Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen: Overview, Questions, Preparation

Q1. Three bags contain a number of red and white balls as follows:

Bag 1: 3 red balls, Bag 2: 2 red balls and 1 white ball, Bag 3: 3 white balls. The probability that bag  $i$ will be chosen and a ball is selected from it is  $\frac{i}{6}$ , where  $i=1, 2, 3$ . What is the probability that:

i. A red ball will be selected?

ii. A white ball is selected?

Sol:

$\begin{array}{l}Giventhat:\\ \text{}\text{}\text{}BagI=3redballsandnowhiteball\\ BagII=2redballsand1whiteball\\ BagIII=noredballand3whiteballs\\ Let{E}_1,E_{2}and{E}_{3}betheeventsofBagII,andBagIIIrespectivelyandaballis\\ drawnfromit.\\ \therefore P\left(E_1\right)=\frac{1}{6},P\left(E_2\right)=\frac{2}{6}andP\left(E_3\right)=\frac{3}{6}\\ \left(i\right)LetEbetheeventthatredballisselected\\ \therefore P\left(E\right)=P\left(E_1\right).P\left(E/E_1\right)+P\left(E_2\right).P\left(E/E_2\right)+P\left(E_3\right).P\left(E/E_3\right)\\ =\frac{1}{6}.\frac{3}{3}+\frac{2}{6}.\frac{2}{3}+\frac{3}{6}.0=\frac{3}{18}+\frac{4}{18}=\frac{7}{18}\\ \left(ii\right)LetFbetheeventthatwhiteballisselected\\ \therefore P\left(F\right)=1-P\left(E\right)\left[P\left(E\right)+P\left(F\right)=1\right]\\ =1-\frac{7}{18}=\frac{11}{18}\\ Hence,therequiredprobabilitiesare\frac{7}{18}and\frac{11}{18}.\end{array}$

Q2. Refer to Question 41 above. If a white ball is selected, what is the probability that it came from

$\mathrm{i.\; B}agII$

$\mathrm{ii.\; B}agIII$

Sol:

$\begin{array}{l}\text{\hspace{0.17em}Referring}toExerciseQ.41,wewillusehere,Bayes\text{'}Theorem\\ \left(i\right)P\left(E_2/F\right)=\frac{P\left(E_2\right).P\left(F/E_2\right)}{P\left(E_1\right).P\left(F/E_1\right)+P\left(E_2\right).P\left(F/E_2\right)+P\left(E_3\right).P\left(F/E_3\right)}\\ =\frac{\frac{2}{6}.\frac{1}{3}}{\frac{1}{6}.0+\frac{2}{6}.\frac{1}{3}+\frac{3}{6}.1}=\frac{\frac{2}{18}}{\frac{2}{18}+\frac{3}{6}}=\frac{2}{18}\times \frac{18}{11}=\frac{2}{11}\\ \left(ii\right)P\left(E_3/F\right)=\frac{P\left(E_3\right).P\left(F/E_3\right)}{P\left(E_1\right).P\left(F/E_1\right)+P\left(E_2\right).P\left(F/E_2\right)+P\left(E_3\right).P\left(F/E_3\right)}\\ =\frac{\frac{3}{6}.1}{\frac{1}{6}.0+\frac{2}{6}.\frac{1}{3}+\frac{3}{6}.1}=\frac{\frac{3}{6}}{\frac{2}{18}+\frac{3}{6}}=\frac{3}{6}\times \frac{18}{11}=\frac{9}{11}\\ Hence,therequiredprobabilitiesare\frac{2}{11}and\frac{9}{11}.\end{array}$

Q3. A shopkeeper sells three types of flower seeds: $A_1$ ,  $A_2$ , and  $A_3$ . They are sold as a mixture where the proportions are 4:4:2, respectively. The germination rates of the three types of seeds are 45%, 60%, and 35%. Calculate the probability:

(i) Of a randomly chosen seed to germinate

(ii) That it will not germinate given that the seed is of type $A_3$

(iii) That it is of type $A_2$ given that a randomly chosen seed does not germinate.

