Ncert Solutions Maths class 12th

18. $P(A)=\frac{3}{5}$

$P(B)=\frac{1}{5}$

As $A$ and $B$ are independent event,

$P(A\cap B)=P(A).P(A)$

$=\frac{3}{5}.\frac{1}{5}$

$=\frac{3}{25}$

17. $P(A/B)=P(B/A)$

$P(A/B)=\frac{P(A\cap B)}{P(B)}$ ___ (i)

$P(B/A)=\frac{P(B\cap A)}{P(A)}$ ___ (ii)

Using (i) and (ii),

$P(A/B)=P(B/A)$

$\Rightarrow \frac{P(A\cap B)}{P(B)}=\frac{P(B\cap A)}{P(A)}$ $[?B\cap A=A\cap B]$

$\Rightarrow P(A)=P(B)$

$\therefore$ The correct answer is $(D).P(A)=P(B)$

16. $P(A)=\frac{1}{2}$ $,P(B)=0$

$P(A/B)=\frac{P(A\cap B)}{P(B)}$

$=\frac{P(A\cap B)}{0}$ = which is not defined

The correct answer is $(C)$ not defined.

15. The sample space of experiment is

$S=\left\{\begin{array}{l}(1,H),(1,T),(2,H),(2,T),(3,1),\\ (3,2),(3,3),(3,4),(3,5),(3,6),\\ (4,H),(4,T),(5,H),(5,T),\\ (6,1),(6,2),(6,3),(6,4),(6,3),\\ (6,5),(6,6)\end{array}\right\}$

Let, $E$ be the event that 'coin shows a tail' and $F$ be the event that 'atleast one die shows a $3$ '

$\Rightarrow E=\{1T,2T,4T,5T\}$

$F=\{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\}$

$\Rightarrow E\cap F=\varnothing \Rightarrow P(E\cap F)=0$

Now,

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{0}{P(F)}=0$

$\Rightarrow P(E/F)=0$

15. When dice is thrown, number of observations in the sample space is $6*6=36$

Let, $A:$ event that the sum of the number on the dice is $4$

$B:$ event that the two number appearing on throwing two dice is different

$A=\{(1,3),(2,2),(3,1)\}$

$B=\left\{\begin{array}{l}(1,2),(1,3),(1,4),(1,5),(1,6)\\ (2,1),(2,3),(2,4),(2,5),(2,6)\\ (3,1),(3,2),(3,4),(3,5),(3,6)\\ (4,1),(4,2),(4,3),(4,5),(4,6)\\ (5,1),(5,2),(5,3),(5,4),(5,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5)\end{array}\right\}$

$(A\cap B)=\{(1,3),(3,1)\}$

$P(B)=\frac{30}{36}$ $,P(A\cap B)=\frac{2}{36}$

$P(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}{\raisebox{1ex}{$30$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}=\frac{2}{30}=\frac{1}{15}$

$\therefore$ Required probability is $\frac{1}{15}$

13. Here, there are two types of questions, True/False or Multiple-Choice Questions. They are divided into easy and difficult.

True/False: 300 easy, 200 difficult

$MCQ$ : 500 easy, 400 difficult

Let,  $E=$ easy question

$M=$ multiple choice question

Total number of questions= $300+200+500+400=1400$

Total number of multiple-choice questions= $500+400=900$

Therefore, probability of selecting an easy $MCQ$ is

$P(E\cap M)=\frac{500}{1400}=\frac{5}{14}$

The probability of setting a $MCQ$ is

$P(M)=\frac{900}{1400}=\frac{9}{14}$

$\therefore$ $P(E/M)=\frac{P(E\cap M)}{P(M)}=\frac{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$14$}\right.}{\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$14$}\right.}=\frac{5}{9}$

$\therefore$ Required probability is $\frac{5}{9}$

12. Let $B$ and $G$ denote boy and girl respectively.

Sample space, $S=\{GG,BB,GB,BG\}$

Let, $E:$ Both are girls

$E=\{G,G\}$

i. Let $,F:$ youngest is a girl

$F=\{GG,B,G\}$

$E\cap F=\{G,G\}$ $,P(E\cap F)=\frac{1}{4}$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=\frac{1}{2}$

ii. Let $A:$ at least one is girl

$A=\{GG,GB,BG\}$

$P(A)=\frac{3}{4}$

$E\cap A=\left\{GG\right\}$

$P(E\cap A)=\frac{1}{4}$

$P(E/A)=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=\frac{1}{3}$

11. The sample space of given condition is

A $S=\{1,2,3,4,5,6\}$

$E=\{1,3,5\}$ $,F=\{2,3\}$ $,G=\{2,3,4,5\}$

$P(E)=\frac{3}{6}=\frac{1}{2}$ $,P(F)=\frac{2}{6}=\frac{1}{3}$ $P(G)=\frac{4}{6}=\frac{2}{3}$

$\mathrm{(i)\; P}(E/F)$ and $P(F/E)$

$P(E\cap F)=\frac{1}{6}$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{1}{2}$

and $P(F/E)=\frac{P(F\cap E)}{P(E)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{1}{3}$

$\mathrm{(ii)\; P}(E/G)$ and $P(G/E)$

$P(E\cap G)=\frac{2}{6}=\frac{1}{3}$

$P(E/G)=\frac{P(E\cap G)}{P(G)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{1}{2}$

$P(G/E)=\frac{P(G\cap E)}{P(E)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{2}{3}$

$\mathrm{(iii)\; P}((E\cup F)/G)$ and $P((E\cap F)/G)$

$(E\cup F)=\{1,2,3,5\}$

$(E\cup F)\cap G=\{1,2,3,5\}\cap \{2,3,4,5\}$

$=\{2,3,5\}$

and $E\cap F=\left\{3\right\}$

$(E\cap F)\cap G=\left\{3\right\}$

$P(E\cup G)=\frac{4}{6}=\frac{2}{3}$

Now,

$P((E\cup F)\cap G)=\frac{3}{6}=\frac{1}{2}$

So,

$P((E\cup F)/G)=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{3}{4}$

$P((E\cap F)/G)=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{3}{12}=\frac{1}{4}$

10. When two die are rolled, the sample space has $6*6=36$

Let $A$ be obtaining a sum greater than

$9=\{(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)\}$

And $B$ be black die result in $5$ ,

$B=\{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}$

$A\cap B=\{(5,5),(5,6)\}$

$\mathrm{a.\; P}(A/E)=\frac{P(A\cap B)}{P(B)}=\frac{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}{\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}=\frac{1}{3}$

b. Let,

$E:$ sum of observation is $8$

$=\{(2,6),(3,5),(4,4),(5,3),(6,2)\}$

$F:$ Red die resulted in number less than $4$

$=\left\{\begin{array}{l}(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)\\ (3,1),(3,2),(3,3),(4,1),(4,2),(4,3)\\ (5,1),(5,2),(5,3),(6,1),(6,2),(6,3)\end{array}\right\}$

$E\cap F=\{(5,3),(6,2)\}$

$P(F)=\frac{18}{36}$ and $P(E\cap F)=\frac{2}{36}$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}{\raisebox{1ex}{$18$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}=\frac{2}{18}=\frac{1}{9}$