NCERT Solutions Class 12 Maths Exercise 13.1

Q.1

\begin{array}{l}\text{G}\text{i}\text{v}\text{e}\text{n}\text{t}\text{h}\text{a}\text{t}\text{E}\text{a}\text{n}\text{d}\text{F}\text{a}\text{r}\text{e}\text{e}\text{v}\text{e}\text{n}\text{t}\text{s}\text{s}\text{u}\text{c}\text{h}\text{t}\text{h}\text{a}\text{t}\text{P}\left(\text{E}\right)=0.6,\text{P}\left(\text{F}\right)=0.3\\ \text{a}\text{n}\text{d}\text{P}(\text{E}\cap \text{F})=0.2,\text{f}\text{i}\text{n}\text{d}\text{P}\left(\text{E}|\text{F}\right)\text{a}\text{n}\text{d}\text{P}\left(\text{F}|\text{E}\right).\end{array}

Ans

$\begin{array}{l}\mathrm{Since},\mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{P}\left(\mathrm{F}\right)}\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{0.2}{0.3}\\ \hspace{0.17em}\hspace{0.17em}=\frac{2}{3}\\ \mathrm{And}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{P}\left(\mathrm{F}|\mathrm{E}\right)=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{P}\left(\mathrm{E}\right)}\\ \hspace{0.17em}\hspace{0.17em}=\frac{0.2}{0.6}\\ \hspace{0.17em}\hspace{0.17em}=\frac{2}{6}=\frac{1}{3}\end{array}$

Q.2

Ans

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{P}(\mathrm{B})\; =\; 0.5\mathrm{and}\mathrm{P}(\mathrm{A}\cap \mathrm{B})\; =\; 0.32\\ \mathrm{So},\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{B}\right)}\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{0.32}{0.5}\\ \hspace{0.17em}\hspace{0.17em}=\frac{16}{25}\end{array}$

Q.3

$\begin{array}{l}\mathrm{If}\mathrm{}\mathrm{P}\left(\mathrm{A}\right)=0.8,\mathrm{P}\left(\mathrm{B}\right)=0.5\mathrm{and}\mathrm{}\mathrm{P}\left(\mathrm{B}|\mathrm{A}\right)=0.4,\mathrm{}\mathrm{find}\\ \left(\mathrm{i}\right)\mathrm{P}(\mathrm{A}\cap \mathrm{B})\left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)\left(\mathrm{iii}\right)\mathrm{P}(\mathrm{A}\cup \mathrm{B})\end{array}$

Ans

$\begin{array}{l}\mathrm{ItisgiventhatP}\left(\mathrm{A}\right)=0.8,\mathrm{P}\left(\mathrm{B}\right)=0.5,\mathrm{andP}\left(\mathrm{B}\right|\mathrm{A})=0.4\\ \left(\mathrm{i}\right)\mathrm{P}\left(\mathrm{B}\right|\mathrm{A})=0.4\\ \therefore \frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{A}\right)}=0.4\\ \Rightarrow \frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{0.8}=0.4\\ \Rightarrow \mathrm{P}(\mathrm{A}\cap \mathrm{B})=0.32\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{A}\right|\mathrm{B})=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{B}\right)}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{0.32}{0.5}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=0.64\\ \left(\mathrm{iii}\right)\mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}(\mathrm{A}\cap \mathrm{B})\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=0.8+0.5-0.32\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=0.98\end{array}$

Q.4 E

Ans

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}2\mathrm{P}\left(\mathrm{A}\right)=\mathrm{P}\left(\mathrm{B}\right)=\frac{5}{13}\Rightarrow \hspace{0.17em}\mathrm{P}\left(\mathrm{A}\right)=\frac{5}{26},\mathrm{\hspace{0.17em}}\mathrm{P}\left(\mathrm{B}\right)=\frac{5}{13}\\ \mathrm{and}\hspace{0.17em}\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{2}{5}.\\ \therefore \mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{B}\right)}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{2}{5}=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\left(\frac{5}{13}\right)}\\ \Rightarrow \mathrm{P}(\mathrm{A}\cap \mathrm{B})=\mathrm{\hspace{0.17em}}\frac{2}{5}\times \frac{5}{13}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{2}{13}\\ \therefore \mathrm{\hspace{0.17em}}\mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}(\mathrm{A}\cap \mathrm{B})\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{5}{26}+\frac{5}{13}-\frac{2}{13}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{15}{26}-\frac{2}{13}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{15-4}{26}\mathrm{\hspace{0.17em}}=\frac{11}{26}\end{array}$

