 # Probability

## Class 12 NCERT

### NCERT

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 ,$P(A|B) =\dfrac{ P( A\cap B)}{ P(A)}\\ \implies P(B|A)= \dfrac{\dfrac{4}{11}}{\dfrac{6}{11}}\\ =\dfrac{4}{6}=\dfrac{2}{3}$

6   A coin is tossed three times, where$\\$ (i) E: head on third toss, F: heads on first two tosses$\\$ (ii) E: at least two heads, F: at most two heads$\\$ (iii)E: at most two tails, F: at least one tail