# NCERT Solutions Class 12 Maths Exercise 13.1

The National Council of Educational Research and Training (NCERT) is a government agency that was established to help India’s schools improve their quality of education. The Government of India created it to preserve the uniformity of the curriculum since different books were being recommended to students in the same standard. The responsibility for creating content for various NCERT textbooks belongs to this body.  The Central Board of Secondary Education (CBSE) and many other State boards follow the NCERT pattern. NCERT Solutions is preferred by CBSE.These questions are predominant in Board examinations, therefore, students appearing for the board examinations are advised to revise the NCERT Solutions thoroughly for better results.

Each chapter has an exercise at the end. Students should solve those exercise questions thoroughly. These NCERT exercise problems are used as the basis for questions in examinations and students are highly recommended to at least practise them once before their examination.

## NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex 13.1) Exercise 13.1

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### NCERT Solutions For Class 12 Maths Chapter 13 Probability Exercise 13.1

Human intellect and logic are fundamentally based on Mathematics, which is essential to understanding the world. Mathematics is a powerful tool for developing mental discipline and promoting logical thinking. Probability is a branch of Mathematics that enables one to foresee the outcomes of future events through calculations. The chapter on Probability is fascinating. One of the most important exercises is Class 12 Maths Chapter 13 Exercise 13.1. The NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.1. deal with basic calculations for a better understanding of the concept. To make studying easier and more effective Extramarks offers notes on Exercise 13.1 Class 12th Maths, past years’ papers, and much more.

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### Topics Covered in Class 12 NCERT Maths of Chapter 13 Exercise 13.1

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### What is Conditional Probability?

The possibility of an event or outcome being contingent on a prior event or outcome is known as Conditional Probability. The probability of the prior event is multiplied by the current likelihood of the subsequent, or conditional, occurrence to determine the conditional probability. The term is present in  NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.1.

### Chapter wise NCERT Solutions for Class 12 Maths

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### Access Other Exercises of Class 12 Maths Chapter 13

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Q.1

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

Ans

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

Q.2

$\text{Compute P(A|B), if P(B) = 0.5and P(A∩B) =0.32.}$

Ans

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

Q.3

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

Ans

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

Q.4 E

$\text{v}\text{a}\text{l}\text{u}\text{a}\text{t}\text{e}\text{}\text{P}\left(\text{A}\cup \text{B}\right),\text{}\text{i}\text{f}\text{}2\text{P}\left(\text{A}\right)=\text{P}\left(\text{B}\right)=\frac{5}{13}\text{}\text{a}\text{n}\text{d}\text{}\text{}\text{P}\left(\text{A}|\text{B}\right)=\frac{2}{5}.$

Ans

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

Q.5

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

Ans

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

Q.6 A coin is tossed three times, where
(i) E: head on third toss, F: heads on first two tosses
(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}= \left\{\mathrm{HHH},\mathrm{HHT},\mathrm{HTH},\mathrm{HTT},\mathrm{THH},\mathrm{THT},\mathrm{TTH},\mathrm{TTT}\right\}\\ \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) \mathrm{ }\mathrm{E}=\left\{\mathrm{HHH},\mathrm{HTH},\mathrm{THH},\mathrm{TTH}\right\}\\ \mathrm{F}=\left\{\mathrm{HHH},\mathrm{HHT}\right\}\\ \therefore \mathrm{E}\cap \mathrm{F}=\left\{\mathrm{HHH}\right\}\\ \mathrm{So},\mathrm{P}\left(\mathrm{E}\right)=\frac{4}{8}=\frac{1}{2},â€‹ \mathrm{P}\left(\mathrm{F}\right)=\frac{2}{8}=\frac{1}{4}\mathrm{and}\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\frac{1}{8}\\ \therefore \mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{P}\left(\mathrm{F}\right)}\\ =\frac{\left(\frac{1}{8}\right)}{\left(\frac{1}{4}\right)}=\frac{1}{2}\\ \left(\mathrm{ii}\right)\right) \mathrm{E}=\left\{\mathrm{HHH},\mathrm{}\mathrm{HHT},\mathrm{}\mathrm{HTH},\mathrm{}\mathrm{THH}\right\}\mathrm{}\\ \mathrm{ }\mathrm{F}=\left\{\mathrm{HHT},\mathrm{}\mathrm{HTH},\mathrm{}\mathrm{HTT},\mathrm{}\mathrm{THH},\mathrm{}\mathrm{THT},\mathrm{}\mathrm{TTH},\mathrm{}\mathrm{TTT}\right\}\mathrm{}\\ \therefore \mathrm{ }\mathrm{E}\cap \mathrm{F}=\left\{\mathrm{HHT},\mathrm{}\mathrm{HTH},\mathrm{}\mathrm{THH}\right\}\\ \mathrm{So},â€‹ \mathrm{P}\left(\mathrm{F}\right)=\frac{7}{8}\mathrm{and}\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\frac{3}{8}\mathrm{}\\ \therefore \mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{P}\left(\mathrm{F}\right)}\\ =\frac{\left(\frac{3}{8}\right)}{\left(\frac{7}{8}\right)}=\frac{3}{7}\\ \left(\mathrm{iii}\right) \mathrm{E}=\left\{\mathrm{HHH},\mathrm{}\mathrm{HHT},\mathrm{}\mathrm{HTT},\mathrm{}\mathrm{HTH},\mathrm{}\mathrm{THH},\mathrm{}\mathrm{THT},\mathrm{}\mathrm{TTH}\right\}\mathrm{}\\ \mathrm{ }\mathrm{F}=\left\{\mathrm{HHT},\mathrm{}\mathrm{HTT},\mathrm{}\mathrm{HTH},\mathrm{}\mathrm{THH},\mathrm{}\mathrm{THT},\mathrm{}\mathrm{TTH},\mathrm{}\mathrm{TTT}\right\}\\ \therefore \mathrm{E}\cap \mathrm{F}=\left\{\mathrm{HHT},\mathrm{HTT},\mathrm{HTH},\mathrm{THH},\mathrm{THT},\mathrm{TTH}\right\}\\ \mathrm{P}\left(\mathrm{F}\right)=\frac{7}{8}\mathrm{and}\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\frac{6}{8}\\ \mathrm{So},\mathrm{â€‹}\mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{P}\left(\mathrm{F}\right)}\\ =\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,
(ii) E: no tail appears, F : no head appears
Determine P (E|F).