Q.5

$\begin{array}{l}\mathrm{If}\mathrm{}\mathrm{P}\left(\mathrm{A}\right)=\frac{6}{11},\mathrm{P}\left(\mathrm{B}\right)=\frac{5}{11}\mathrm{and}\mathrm{}(\mathrm{A}\cup \mathrm{B}=\frac{7}{11},\mathrm{find}\left(\mathrm{i}\right)\mathrm{P}(\mathrm{A}\cap \mathrm{B})\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)\left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{B}|\mathrm{A}\right).\end{array}$

Ans

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\\ \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{6}}{11},\mathrm{P}\left(\mathrm{B}\right)=\frac{\mathrm{5}}{11}\mathrm{and}\mathrm{P}(\mathrm{A}\cup \mathrm{B})=\frac{\mathrm{7}}{11}\\ \left(\mathrm{i}\right)\mathrm{\hspace{0.17em}}\mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}(\mathrm{A}\cap \mathrm{B})\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{7}}{11}\mathrm{\hspace{0.17em}}=\frac{\mathrm{6}}{11}+\frac{\mathrm{5}}{11}-\mathrm{P}(\mathrm{A}\cap \mathrm{B})\\ \hspace{0.17em}\hspace{0.17em}=\frac{11}{11}-\mathrm{P}(\mathrm{A}\cap \mathrm{B})\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{P}(\mathrm{A}\cap \mathrm{B})\mathrm{\hspace{0.17em}}=\frac{11}{11}-\frac{7}{11}\mathrm{\hspace{0.17em}}=\frac{4}{11}\\ \left(\mathrm{ii}\right)\mathrm{Since},\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{B}\right)}\hspace{0.17em}\hspace{0.17em}=\frac{\left(\frac{4}{11}\right)}{\left(\frac{\mathrm{5}}{11}\right)}\\ \hspace{0.17em}\hspace{0.17em}=\frac{4}{5}\\ \left(\mathrm{iii}\right)\mathrm{Since},\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{P}\left(\mathrm{B}|\mathrm{A}\right)=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{A}\right)}\mathrm{\hspace{0.17em}}=\frac{\left(\frac{4}{11}\right)}{\left(\frac{\mathrm{6}}{11}\right)}\\ \hspace{0.17em}\hspace{0.17em}=\frac{4}{6}=\frac{2}{3}\end{array}$

Q.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
Determine P (E|F).

