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

Solution :

If a coin is tossed three times, then the sample space $S$ is$\\$ $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$$\\$ It can be seen that the sample space has $8$ elements.$\\$ $(i)E = \{HHH, HTH, THH, TTH\}\\ F = \{HHH, HHT\}\\ \therefore E \cap F = \{HHH\}\\ P(F)=\dfrac{2}{8}=\dfrac{1}{4}$ and $ P(E \cap F)=\dfrac{1}{8} P(E|F)=\dfrac{P(E\cap F)}{P(F)}=\dfrac{\dfrac{1}{8}}{\dfrac{1}{4}}=\dfrac{4}{8}=\dfrac{1}{2}\\ (ii) E=\{HHH, HHT, HTH, THH\}\\ F = \{HHT, HTH, HTT, THH, THT, TTH, TTT\}\\ \therefore E \cap F = \{HHT, HTH, THH\}$ Clearly,$P(E \cap F)=\dfrac{3}{8}$ and $ P(F)=\dfrac{7}{8}\\ P(E|F)=\dfrac{P(E\cap F)}{P(F)}=\dfrac{\dfrac{3}{8}}{\dfrac{7}{8}}=\dfrac{3}{7}\\ (iii) E = \{HHH, HHT, HTT, HTH, THH, THT, TTH\}\\ F = \{HHT, HTT, HTH, THH, THT, TTH, TTT\}\\ \therefore E \cap F = \{HHT, HTT, HTH, THH, THT, TTH\}\\ P(F)=\dfrac{7}{8}$ and $ P(E \cap F) =\dfrac{6}{8}$$\\$ Therefore, $ P(E|F)=\dfrac{P(E\cap F)}{P(F)}=\dfrac{\dfrac{6}{8}}{\dfrac{7}{8}}=\dfrac{6}{7}$$\\$

7   Two coins are tossed once, where$\\$ (i) E: tail appears on one coin, F: one coin shows head$\\$ (ii) E: not tail appears, F: no head appears

Solution :

If two coins are tossed once, then the sample space $S$ is $\\$ $S = \{HH, HT, TH, TT\}\\ (i) E = \{HT, TH\}\\ F = \{HT, TH\}\\ \therefore E \cap F = \{HT, TH\}\\ P(F)=\dfrac{2}{8}=\dfrac{1}{4}\\ P(E\cap F)=\dfrac{2}{8}=\dfrac{1}{4}\\ \therefore P(E|F)=\dfrac{P(E\cap F)}{P(F)}=\dfrac{2}{2}=1\\ (ii)E=\{HH\}\\ F=\{TT\}\\ \therefore E\cap F=\phi \\ P(F=1) \text{and} P(E\cap F)=0\\ \therefore P(E|F)=\dfrac{P(E\cap F)}{P(F)}=\dfrac{0}{1}=0$

8   A die is thrown three times,$\\$ $E: 4$ appears on the third toss, $F: 6$ and $5$ appears respectively on first two tosses

Solution :

If $a$ die is thrown three times, then the number of elements in the sample space will be$ 6*6*6=216$$\\$ $E=\begin{Bmatrix} (1,14),(1,2,4),.....(1,6,4)\\[0.3em] ( 2,1, 4),( 2, 2, 4 ) ,.... ( 2, 6, 4 )\\[0.3em] (3,1, 4) ,( 3, 2, 4) ,....( 3, 6, 4)\\[0.3em] (4,1, 4 ) , (4, 2, 4 ) ,....(4, 6, 4)\\[0.3em] (5,1, 4) , (5, 2, 4) ,.... (5, 6, 4)\\[0.3em] ( 6,1, 4) , (6, 2, 4) ,.... (6, 6, 4 )\end{Bmatrix}$$\\$ $F = \{(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5),(6,5,6)\}\\ \therefore E \cap F = \{(6,5,4)\}\\ P(F)=\dfrac{6}{216} \text{and} P(E\cap F)=\dfrac{1}{216}\\ \therefore P(E|F)=\dfrac{P(E \cap F)}{P(F)}=\dfrac{\dfrac{1}{216}}{\dfrac{6}{216}}=\dfrac{1}{6}$

9   Mother, father and son line up at random for a family picture$\\$ E: son on one end, F: father in middle

Solution :

If mother $(M)$, father $(F)$, and son $(S)$ line up for the family picture, then the sample space will be$\\$ $S = \{MFS, MSF, FMS, FSM, SMF, SFM\}\\ \Rightarrow E = \{MFS, FMS, SMF, SFM\}\\ F = \{MFS, SFM\}\\ \therefore E \cap F = \{MFS, SFM\}\\ P(E \cap F)=\dfrac{2}{6}=\dfrac{1}{3}\\ P(F)=\dfrac{2}{6}=\dfrac{1}{3}\\ \therefore P(E|F)=\dfrac{P(E\cap F)}{P(F)}=\dfrac{\dfrac{1}{3}}{\dfrac{1}{3}}=1$

10   A black and a red dice are rolled.$\\$ a) Find the conditional probability of obtaining a sum greater than $9$, given that the black die resulted in a $5$.$\\$ b) Find the conditional probability of obtaining the sum $8$, given that the red die resulted in a number less than $4.$

Solution :

Let the first observation be from the black die and second from the red die.$\\$ When two dice (one black and another red) are rolled, the sample space $S$ has $6 * 6 = 36$ number of elements.$\\$ (a) Let$\\$ A: Obtaining a sum greater than $9= \{(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)\}$$\\$ B: Black die results in a $5 = \{(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\}\\ \therefore A \cap B=\{( 5, 5) , (5,6 )\}$$\\$ The conditional probability of obtaining a sum greater than $9$, given that the black die resulted in a $5$, is given by $P ( A | B )$$\\$ $\therefore P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{\dfrac{2}{36}}{\dfrac{6}{35}} =\dfrac{2}{6}=\dfrac{1}{3}$$\\$ .(b) E: Sum of the observations is $8$.$\\$ $= \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\}$$\\$ F: Red die resulted in a number less than $4$$\\$ $=\begin{Bmatrix} (1,1),(1,2),(1,3),(2,1),(2,2),(2,3),\\ (3,1),(3,2),(3,3),(4,1),(4,2),(4,3),\\ (5,1),(5,2),(5,3),(6,1),(6,2),(6,3)\end{Bmatrix}$$\\$ $\therefore E\cap F=\{(5,3),(6,2)\}\\ P(F)=\dfrac{18}{36}\text{and} P(E \cap F) =\dfrac{2}{36}$$\\$ The conditional probability of obtaining the sum equal to $8$, given that the red die resulted in a number less than $4$, is given by $P(E | F)$ .$\\$ Therefore , $ P(E|F)=\dfrac{P(E\cap F)}{P(F)}=\dfrac{\dfrac{2}{36}}{\dfrac{18}{36}}=\dfrac{2}{18}=\dfrac{1}{9}$