Ans

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

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

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

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

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

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}\left(\mathrm{one}\mathrm{black}\mathrm{and}\mathrm{another}\mathrm{red}\right)\mathrm{are}\mathrm{rolled},\\ \mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{S}\mathrm{has}6 × 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\\ =\left\{\left(4, 6\right), \left(5, 5\right), \left(5, 6\right), \left(6, 4\right), \left(6, 5\right), \left(6, 6\right)\right\}\\ \mathrm{B}:\mathrm{Black}\mathrm{die}\mathrm{results}\mathrm{in}\mathrm{a}5.\\ =\left\{\left(5, 1\right), \left(5, 2\right), \left(5, 3\right), \left(5, 4\right), \left(5, 5\right), \left(5, 6\right)\right\}\\ \therefore \mathrm{ }\mathrm{A}\cap \mathrm{B}=\left\{\left(5, 5\right), \left(5, 6\right)\right\}\\ \mathrm{P}\left(\mathrm{B}\right)=\frac{6}{36}\mathrm{and}\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\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}|\mathrm{B}\right).\\ \mathrm{ }\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\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)\right)\mathrm{E}:\mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{observations}\mathrm{is}8.\\ =\left\{\left(2, 6\right), \left(3, 5\right), \left(4, 4\right), \left(5, 3\right), \left(6, 2\right)\right\}\\ \mathrm{F}:\mathrm{Red}\mathrm{die}\mathrm{resulted}\mathrm{in}\mathrm{a}\mathrm{number}\mathrm{less}\mathrm{than}4.\\ =\left\{\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(2,1\right),\left(2,2\right),\left(2,3\right)\\ \left(3,1\right),\left(3,2\right),\left(3,3\right),\left(4,1\right),\left(4,2\right),\left(4,3\right)\\ \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(6,1\right),\left(6,2\right),\left(6,3\right)\end{array}\right\}\\ \mathrm{and}\mathrm{ }\mathrm{E}\cap \mathrm{F}=\left\{\left(5,3\right),\left(6,2\right)\right\}\\ \therefore \mathrm{ }\mathrm{P}\left(\mathrm{F}\right)=\frac{18}{36}\mathrm{and}\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\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}|\mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)}{\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}= \left\{1, 2, 3, 4, 5, 6\right\}\\ \mathrm{then}\mathrm{given}\mathrm{events}\mathrm{are}\mathrm{E}=\left\{1,3,5\right\},\mathrm{}\mathrm{F}=\left\{2,3\right\}\mathrm{}\mathrm{and}\mathrm{}\mathrm{G}=\left\{2,3,4,5\right\}\\ \mathrm{So},\mathrm{E}\cap \mathrm{F}=\left\{3\right\},\mathrm{then}\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\frac{1}{6}, \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}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{P}\left(\mathrm{F}\right)}\\ \mathrm{ }=\frac{\left(\frac{1}{6}\right)}{\left(\frac{2}{6}\right)}=\frac{1}{2}\\ \mathrm{P}\left(\mathrm{F}|\mathrm{E}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{P}\left(\mathrm{E}\right)}\\ \mathrm{ }=\frac{\left(\frac{1}{6}\right)}{\left(\frac{3}{6}\right)}=\frac{1}{3}\\ \left(\mathrm{ii}\right) \mathrm{ }\mathrm{E}\cap \mathrm{G}=\left\{5\right\}\\ \mathrm{P}\left(\mathrm{E}\cap \mathrm{G}\right)=\frac{1}{6}\\ \mathrm{ }\mathrm{P}\left(\mathrm{E}|\mathrm{G}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{G}\right)}{\mathrm{P}\left(\mathrm{G}\right)}\\ =\frac{\left(\frac{1}{6}\right)}{\left(\frac{4}{6}\right)}=\frac{1}{4}\\ \therefore \mathrm{P}\left(\mathrm{G}|\mathrm{E}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{G}\right)}{\mathrm{P}\left(\mathrm{E}\right)}\\ =\frac{\left(\frac{1}{6}\right)}{\left(\frac{3}{6}\right)}=\frac{1}{3}\\ \left(\mathrm{iii}\right) \mathrm{ }\left(\mathrm{E}\cup \mathrm{F}\right)=\left\{1,2,3,5\right\}\\ \left(\mathrm{E}\cup \mathrm{F}\right)\cap \mathrm{G}=\left\{2,3,5\right\}\\ \mathrm{P}\left(\left(\mathrm{E}\cup \mathrm{F}\right)\cap \mathrm{G}\right)=\frac{3}{6}=\frac{1}{2}\\ \mathrm{ }\mathrm{P}\left\{\left(\mathrm{E}\cup \mathrm{F}\right)|\mathrm{G}\right\}=\frac{\mathrm{P}\left(\left(\mathrm{E}\cup \mathrm{F}\right)\cap \mathrm{G}\right)}{\mathrm{P}\left(\mathrm{G}\right)}\\ =\frac{\left(\frac{1}{2}\right)}{\left(\frac{4}{6}\right)}=\frac{3}{4}\\ \left(\mathrm{E}\cap \mathrm{F}\right)\cap \mathrm{G}=\left\{3\right\}\\ \mathrm{ }\mathrm{P}\left(\left(\mathrm{E}\cap \mathrm{F}\right)\cap \mathrm{G}\right)=\frac{1}{6}\\ \therefore \mathrm{ }\mathrm{P}\left(\left(\mathrm{E}\cap \mathrm{F}\right)|\mathrm{G}\right)=\frac{\mathrm{P}\left(\left(\mathrm{E}\cap \mathrm{F}\right)\cap \mathrm{G}\right)}{\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}= \left\{\left(\mathrm{B},\mathrm{B}\right), \left(\mathrm{B},\mathrm{G}\right), \left(\mathrm{G},\mathrm{B}\right), \left(\mathrm{G},\mathrm{G}\right)\right\}\\ \mathrm{Let}\mathrm{A}=\mathrm{both}\mathrm{children}\mathrm{are}\mathrm{girls}=\left\{\left(\mathrm{G},\mathrm{G}\right)\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}\\ \mathrm{ }\mathrm{B}=\left\{\left(\mathrm{B},\mathrm{G}\right),\left(\mathrm{G},\mathrm{G}\right)\right\}\\ ⇒\mathrm{A}\cap \mathrm{B}=\left\{\left(\mathrm{G},\mathrm{G}\right)\right\}\\ \therefore \mathrm{P}\left(\mathrm{B}\right)=\frac{2}{4}=\frac{1}{2}\mathrm{and}\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\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}|\mathrm{B}\right)\\ =\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\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}.\\ \mathrm{E}=\left\{\left(\mathrm{B},\mathrm{G}\right),\left(\mathrm{G},\mathrm{B}\right),\left(\mathrm{G},\mathrm{G}\right)\right\}\\ \mathrm{Then}, \mathrm{A}\cap \mathrm{E}=\left\{\left(\mathrm{G},\mathrm{G}\right)\right\}\\ \therefore \mathrm{P}\left(\mathrm{A}\cap \mathrm{E}\right)=\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}|\mathrm{E}\right)\\ =\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{E}\right)}{\mathrm{P}\left(\mathrm{E}\right)}\\ =\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.

