Ncert Solutions Maths class 12th

24. Given,

$P(A)=0.3$

$P(B)=0.4$

$A$ and $B$ are independent

$\mathrm{(i)\; P}(A\cap B)=P(A).P(B)$

$\begin{array}{l}=0.3*0.4\\ =0.12\end{array}$

$\mathrm{(ii)\; P}(A\cup B)=P(A)+P(B)-P(A\cap B)$

$\begin{array}{l}P(A\cup B)=0.3+0.4-0.12\\ =0.58\end{array}$

$\mathrm{(iii)\; P}(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{0.12}{0.4}=0.3$

$\mathrm{(iv)\; P}(B/A)=\frac{P(B\cap A)}{P(A)}=\frac{0.12}{0.3}=0.4$

23. Given,

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

$P(A\cup B)=\frac{3}{5}$

$P(B)=p$

(i) We know that,

when $A$ and $B$ are mutually exclusive

$(A\cap B)=\varnothing$ $,P(A\cap B)=0$

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

$\Rightarrow \frac{3}{5}=\frac{1}{2}+p$

$\Rightarrow p=\frac{3}{5}-\frac{1}{2}=\frac{6-5}{10}=\frac{1}{10}$

(ii) when $A$ and $B$ are independent

$\begin{array}{l}P(A\cap B)=P(A).P(B)\\ P(A\cap B)=\frac{1}{2}*p\end{array}$

Now,

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

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

$\Rightarrow \frac{3}{5}-\frac{1}{2}=\frac{2p-p}{2}$

$\Rightarrow \frac{6-5}{10}=\frac{p}{2}$

$\Rightarrow p=\frac{1*2}{10}=\frac{2}{10}=\frac{1}{5}$

$$22 . $P(F)=\frac{3}{10}$

$P(E\cap F)=\frac{1}{5}$

$P(E).P(F)=\frac{3}{5}*\frac{3}{10}$

$=\frac{9}{50}\ne P(E\cap F)$

Here,  $P(E).P(F)\ne P(E\cap F)$ . $A$ and $B$ are not independent.

21. When die is thrown, the sample space

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

Let,  $A$ =the number is even $=\{2,4,6\}$

$B$ =the number is red $=\{1,2,3\}$

$\begin{array}{l}P(A)=\frac{3}{6}=\frac{1}{2}\\ P(B)=\frac{3}{6}=\frac{1}{2}\\ A\cap B=\left\{2\right\}\\ P(A\cap B)=\frac{1}{6}\\ \therefore P(A).P(B)=\frac{1}{2}*\frac{1}{2}=\frac{1}{4}\ne P(A\cap B)\end{array}$

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

20. The sample space of given condition are

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

Let $A$ = head appear on the coin

$B$ = $3$ on the die

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

$P(A)=\frac{6}{12}=\frac{1}{2}$

$B=\{(H,3),(T,3)\}$

$P(B)=\frac{2}{12}=\frac{1}{6}$

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

$i.e.,P(A\cap B)=\frac{1}{12}$

$P(A).P(B)=\frac{1}{2}*\frac{1}{6}=\frac{1}{12}=P(A\cap B)$

Therefore, $A$ and $B$ are independent.

19. Let $A$ ,  $B$ and $C$ be the respective events that the first, second and third drawn orange is good.

$P(A)=$ Probability that first drawn orange is good= $\frac{12}{15}$

The orange is not replaced;

$P(B)=$ Probability of getting second orange is good= $\frac{11}{14}$

Similarity, probability of getting third orange is good,  $P(C)=\frac{10}{13}$

$\therefore$ Probability of getting all orange good= $\frac{12}{15}*\frac{11}{14}*\frac{10}{13}$ $=\frac{44}{91}$

Therefore, probability that will approve for sale $=\frac{44}{91}$

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.