(i) (a): \(=\frac{4\left({ }^{13} \mathrm{C}_{4}\right)}{{ }^{52} \mathrm{C}_{4}}\)
As we know there are four suits,
Favourable cases are: ¹³C₄ + ¹³C₄ + ¹³C₄ + ¹³C₄  
= 4(¹³C₄)
Required probabilty \(=\frac{4\left({ }^{13} \mathrm{C}_{4}\right)}{{ }^{52} \mathrm{C}_{4}}\) 
(ii) (d): \(\frac{15229}{54145}\) 
Since, there are 4 aces in the pack of 52 cards, therefore, the no.of ways of drawing 4 cards so that no card is an ace = ⁴⁸C₄. 
Probability of four cards so that none is an ace
\(=\frac{{ }^{48} C_{4}}{{ }^{52} C_{4}} \)
\(=\frac{48 \times 47 \times 46 \times 45}{52 \times 51 \times 50 \times 49} \)
\(=\frac{38916}{54145}\)
Thus, required probability = \( 1-\frac{38916}{54145} \)
\(= \frac{15229}{54145}\)
(iii) (d): \(\frac{2197}{20825}\) 
Favourable cases to draw one card from each suit is:¹³C₁ x ¹³C₁ x ¹³C₁ x ¹³C₁  = (¹³C1)₄
Required probability \( =\frac{\left({ }^{13} C_{1}\right)^{4}}{{ }^{52} C_{4}} \)
\(=\frac{13 \times 13 \times 13 \times 13}{\frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1}} \)
\(=\frac{2197}{20825}\) 
(iv) (a): \(2\left(\frac{{ }^{26} C_{4}}{{ }^{52} C_{4}}\right)\)
Total favourable cases of drawing all cards of same colour  = ²⁶C4 x ²⁶C₄  = 2(²⁶C₄)
Required probability = \(\frac{2\left({ }^{26} \mathrm{C}_{4}\right)}{{ }^{52} \mathrm{C}_{4}}\)
(v) (a): \(\frac{1}{270725}\)
Required probability \( =\frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} \times \frac{1}{49} \)
\(=\frac{1}{270725} \)

(i) (a): \(\frac{1}{26}\)
Favourable cases (2, 6), (6, 2), (4, 4), (5, 3), (3, 5)
Probability of getting the sum as eight =\(\frac{5}{36}\) 
(ii) (d): 0
As the sum of numbers on two dice is 13 will not be possible, therefore zero probability.
(iii) (a): 1
As the sum on the two dice will always be less then or equal to 12, therefore probability = 1
(iv) (c):  \(\frac{1}{6}\) 
Favourable events of getting sum = 7 i.e., (1, 6), (6, 1), (5, 2), (4, 3), (3, 4), (2, 5)
Required probability \(=\frac{6}{36}=\frac{1}{6}\)
(v) (d): \(\frac{5}{18}\)
Favourable events = (6, 3), (3, 6), (5, 4), (4, 5), (5, 5),(5, 6), (6, 5), (6, 6), (4, 6), (6, 4)
Required probability = \(\frac{10}{36}=\frac{5}{18}\)

(i) (a): \(\frac{4}{13}\)
Favourable events are :
Total red cards = 26 ( includes two kings)
Two black kings = total cards =28
Total cards left (favourable) = 52 – 28 = 24
Required probability = \(\frac{24}{52}=\frac{4}{13}\) 
(ii) (a): \(\frac{1}{13}\)
Total number of aces = 4
Required probability \(=\frac{4}{52}=\frac{1}{13}\)
(iii) (b): \(\frac{1}{2}\)
Total number of red cards = 26
Required probability \(=\frac{26}{52}=\frac{1}{2}\) 
(iv) (a): \(\frac{1}{26}\)
There are two red colour ace cards. Therefore, the required probability is
\(\frac{2}{52}=\frac{1}{26}\)
(v) (d): \(\frac{24}{169}\)
When one ace is there, then X= 1
Required condition:
(i) ace is selected first and then no ace (ii) no ace is selected first and an ace is selected
\( P(X=1)=\frac{4}{52} \times \frac{48}{52}+\frac{48}{52} \times \frac{4}{52} \)
\(\Rightarrow \ P(X=1)=\frac{24}{169}\)

(i) (d): \(\frac{2}{5}\)
Probability of Varun hitting the target = \(\frac{4}{5}\)
Probability of Abhay hitting the target = \(\frac{3}{4}\)
Probability of Rohit to hit the target =\(\frac{2}{3}\)
If all may hit the target, then required probabilty
\(=\frac{4}{5} \times \frac{3}{4} \times \frac{2}{3}=\frac{2}{5}\)
(ii) (a): \(\frac{1}{10}\)
Required probability = \(=\left(1-\frac{4}{5}\right) \times \frac{3}{4} \times \frac{2}{3}\)
\(\frac{1}{10}\)
(iii) (d): \(\frac{13}{30}\)
Required probability
\( =\frac{4}{5} \times \frac{3}{4} \times\left(1-\frac{2}{3}\right)+\frac{4}{5} \times\left(1-\frac{3}{4}\right) \times \frac{2}{3}+\left(1-\frac{4}{5}\right) \times \frac{3}{4} \times \frac{2}{3} \)
\(=\frac{4}{5} \times \frac{3}{4} \times \frac{1}{3}+\frac{4}{5} \times \frac{1}{4} \times \frac{2}{3}+\frac{1}{5} \times \frac{3}{4} \times \frac{2}{3} \)
(iv) (d): \(\frac{1}{60}\)
For none of them will hit the target, we have to the find when they will not hit the target individually.
\( =\left(1-\frac{4}{5}\right) \times\left(1-\frac{3}{4}\right) \times\left(1-\frac{2}{3}\right) \)
\(=\frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \)
\(=\frac{1}{60}\)
(v) (b): False
Total probability can’t be more than one. Hence false.