Ans

$\begin{array}{l}\mathrm{If}\mathrm{a}\mathrm{coin}\mathrm{is}\mathrm{tossed}\mathrm{three}\mathrm{times},\mathrm{then}\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{S}\mathrm{is}\\ \mathrm{S}=\; \{\mathrm{HHH},\mathrm{HHT},\mathrm{HTH},\mathrm{HTT},\mathrm{THH},\mathrm{THT},\mathrm{TTH},\mathrm{TTT}\}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{seen}\mathrm{that}\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{has}8\mathrm{elements}.\\ \left(\mathrm{i}\right)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{E}=\{\mathrm{HHH},\mathrm{HTH},\mathrm{THH},\mathrm{TTH}\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{F}=\{\mathrm{HHH},\mathrm{HHT}\}\\ \therefore \mathrm{E}\cap \mathrm{F}=\left\{\mathrm{HHH}\right\}\\ \mathrm{So},\mathrm{P}\left(\mathrm{E}\right)=\frac{4}{8}=\frac{1}{2},​\hspace{0.17em}\mathrm{P}\left(\mathrm{F}\right)=\frac{2}{8}=\frac{1}{4}\mathrm{and}\mathrm{P}(\mathrm{E}\cap \mathrm{F})=\frac{1}{8}\\ \therefore \mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{P}\left(\mathrm{F}\right)}\\ =\frac{\left(\frac{1}{8}\right)}{\left(\frac{1}{4}\right)}=\frac{1}{2}\\ \left(\mathrm{ii}\right))\hspace{0.17em}\hspace{0.17em}\mathrm{E}=\{\mathrm{HHH},\mathrm{}\mathrm{HHT},\mathrm{}\mathrm{HTH},\mathrm{}\mathrm{THH}\}\mathrm{}\\ \mathrm{\hspace{0.17em}}\mathrm{F}=\{\mathrm{HHT},\mathrm{}\mathrm{HTH},\mathrm{}\mathrm{HTT},\mathrm{}\mathrm{THH},\mathrm{}\mathrm{THT},\mathrm{}\mathrm{TTH},\mathrm{}\mathrm{TTT}\}\mathrm{}\\ \therefore \mathrm{\hspace{0.17em}}\mathrm{E}\cap \mathrm{F}=\{\mathrm{HHT},\mathrm{}\mathrm{HTH},\mathrm{}\mathrm{THH}\}\\ \mathrm{So},​\hspace{0.17em}\mathrm{P}\left(\mathrm{F}\right)=\frac{7}{8}\mathrm{and}\mathrm{P}(\mathrm{E}\cap \mathrm{F})=\frac{3}{8}\mathrm{}\\ \therefore \mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{P}\left(\mathrm{F}\right)}\\ =\frac{\left(\frac{3}{8}\right)}{\left(\frac{7}{8}\right)}=\frac{3}{7}\\ \left(\mathrm{iii}\right)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{E}=\{\mathrm{HHH},\mathrm{}\mathrm{HHT},\mathrm{}\mathrm{HTT},\mathrm{}\mathrm{HTH},\mathrm{}\mathrm{THH},\mathrm{}\mathrm{THT},\mathrm{}\mathrm{TTH}\}\mathrm{}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{F}=\{\mathrm{HHT},\mathrm{}\mathrm{HTT},\mathrm{}\mathrm{HTH},\mathrm{}\mathrm{THH},\mathrm{}\mathrm{THT},\mathrm{}\mathrm{TTH},\mathrm{}\mathrm{TTT}\}\\ \therefore \mathrm{E}\cap \mathrm{F}=\{\mathrm{HHT},\mathrm{HTT},\mathrm{HTH},\mathrm{THH},\mathrm{THT},\mathrm{TTH}\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{P}\left(\mathrm{F}\right)=\frac{7}{8}\mathrm{and}\mathrm{P}(\mathrm{E}\cap \mathrm{F})=\frac{6}{8}\\ \mathrm{So},\mathrm{​}\mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{P}\left(\mathrm{F}\right)}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\left(\frac{6}{8}\right)}{\left(\frac{7}{8}\right)}=\frac{6}{7}\end{array}$

Q.7 Two coins are tossed once, where
(i) E: tail appears on one coin,
F: one coin shows head
(ii) E: no tail appears, F : no head appears
Determine P (E|F).

Ans

$\begin{array}{l}\mathrm{If}\mathrm{two}\mathrm{coins}\mathrm{are}\mathrm{tossed}\mathrm{once},\mathrm{then}\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{S}\mathrm{is}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{S}=\; \{\mathrm{HH},\mathrm{HT},\mathrm{TH},\mathrm{TT}\}\\ \left(\mathrm{i}\right)\mathrm{E}=\; \{\mathrm{HT},\mathrm{TH}\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{F}=\; \{\mathrm{HT},\mathrm{TH}\}\\ \therefore \mathrm{E}\cap \mathrm{F}=\{\mathrm{HT},\mathrm{TH}\}\\ \mathrm{P}\left(\mathrm{F}\right)=\frac{2}{4}=\frac{1}{2}\mathrm{and}\mathrm{P}(\mathrm{E}\cap \mathrm{F})=\frac{2}{4}=\frac{1}{2}\\ \therefore \mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{P}\left(\mathrm{F}\right)}\\ =\frac{\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)}=1\\ \left(\mathrm{ii}\right)\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{E}=\left\{\mathrm{HH}\right\},\mathrm{F}=\left\{\mathrm{TT}\right\}\\ \mathrm{and}\mathrm{E}\cap \mathrm{F}=\mathrm{Ï•}\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{P}\left(\mathrm{F}\right)=\frac{1}{4},\mathrm{\hspace{0.17em}}\mathrm{P}(\mathrm{E}\cap \mathrm{F})=0\\ \mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{\hspace{0.17em}}\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{\hspace{0.17em}}\mathrm{P}\left(\mathrm{F}\right)}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{0}{\left(\frac{1}{4}\right)}=0\end{array}$

Q.8 A die is thrown three times,
E: 4 appears on the third toss,
F: 6 and 5 appears respectively on first two tosses
Determine P (E|F).

