Answer: Option B. -> $\frac{7}{15}$
:
B

The number of ways to arrange 7 white an 3 black balls in a row
$\frac{10!}{7!.3!}=\frac{\mathrm{10.9.8}}{\mathrm{1.2.3}}=120$
Numbers of blank places between 7 balls are 6. There is 1 place before first ball and 1 place after last ball. Hence total number of places are 8.
Hence 3 black balls are arranged on these 8 places so that no two black balls are together in number of ways.
${=}^8{C}_3=\frac{8\times 7\times 6}{1\times 2\times 3}=56$
So required probability = $\frac{56}{120}=\frac{7}{15}$

Answer: Option D. -> None of these 
:
D

Let the events are
$R_1$ = A red ball is drawn from urn A and placed in B
$B_1$ = A black ball is drawn from urn A and placed in B
$R_2$ = A red ball is drawn from urn A and placed in A
$B_2$ = A black ball is drawn from urn A and placed in B
R = A red ball is drawn in the second attempt from A
Then the required probability
$=P\left(R_1{R}_{2}R\right)+\left(R_1{B}_{2}R\right)+P\left(B_1{R}_{2}R\right)+P\left(B_1{B}_{2}R\right)$
$=P\left(R_1\right)P\left(R_2\right)P\left(R\right)+P\left(R_1\right)P\left(B_2\right)P\left(R\right)+P\left(B_1\right)P\left(R_2\right)P\left(R\right)+P\left(B_1\right)P\left(B_2\right)P\left(R\right)$

$=\frac{6}{10}\times \frac{5}{11}\times \frac{6}{10}\times \frac{6}{10}\times \frac{6}{11}\times \frac{5}{10}\times \frac{4}{10}\times \frac{4}{11}\times \frac{7}{10}+\frac{4}{10}\times \frac{7}{11}\times \frac{6}{10}$
$=\frac{32}{55}$

Answer: Option D. -> $3:2$
:
D

Let p be the probability of the other event, then the probability of the first event is $\frac{2}{3}$ p. Since two events are totally exclusive, we have $p+\left(\frac{2}{3}\right)p=1\Rightarrow p=\frac{3}{5}$
Hence odds in favour of the other are 3 : 5 – 3, i.e, 3 : 2.

Answer: Option D. -> All of the above
:
D
We have $P(A\cup B)\ge$           max. $\left\{P\right(A),P(B\left)\right\}=\frac{2}{3}$
$P(A\cap B)\le$               min.  $\left\{P\right(A),P(B\left)\right\}=\frac{1}{2}$
and $P(A\cap B)=P\left(A\right)+P\left(B\right)-P(A\cup B)\ge P\left(A\right)+P\left(B\right)-1=\frac{1}{6}$
$\Rightarrow \frac{1}{6}\le P(A\cap B)\le \frac{1}{2}$
$P(A^{\prime }\cap B)=P\left(B\right)-P(A\cap B)$
Hence $\frac{2}{3}-\frac{1}{2}\le P(A^{\prime }\cap B)\le \frac{2}{3}-\frac{1}{6}$
$\Rightarrow \frac{1}{6}\le P(A^{\prime }\cap B)\le \frac{1}{2}$

Answer: Option C. -> $\frac{3}{7}$
:
C
A leap year consists of 366 days comprising of 52 weeks and 2 days. There are 7 possibilities for these 2 extra days.
(i) Sunday, Monday                       
(ii) Monday, Tuesday                    
(iii) Tuesday, Wednesday            
(iv) Wednesday, Thursday              
(v) Thursday, Friday                     
(vi) Friday, Saturday             
(vii) Saturday, Sunday
Let us consider two events :
A : the leap year contains 53 Sundays
B : the leap year contains 53 Mondays.
Then we have $P\left(A\right)=\frac{2}{7},P\left(B\right)=\frac{2}{7},P(A\cap B)=\frac{1}{7}$
$\therefore$ Required probability = $P(A\cup B)$
$=P\left(A\right)+P\left(B\right)-P(A\cup B)=\frac{2}{7}+\frac{2}{7}-\frac{1}{7}=\frac{3}{7}$

Answer: Option A. -> $\frac{32}{55}$
:
A

Let the events are
$R_1$ = A  red ball is drawn from urn A and placed in B
$B_1$ = A  black ball is drawn from urn A and placed in B
$R_2$ = A  red ball is drawn from urn B and placed in A
$B_2$ = A  black ball is drawn from urn B and placed in A
R = A  red ball is drawn in the second attempt from A
Then the required probability
= $P\left(R_1{R}_{2}R\right)+\left(R_1{B}_{2}R\right)+P\left(B_1{R}_{2}R\right)+P\left(B_1{B}_{2}R\right)$
= $P\left(R_1\right)P\left(R_2\right)P\left(R\right)+P\left(R_1\right)P\left(B_2\right)P\left(R\right)+P\left(B_1\right)P\left(R_2\right)P\left(R\right)+P\left(B_1\right)P\left(B_2\right)P\left(R\right)$
$=\frac{6}{10}\times \frac{5}{11}\times \frac{6}{10}+\frac{6}{10}\times \frac{6}{11}\times \frac{5}{10}+\frac{4}{10}\times \frac{4}{11}\times \frac{7}{10}+\frac{4}{10}\times \frac{7}{11}\times \frac{6}{10}$
$=\frac{32}{55}$

Answer: Option D. -> $3:2$
:
D

Let p be the probability of the other event, then the probability of the first event is $\frac{2}{3}$ p. Since two events are totally exclusive, we have $p+\left(\frac{2}{3}\right)p=1\Rightarrow p=\frac{3}{5}$
Hence odds in favour of the other are 3 : 5 – 3, i.e, 3 : 2.

Answer: Option A. -> 8
:
A

Let n be the least number of bombs required and x the number of bombs that hit the bridge. Then x follows a binomial distribution with parameter n and $p=\frac{1}{2}$
Now, $P(X\ge 2)>0.9\Rightarrow 1-P(X<2)>0.9$
$\Rightarrow P(X=0)+P(X=1)<0.1$
$\Rightarrow {}^n{C}_0{\left(\frac{1}{2}\right)}^n+{}^n{C}_1{\left(\frac{1}{2}\right)}^{n-1}\left(\frac{1}{2}\right)<0.1\Rightarrow 10(n+1)<2^n$
This gives $n\ge 8$

Answer: Option B. -> 0.2
:
B
$P(\overline{A}\cap \overline{B})=P(\overline{A}\cup \overline{B})-1-P(A\cup B)$
Since A and B are mutually exclusive,
$P(A\cup B)=P\left(A\right)+P\left(B\right)$
Hence required probability $=1–(0.5+0.3)=0.2$

Answer: Option A. -> $\frac{1}{4}$
:
A
Since m and n are selected between 1 and 100, hence sample space = $100\times 100.$
Also $7^1=7,7^2=49,7^3=343,7^4=2401,7^5=16807$ etc. Hence 1, 3, 7 and 9 will be the last digits in the powers of 7. Hence for favorable cases
$nm\to \downarrow 1,11,21,\mathrm{3....}1,1002,12,22,\mathrm{3...}2,100...........................................100,1100,2100,3100,100$
For  m = 1; n = 3, 7, 11….. 97
$\therefore$ Favorable cases = 25
Similarly for every m, favorable n are 25.
$\therefore$ Total favourable cases = $100\times 25$
Hence required probability = $\frac{100\times 25}{100\times 100}=\frac{1}{4}$