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

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.

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

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

$\begin{array}{l}\text{The sample space of the experiment is given below:}\\ \text{S}=\left\{\begin{array}{l}\left(1,H\right),\left(1,T\right),\left(2,H\right),\left(2,T\right),\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\\ \left(4,H\right),\left(4,T\right),\left(5,H\right),\left(5,T\right),\left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right),\end{array}\right\}\\ \text{Let A}=\text{the coin shows a tail}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left\{\left(1,T\right),\left(2,T\right),\left(4,T\right),\left(5,T\right)\right\}\text{and}\\ \text{B}=\text{at least one die shows 3}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left\{\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\left(6,3\right)\right\}\\ A\cap B=\mathrm{Ï•}⇒P\left(A\cap B\right)=0\\ P\left(B\right)=P\left\{\left(3,1\right)\right\}+P\left\{\left(3,1\right)\right\}+P\left\{\left(3,1\right)\right\}+P\left\{\left(3,1\right)\right\}+P\left\{\left(3,1\right)\right\}+P\left\{\left(3,1\right)\right\}\\ +\text{\hspace{0.17em}}P\left\{\left(3,1\right)\right\}\\ =\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{7}{36}\\ \text{Probability of the event that the coin shows a tail, given that at}\\ \text{least one die shows 3}\\ =\text{P(A|B)}\\ =\frac{P\left(A\cap B\right)}{P\left(B\right)}\\ =\frac{0}{\left(\frac{7}{36}\right)}=0\\ \text{Thus, the required probability is 0}\text{.}\end{array}$

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

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

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

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

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