Ans

If a die is thrown three times, then the number of elements in the sample space will be 6× 6 × 6 = 216

$\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{E}=\left\{\begin{array}{l}(1,1,4),(1,2,4),(1,3,4),...\hspace{0.17em}\hspace{0.17em}(1,6,4)\\ (2,1,4),(2,2,4),(2,3,4),...(2,6,4)\\ (3,1,4),(3,2,4),(3,3,4),...(3,6,4)\\ (4,1,4),(4,2,4),(4,3,4),...(4,6,4)\\ (5,1,4),(5,2,4),(5,3,4),...(5,6,4)\\ (6,1,4),(6,2,4),(6,3,4),...(6,6,4)\end{array}\right\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{F}=\{(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)\}\\ \therefore \mathrm{E}\cap \mathrm{F}=\left\{(6,5,4)\right\}\\ \mathrm{So},\mathrm{P}\left(\mathrm{F}\right)=\frac{6}{216}\mathrm{and}\mathrm{P}(\mathrm{E}\cap \mathrm{F})=\frac{1}{216}\\ \mathrm{Then},\hspace{0.17em}\hspace{0.17em}\mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{P}\left(\mathrm{F}\right)}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{\left(\frac{1}{216}\right)}{\left(\frac{6}{216}\right)}=\frac{1}{6}\end{array}$

Q.9 Mother, father and son line up at random for a family picture
E: son on one end,
F: father in middle
Determine P (E|F).

Ans

$\begin{array}{l}\mathrm{If}\mathrm{mother}(\mathrm{M}),\mathrm{father}(\mathrm{F}),\mathrm{and}\mathrm{son}\left(\mathrm{S}\right)\mathrm{line}\mathrm{up}\mathrm{for}\mathrm{the}\mathrm{family}\\ \mathrm{picture},\mathrm{then}\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{will}\mathrm{be}\\ \mathrm{S}=\{\mathrm{MFS},\mathrm{}\mathrm{MSF},\mathrm{}\mathrm{FMS},\mathrm{}\mathrm{FSM},\mathrm{}\mathrm{SMF},\mathrm{}\mathrm{SFM}\}\mathrm{}\\ \mathrm{E}=\{\mathrm{MFS},\mathrm{}\mathrm{FMS},\mathrm{}\mathrm{SMF},\mathrm{}\mathrm{SFM}\}\\ \mathrm{F}=\{\mathrm{MFS},\mathrm{SFM}\}\\ \mathrm{So},\mathrm{E}\cap \mathrm{F}=\{\mathrm{MFS},\mathrm{SFM}\}\\ \mathrm{P}\left(\mathrm{F}\right)=\frac{2}{6}\mathrm{and}\mathrm{P}(\mathrm{E}\cap \mathrm{F})=\frac{2}{6}\\ \therefore \mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{P}\left(\mathrm{F}\right)}\\ =\frac{\left(\frac{2}{6}\right)}{\left(\frac{2}{6}\right)}=1\mathrm{}\end{array}$

Q.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.

