Probability

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,

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

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

$=\frac{1}{6}$

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

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

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 $P(A)P(B^{\prime })+P(B)P(A^{\prime })=\frac{1}{2}*\frac{2}{3}+\frac{1}{2}*\frac{1}{3}$

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

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 $=\frac{1}{2}*\frac{1}{2}*\frac{1}{2}=\frac{1}{8}$

So, Probability of getting an odd number at least once

$=1-$ probability of getting an odd number in no throw

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

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

$=\frac{7}{8}$

$\therefore$ Probability of getting an odd number at least once $=\frac{7}{8}$

27. Given,

$P(A)=0.3$

$P(B)=0.6$

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

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

$=0.3*0.6$

$=0.18$

$\mathrm{ii.\; P}$ ( $A$ 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,

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

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

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

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

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

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

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

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

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

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

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.