Sol:

$\begin{array}{l}Giventhat{A}_1:A_2:A_3=4:4:2\\ \therefore P\left(A_1\right)=\frac{4}{10},P\left(A_2\right)=\frac{4}{10}andP\left(A_3\right)=\frac{2}{10}where{A}_1,A_{2}and{A}_{3}arethethreetypesofseeds.\\ LetEbetheeventsthataseedand\bar{E}betheeventsthataseeddoesnot\\ \therefore P\left(\frac{E}{A_1}\right)=\frac{45}{100},P\left(\frac{E}{A_2}\right)=\frac{60}{100}andP\left(\frac{E}{A_3}\right)=\frac{35}{100}\\ andP\left(\frac{\bar{E}}{A_1}\right)=\frac{55}{100},P\left(\frac{\bar{E}}{A_2}\right)=\frac{40}{100}andP\left(\frac{\bar{E}}{A_3}\right)=\frac{65}{100}\\ \left(i\right)P\left(E\right)=P\left(A_1\right).P\left(\frac{E}{A_1}\right)+P\left(A_2\right).P\left(\frac{E}{A_2}\right)+P\left(A_3\right).P\left(\frac{E}{A_3}\right)\\ =\frac{4}{10}.\frac{45}{100}+\frac{4}{10}.\frac{60}{100}+\frac{2}{10}.\frac{35}{100}\\ =\frac{180}{1000}+\frac{240}{1000}+\frac{70}{1000}=\frac{490}{1000}=0.49\\ \left(ii\right)P\left(\bar{E}/A_3\right)=1-P\left(E/A_3\right)=1-\frac{35}{100}=\frac{65}{100}=0.65\\ \left(iii\right),Bayes\text{'}Theorem,weget\\ P\left(A_2/\bar{E}\right)=\frac{P\left(A_2\right).P\left(\bar{E}/A_2\right)}{P\left(A_1\right).P\left(\bar{E}/A_1\right)+P\left(A_2\right).P\left(\bar{E}/A_2\right)+P\left(A_3\right).P\left(\bar{E}/A_3\right)}\\ =\frac{\frac{4}{10}.\frac{40}{100}}{\frac{4}{10}.\frac{55}{100}+\frac{4}{10}.\frac{40}{100}+\frac{2}{10}.\frac{65}{100}}=\frac{\frac{160}{1000}}{\frac{220}{1000}+\frac{160}{1000}+\frac{130}{1000}}\\ =\frac{160}{220+160+130}=\frac{160}{510}=\frac{16}{51}=0.314\\ Hence,therequiredprobabilityis\frac{16}{51}or0.314\end{array}$

Sol:

$\begin{array}{l}Let{E}_1:TheeventthatthelettercomesfromTATANAGAR\\ and{E}_2:TheeventthatthelettercomesfromCALCUTTA\\ Also,E_3:Theeventthatontheletter,twolettersTAarevisible.\\ \therefore P\left(E_1\right)=\frac{1}{2},P\left(E_2\right)=\frac{1}{2}andP\left(\frac{E_3}{E_1}\right)=\frac{2}{8}andP\left(\frac{E_3}{E_2}\right)=\frac{1}{7}\\ \left[\because ForTATANAGAR,thetwo\text{\hspace{0.17em}consecutive}lettersvisibleareTA,AT,TA,AN,NA,AG,GA,AR\right]\\ \therefore P\left(E_3/E_1\right)=\frac{2}{8}\\ and\left[ForCALCUTTA,thetwo\text{\hspace{0.17em}consecutive}lettersvisibleareCA,AL,LC,CU,UT,TTandTA\right]\\ So,P\left(E_3/E_2\right)=\frac{1}{7}\\ Now\text{\hspace{0.17em}using}Bayes\text{'}Therorm,wehave\\ \therefore P\left(E_1/E_3\right)=\frac{P\left(E_1\right).P\left(E_3/E_1\right)}{P\left(E_1\right).P\left(E_3/E_1\right)+P\left(E_2\right).P\left(E_3/E_2\right)}\\ =\frac{\frac{1}{2}.\frac{2}{8}}{\frac{1}{2}.\frac{2}{8}+\frac{1}{2}.\frac{1}{7}}=\frac{\frac{1}{8}}{\frac{1}{8}+\frac{1}{14}}=\frac{\frac{1}{8}}{\frac{7+4}{56}}=\frac{7}{11}\\ Hence,therequiredprobabilityis\frac{7}{11}.\end{array}$