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{first}\mathrm{observation}\mathrm{be}\mathrm{from}\mathrm{the}\mathrm{black}\mathrm{die}\mathrm{and}\mathrm{second}\mathrm{from}\\ \mathrm{the}\mathrm{red}\mathrm{die}.\mathrm{When}\mathrm{two}\mathrm{dice}(\mathrm{one}\mathrm{black}\mathrm{and}\mathrm{another}\mathrm{red})\mathrm{are}\mathrm{rolled},\\ \mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{S}\mathrm{has}6\; \times \; 6=36\mathrm{number}\mathrm{of}\mathrm{elements}.\\ \left(\mathrm{a}\right)\mathrm{Let}\\ \mathrm{A}:\mathrm{Obtaining}\mathrm{a}\mathrm{sum}\mathrm{greater}\mathrm{than}9\\ =\{(4,\; 6),\; (5,\; 5),\; (5,\; 6),\; (6,\; 4),\; (6,\; 5),\; (6,\; 6)\}\\ \mathrm{B}:\mathrm{Black}\mathrm{die}\mathrm{results}\mathrm{in}\mathrm{a}5.\\ =\{(5,\; 1),\; (5,\; 2),\; (5,\; 3),\; (5,\; 4),\; (5,\; 5),\; (5,\; 6)\}\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{A}\cap \mathrm{B}=\{(5,\; 5),\; (5,\; 6)\}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{P}\left(\mathrm{B}\right)=\frac{6}{36}\mathrm{and}\mathrm{P}(\mathrm{A}\cap \mathrm{B})=\frac{2}{36}\mathrm{}\\ \mathrm{The}\mathrm{conditional}\mathrm{probability}\mathrm{of}\mathrm{obtaining}\mathrm{a}\mathrm{sum}\mathrm{greater}\mathrm{than}9,\\ \mathrm{given}\mathrm{that}\mathrm{the}\mathrm{black}\mathrm{die}\mathrm{resulted}\mathrm{in}\mathrm{a}5,\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{P}\left(\mathrm{A}\right|\mathrm{B}).\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{B}\right)}\\ =\frac{\left(\frac{2}{36}\right)}{\left(\frac{6}{36}\right)}\\ =\frac{2}{6}=\frac{1}{3}\\ \left(\mathrm{b}\right))\mathrm{E}:\mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{observations}\mathrm{is}8.\\ =\{(2,\; 6),\; (3,\; 5),\; (4,\; 4),\; (5,\; 3),\; (6,\; 2)\}\\ \mathrm{F}:\mathrm{Red}\mathrm{die}\mathrm{resulted}\mathrm{in}\mathrm{a}\mathrm{number}\mathrm{less}\mathrm{than}4.\\ =\left\{\begin{array}{l}(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{array}\right\}\\ \mathrm{and}\mathrm{\hspace{0.17em}}\mathrm{E}\cap \mathrm{F}=\{(5,3),(6,2)\}\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{P}\left(\mathrm{F}\right)=\frac{18}{36}\mathrm{and}\mathrm{P}(\mathrm{E}\cap \mathrm{F})=\frac{2}{36}\\ \mathrm{The}\mathrm{conditional}\mathrm{probability}\mathrm{of}\mathrm{obtaining}\mathrm{the}\mathrm{sum}\mathrm{equal}\mathrm{to}8,\\ \mathrm{given}\mathrm{that}\mathrm{the}\mathrm{red}\mathrm{die}\mathrm{resulted}\mathrm{in}\mathrm{a}\mathrm{number}\mathrm{less}\mathrm{than}4,\\ \mathrm{P}\left(\mathrm{E}\right|\mathrm{F})=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{P}\left(\mathrm{F}\right)}=\frac{\frac{2}{36}}{\frac{18}{36}}=\frac{1}{9}\end{array}$

Q.11 A fair die is rolled. Consider events
E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}
Find
(i) P(E|F) and P (F|E)
(ii) P(E|G) and P(G|E)
(iii) P((E ∪ F)|G) and P((E ∩ F)|G)

