P(B)-P(A\cap B)=\dfrac{7}{11} \\ \implies \dfrac{6}{11}+\dfrac{ 5}{11}- P(A\cap B)=\dfrac{ 7}{11}\\ \implies P(A\cap B)=\dfrac{11}{11}-\dfrac{7}{11}=\dfrac{4}{11}$$\\$ (ii) It is known that ,$P(A|B) =\dfrac{P(A\cup B)}{P(B)}$$\\$ $\implies P( A|B)=\dfrac{\dfrac{4}{11}}{\dfrac{5}{11}}=\dfrac{4}{5}$$\\$ (iii) It is known that ,$ P(A|B) =\dfrac{ P( A\cap B)}{ P(A)}\\ \implies P(B|A)= \dfrac{\dfrac{4}{11}}{\dfrac{6}{11}}\\ =\dfrac{4}{6}=\dfrac{2}{3}$