Ncert Solutions Maths class 12th

$$34. $P(A^{\prime }{B}^{\prime })=[1-P(A)][1-P(B)]$

$\Rightarrow P(A^{\prime }\cap B^{\prime })=1-P(A)-P(B)+P(A)P(B)$

$\Rightarrow -[P(A)+P(B)-P(A\cap B)]=-P(A)-P(B)+P(A)(B)$

$\Rightarrow -P(A)-P(B)+P(A\cap B)=-P(A)-P(B)+P(A)P(B)$

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

Therefore, it shows A and B are independant

33. The outcome of rolling $2$ dice is $36$ . The only prime number is $2$ .

Let,  $E=$ event of getting an even prime number on each die

$E=\{(2,2)\}$ therefore

$P(E)=\frac{1}{36}$

Therefore, the correct answer is $(D)$ $\frac{1}{36}$

32. Let,

$P(H)=$ denote the student who read Hindi newspaper

$P(E)=$ denote the student who read English newspaper

Then, $P(H)=60\mathrm{\%}=\frac{6}{10}=\frac{3}{5}$

$P(E)=40\mathrm{\%}=\frac{4}{10}=\frac{2}{5}$

$\mathrm{i.\; P}(H\cap E)=20\mathrm{\%}=\frac{2}{10}=\frac{1}{5}$

$\Rightarrow P(H\cup E{)}^{\prime }=1-[P(H)+P(E)-P(H\cap E)]$

$=1-\left[\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\right]$

$=1-\frac{3+2-1}{5}$

$=1-\frac{4}{5}=\frac{1}{5}$

ii. Probability of randomly chosen student that reads English newspaper, if she reads Hindi newspaper, is given by

$P(E/H)=\frac{P(E\cap H)}{P(H)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}=\frac{1}{3}$

iii. Probability that randomly chosen student reads Hindi newspaper

31. i. In a deck of $52$ cards,

$13$ are spades and $4$ are aces.

$\therefore P(E)=P$ (the card drawn is spade) $=\frac{13}{52}=\frac{1}{4}$

$\therefore P(F)=P$ (the card drawn is an ace) $=\frac{4}{52}=\frac{1}{13}$

$\therefore P(E\cap F)=P$ (the card drawn is spade and an ace) $=\frac{1}{52}$

$P(E).P(F)=\frac{1}{4}\times \frac{1}{13}=\frac{1}{52}=P(E\cap F)$

$\Rightarrow P(E).P(F)=P(E\cap F)$

$E$ and $F$ are independent.

ii. In a deck of $52$ cards,

$26$ cards are black and $4$ a

30. Let, $P(A)=$ probability of solving problem by $A=\frac{1}{2}$

$P(B)=$ probability of solving problem by $B=\frac{1}{3}$

Since, the problem is solved independently by $A$ and $B$ , then,

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

$\Rightarrow P(A\cap B)=\frac{1}{2}\times \frac{1}{3}$

$=\frac{1}{6}$

$P(A^{\prime })=1-P(A)=1-\frac{1}{2}=\frac{1}{2}$

$P(B^{\prime })=1-P(B)=1-\frac{1}{3}=\frac{2}{3}$

i. Probability of the problem is solved

$\Rightarrow P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$\Rightarrow P(A\cup B)=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$

$=\frac{3+2-1}{6}$

$=\frac{4}{6}=\frac{2}{3}$

ii. Probability that exactly one of the solved problem is given by $P(A)P(B^{\prime })+P(B)P(A^{\prime })=\frac{1}{2}\times \frac{2}{3}+\frac{1}{2}\times \frac{1}{3}$

$=\frac{2}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$

29. Total number of balls $=18$

Number of red balls $=8$

Number of black balls $=10$

i. Probability of getting red ball at first draw $=\frac{8}{18}=\frac{4}{9}$

The red ball is replaced after the first draw

$\therefore$ Probability of getting red ball in second draw $=\frac{8}{18}=\frac{4}{9}$

$\therefore$ Probability of getting both the balls red $=\frac{4}{9}\times \frac{4}{9}=\frac{16}{81}$

ii. Probability o

28. The sample space of given condition is

$S=\{1,2,3,4,5,6\}$

Let,  $P(A)=$ probability of getting an odd number in first throw

$\Rightarrow P(A)=\frac{3}{6}=\frac{1}{2}$

$P(B)=$ probability of getting an even number

$\Rightarrow P(B)=\frac{3}{6}=\frac{1}{2}$

Probability of getting an even number in three times $=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}$

So, Probability of getting an odd number at least once

$=1-$ probability of getting an odd

27. Given,

$P(A)=0.3$

$P(B)=0.6$

$\mathrm{i.\; A}$ and $B$ are independent

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

$=0.3\times 0.6$

$=0.18$

$\mathrm{ii.\; P}$ ( $A$ and not $B$ )

$\Rightarrow P(A\cap B^{\prime })=P(A)-P(A\cap B)$

$=0.3-0.18$

$=0.12$

$\mathrm{iii.\; P}$ ( $A$ or $B$ )

$\Rightarrow P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$=0.3+0.6-0.18$

$=0.72$

$\mathrm{iv.\; P}$ (neither $A$ nor $B$ )

$\Rightarrow P(A^{\prime }\cap B^{\prime })=P(A\cup B{)}^{\prime }$

$=1-P(A\cup B)$

$=1-0.72$

$=0.28$

26. Given,

$P(A)=\frac{1}{2}$

$P(B)=\frac{7}{12}$

$P$ (not $A$ or not $B$ )= $\frac{1}{4}$

$\Rightarrow P(A^{\prime }\cup B^{\prime })=P(A\cap B{)}^{\prime }$

$\Rightarrow P(A\cap B{)}^{\prime }=\frac{1}{4}$

$\Rightarrow 1-P(A\cap B)=\frac{1}{4}$

$\Rightarrow P(A\cap B)=1-\frac{1}{4}$

$\Rightarrow \frac{4-1}{4}$

$\Rightarrow \frac{3}{4}$

Now,

$P(A).P(B)=\frac{1}{2}\times \frac{7}{12}$

$=\frac{7}{24}\ne \frac{3}{4}$

$\Rightarrow P(A).P(B)\ne P(A\cap B)$

Therefore, $A$ and $B$ are not independent.

25. Given,

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

$P(B)=\frac{1}{2}$

$P(A\cap B)=\frac{1}{8}$ Here,  $P$  (not $A$ not $B$ ) $=P(A^{\prime }\cap B^{\prime })=P(A\cup B{)}^{\prime }$

$\Rightarrow 1-P(A\cup B)$

$\Rightarrow 1-\left[P(A)+P(B)-P(A\cap B)\right]$

$\Rightarrow 1-\left[\frac{1}{4}+\frac{1}{2}-\frac{1}{8}\right]$

$\Rightarrow 1-\left[\frac{2+4-1}{8}\right]$

$\Rightarrow 1-\frac{5}{8}=\frac{8-5}{8}$

$\Rightarrow \frac{3}{8}$