Ans

$\begin{array}{l}\mathrm{When}\mathrm{a}\mathrm{fair}\mathrm{die}\mathrm{is}\mathrm{rolled},\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{S}\mathrm{will}\mathrm{be}\\ \mathrm{S}=\; \{1,\; 2,\; 3,\; 4,\; 5,\; 6\}\\ \mathrm{then}\mathrm{given}\mathrm{events}\mathrm{are}\mathrm{E}=\{1,3,5\},\mathrm{}\mathrm{F}=\{2,3\}\mathrm{}\mathrm{and}\mathrm{}\mathrm{G}=\{2,3,4,5\}\\ \mathrm{So},\mathrm{E}\cap \mathrm{F}=\left\{3\right\},\mathrm{then}\mathrm{P}(\mathrm{E}\cap \mathrm{F})=\frac{1}{6},\hspace{0.17em}\mathrm{P}\left(\mathrm{E}\right)=\frac{3}{6},\mathrm{}\\ \mathrm{P}\left(\mathrm{F}\right)=\frac{2}{6}\mathrm{and}\mathrm{P}\left(\mathrm{G}\right)=\frac{4}{6}\\ \left(\mathrm{i}\right)\mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{P}\left(\mathrm{F}\right)}\\ \mathrm{\hspace{0.17em}}=\frac{\left(\frac{1}{6}\right)}{\left(\frac{2}{6}\right)}=\frac{1}{2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{P}\left(\mathrm{F}|\mathrm{E}\right)=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{P}\left(\mathrm{E}\right)}\\ \mathrm{\hspace{0.17em}}=\frac{\left(\frac{1}{6}\right)}{\left(\frac{3}{6}\right)}=\frac{1}{3}\\ \left(\mathrm{ii}\right)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{E}\cap \mathrm{G}=\left\{5\right\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{P}(\mathrm{E}\cap \mathrm{G})=\frac{1}{6}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{P}\left(\mathrm{E}|\mathrm{G}\right)=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{G})}{\mathrm{P}\left(\mathrm{G}\right)}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\left(\frac{1}{6}\right)}{\left(\frac{4}{6}\right)}=\frac{1}{4}\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{P}\left(\mathrm{G}|\mathrm{E}\right)=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{G})}{\mathrm{P}\left(\mathrm{E}\right)}\\ =\frac{\left(\frac{1}{6}\right)}{\left(\frac{3}{6}\right)}=\frac{1}{3}\\ \left(\mathrm{iii}\right)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(\mathrm{E}\cup \mathrm{F})=\{1,2,3,5\}\\ (\mathrm{E}\cup \mathrm{F})\cap \mathrm{G}=\{2,3,5\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{P}((\mathrm{E}\cup \mathrm{F})\cap \mathrm{G})=\frac{3}{6}=\frac{1}{2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{P}\left\{(\mathrm{E}\cup \mathrm{F})|\mathrm{G}\right\}=\frac{\mathrm{P}((\mathrm{E}\cup \mathrm{F})\cap \mathrm{G})}{\mathrm{P}\left(\mathrm{G}\right)}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\left(\frac{1}{2}\right)}{\left(\frac{4}{6}\right)}=\frac{3}{4}\\ (\mathrm{E}\cap \mathrm{F})\cap \mathrm{G}=\left\{3\right\}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{P}((\mathrm{E}\cap \mathrm{F})\cap \mathrm{G})=\frac{1}{6}\\ \therefore \mathrm{\hspace{0.17em}}\mathrm{P}\left((\mathrm{E}\cap \mathrm{F})|\mathrm{G}\right)=\frac{\mathrm{P}((\mathrm{E}\cap \mathrm{F})\cap \mathrm{G})}{\mathrm{P}\left(\mathrm{G}\right)}\\ =\frac{\left(\frac{1}{6}\right)}{\left(\frac{4}{6}\right)}=\frac{1}{4}\end{array}$

Q.12 Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{B}\mathrm{and}\mathrm{G}\mathrm{represent}\mathrm{the}\mathrm{boy}\mathrm{and}\mathrm{the}\mathrm{girl}\mathrm{child}\mathrm{respectively}.\mathrm{If}\\ \mathrm{a}\mathrm{family}\mathrm{has}\mathrm{two}\mathrm{children},\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{will}\mathrm{be}\\ \mathrm{S}=\; \{(\mathrm{B},\mathrm{B}),\; (\mathrm{B},\mathrm{G}),\; (\mathrm{G},\mathrm{B}),\; (\mathrm{G},\mathrm{G})\}\\ \mathrm{Let}\mathrm{A}=\mathrm{both}\mathrm{children}\mathrm{are}\mathrm{girls}=\left\{\right(\mathrm{G},\mathrm{G}\left)\right\}\\ \left(\mathrm{i}\right)\mathrm{Let}\mathrm{B}\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{that}\mathrm{the}\mathrm{youngest}\mathrm{child}\mathrm{is}\mathrm{a}\mathrm{girl},\mathrm{then}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{B}=\left\{\right(\mathrm{B},\mathrm{G}),(\mathrm{G},\mathrm{G}\left)\right\}\\ \Rightarrow \mathrm{A}\cap \mathrm{B}=\left\{\right(\mathrm{G},\mathrm{G}\left)\right\}\\ \therefore \mathrm{P}\left(\mathrm{B}\right)=\frac{2}{4}=\frac{1}{2}\mathrm{and}\mathrm{P}(\mathrm{A}\cap \mathrm{B})=\frac{1}{4}\\ \mathrm{The}\mathrm{conditional}\mathrm{probability}\mathrm{that}\mathrm{both}\mathrm{are}\mathrm{girls},\mathrm{given}\mathrm{that}\mathrm{the}\\ \mathrm{youngest}\mathrm{child}\mathrm{is}\mathrm{a}\mathrm{girl}=\mathrm{P}\left(\mathrm{A}\right|\mathrm{B})\\ =\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{B}\right)}\\ =\frac{\left(\frac{1}{4}\right)}{\left(\frac{1}{2}\right)}=\frac{1}{2}\\ \mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{is}\frac{1}{2}.\\ \left(\mathrm{ii}\right)\mathrm{Let}\mathrm{E}\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{that}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{child}\mathrm{is}\mathrm{a}\mathrm{girl}.\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{E}=\{(\mathrm{B},\mathrm{G}),(\mathrm{G},\mathrm{B}),(\mathrm{G},\mathrm{G})\}\\ \mathrm{Then},\hspace{0.17em}\hspace{0.17em}\mathrm{A}\cap \mathrm{E}=\left\{(\mathrm{G},\mathrm{G})\right\}\\ \therefore \mathrm{P}(\mathrm{A}\cap \mathrm{E})=\frac{1}{4}\mathrm{and}\mathrm{P}\left(\mathrm{E}\right)=\frac{3}{4}\\ \mathrm{The}\mathrm{conditional}\mathrm{probability}\mathrm{that}\mathrm{both}\mathrm{are}\mathrm{girls},\mathrm{given}\mathrm{that}\mathrm{at}\mathrm{least}\\ \mathrm{one}\mathrm{child}\mathrm{is}\mathrm{a}\mathrm{girl}=\mathrm{P}\left(\mathrm{A}\right|\mathrm{E})\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{E})}{\mathrm{P}\left(\mathrm{E}\right)}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\left(\frac{1}{4}\right)}{\left(\frac{3}{4}\right)}=\frac{1}{3}\\ \mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{is}\frac{1}{3}.\end{array}$

Q.13 An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Ans

Number of easy True/False questions = 300
Number of easy multiple choice questions = 500
Number of difficult True/False questions = 200
Number of difficult multiple choice questions = 400
Total number of multiple choice questions = 900
Total questions = 1400
Let the notations of questions are as follows:
E = Easy questions, D = Difficult questions, M = Multiple choice questions and T = True/False questions.

$\begin{array}{l}\mathrm{So},\mathrm{Probability}\mathrm{of}\mathrm{easy}\mathrm{and}\mathrm{multiple}\mathrm{choice}\mathrm{questions}\\ \mathrm{P}(\mathrm{E}\cap \mathrm{M})\mathrm{\hspace{0.17em}}=\frac{500}{1400}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{5}{14}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Probability}\mathrm{of}\mathrm{multiple}\mathrm{choice}\mathrm{question},\mathrm{P}\left(\mathrm{M}\right)=\frac{900}{1400}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{9}{14}\\ \mathrm{Probability}\mathrm{of}\mathrm{a}\mathrm{randomly}\mathrm{selected}\mathrm{an}\mathrm{easy}\mathrm{question},\mathrm{given}\mathrm{that}\\ \mathrm{it}\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{choice}\mathrm{question}=\mathrm{P}\left(\mathrm{E}|\mathrm{M}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{M})}{\mathrm{P}\left(\mathrm{M}\right)}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\left(\frac{5}{14}\right)}{\left(\frac{9}{14}\right)}=\frac{5}{9}\\ \mathrm{Thus},\mathrm{required}\mathrm{probability}\mathrm{is}\frac{5}{9}.\end{array}$

Q.14 Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

Ans

When two dice are thrown, sample space observations = 6 × 6 = 36
Let A = the sum of the numbers on the dice is 4 and
B = the two numbers appearing on throwing the two dice are different.

$\begin{array}{l}\therefore \mathrm{A}=\{(1,3),(3,1)\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{B}=\left\{\begin{array}{l}(1,2),(1,3),(1,4),(1,5),(1,6)\\ (2,1),(2,3),(2,4),(2,5),(2,6)\\ (3,1),(3,2),(3,4),(3,5),(3,6)\\ (4,1),(4,2),(4,3),(4,5),(4,6)\\ (5,1),(5,2),(5,3),(5,4),(5,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5)\end{array}\right\}\\ \mathrm{Then},\\ \mathrm{A}\cap \mathrm{B}=\{(1,3),(3,1)\}\\ \mathrm{P}\left(\mathrm{B}\right)=\frac{30}{36}=\frac{5}{6}\mathrm{and}\mathrm{P}(\mathrm{A}\cap \mathrm{B})=\frac{2}{30}=\frac{1}{18}\\ \mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{numbers}\mathrm{on}\mathrm{the}\mathrm{dice}\mathrm{is}4,\\ \mathrm{given}\mathrm{that}\mathrm{the}\mathrm{two}\mathrm{numbers}\mathrm{appearing}\mathrm{on}\mathrm{throwing}\mathrm{the}\mathrm{two}\mathrm{dice}\\ \mathrm{are}\mathrm{different}=\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)\\ =\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{B}\right)}\\ =\frac{\left(\frac{1}{18}\right)}{\left(\frac{5}{6}\right)}=\frac{1}{15}\end{array}$

Q.15 Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.

Ans

$\begin{array}{l}\text{The sample space of the experiment is given below:}\\ \text{S}=\left\{\begin{array}{l}(1,H),(1,T),(2,H),(2,T),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),\\ (4,H),(4,T),(5,H),(5,T),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6),\end{array}\right\}\\ \text{Let A}=\text{the coin shows a tail}\\ =\{(1,T),(2,T),(4,T),(5,T)\}\text{and}\\ \text{B}=\text{at least one die shows 3}\\ =\{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,3)\}\\ A\cap B=\mathrm{Ï•}\Rightarrow P(A\cap B)=0\\ P\left(B\right)=P\left\{(3,1)\right\}+P\left\{(3,1)\right\}+P\left\{(3,1)\right\}+P\left\{(3,1)\right\}+P\left\{(3,1)\right\}+P\left\{(3,1)\right\}\\ +P\left\{(3,1)\right\}\\ =\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{7}{36}\\ \text{Probability of the event that the coin shows a tail, given that at}\\ \text{least one die shows 3}\\ =\text{P(A|B)}\\ =\frac{P(A\cap B)}{P\left(B\right)}\\ =\frac{0}{\left(\frac{7}{36}\right)}=0\\ \text{Thus, the required probability is 0}\text{.}\end{array}$

Q.16 If P (A) =1/2, P (B) = 0, then P (A|B) is
(A) 0 (B) 1/2
(C) not defined (D) 1

Ans

$\begin{array}{l}\mathrm{Given}\mathrm{that}\mathrm{P}\left(\mathrm{A}\right)=\frac{1}{2},\mathrm{P}\left(\mathrm{B}\right)=0,\mathrm{}\mathrm{then}\mathrm{}\\ \mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{B}\right)}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{0}\\ \mathrm{Therefore},\mathrm{P}\left(\mathrm{A}\right|\mathrm{B})\mathrm{is}\mathrm{not}\mathrm{defined}.\\ \mathrm{Thus},\mathrm{the}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{C}\mathrm{.}\end{array}$

Q.17 If A and B are events such that P (A|B) = P(B|A), then
(A) A ⊂ B but A ≠ B (B) A = B
(C) A ∩ B = Φ (D) P (A) = P(B)

Ans

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\mathrm{P}\left(\mathrm{B}|\mathrm{A}\right)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{B}\right)}=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{A}\right)}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{1}{\mathrm{P}\left(\mathrm{B}\right)}=\frac{1}{\mathrm{P}\left(\mathrm{A}\right)}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{P}\left(\mathrm{A}\right)=\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{Thus},\mathrm{option}\mathrm{D}\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$