Total number of blocks in the kit = 120
Number of red blocks = 40
Number of blue blocks = 25
Number of green blocks = 30
Number of yellow blocks = 120 - (40 + 25 + 30)
= 120 - 95 = 25
(i) (d): P(block is red) \(\begin{equation} =\frac{40}{120}=\frac{1}{3} \end{equation}\)
(ii) (c): P(block is not yellow) = 1 - P(block is yellow)
\(\begin{equation} =1-\frac{25}{120}=1-\frac{5}{24}=\frac{19}{24} \end{equation}\)
(iii) (c) : P(block is green) = \(\begin{equation} \frac{30}{120}=\frac{1}{4} \end{equation}\)
(iv) (b) : P(block is yellow) = \(\begin{equation} \frac{5}{24} \end{equation}\)
(v) (b): P(block is not blue) = 1 - P(block is blue)
\(\begin{equation} =1-\frac{25}{120}=1-\frac{5}{24}=\frac{19}{24} \end{equation}\)

Sample space (5) = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
\(\Rightarrow\)n(5) = 8
(i) (c): Let A be the event of getting atmost one tail.
 A = {HHH, HHT, HTH, THH}
\(\Rightarrow\)n(A) = 4
Required probability = \(\begin{equation} \frac{4}{8}=\frac{1}{2} \end{equation}\)
(ii) (d): Let B be the event of getting exactly 1 head. B = {HTT, THT, TTH}
\(\Rightarrow\) n(B) = 3
Required probability = \(\begin{equation} \frac{3}{8} \end{equation}\)
(iii) (d) : Let C be the event of getting exactly 3 tails.
C = {TTT} \(\Rightarrow\) n( C) = 1
Required probability = \(\begin{equation} \frac{1}{8} \end{equation}\)\(\begin{equation} \frac{1}{8} \end{equation}\)\(\begin{equation} \frac{1}{8} \end{equation}\)\(\begin{equation} \frac{1}{8} \end{equation}\)
(iv) (b) : Let D be the event of getting atmost 3 heads.
D = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
\(\Rightarrow\)n(D) = 8
Required probability = \(\begin{equation} \frac{8}{8}=1 \end{equation}\)
(v) (c): Let E be the event of getting atleast two heads.
E = {HHT, HTH, THH, HHH}
\(\Rightarrow\)n(E) = 4
Required probability = \(\begin{equation} \frac{n(E)}{n(S)}=\frac{4}{8}=\frac{1}{2} \end{equation}\)

Total time = 3 mins = 3 x 60 sees = 180 secs
(i) (a): Required probability = \(\begin{equation} \frac{30}{180}=\frac{1}{6} \end{equation}\)
(ii) (a): Required probability = \(\begin{equation} \frac{45}{180}=\frac{1}{4} \end{equation}\)
(iii) (d) : P(music will stop within 2 mins) = \(\begin{equation} \frac{120}{180}=\frac{2}{3} \end{equation}\)
P(music will stop after 2 mins) = \(\begin{equation} 1-\frac{2}{3}=\frac{1}{3} \end{equation}\)
(iv) (b) : Required probability = 1 - P(music will stop within first 60 secs)
\(\begin{equation} =1-\frac{60}{180}=1-\frac{1}{3}=\frac{2}{3} \end{equation}\)
(v) (b): Required probability = \(\begin{equation} \frac{82}{180}=\frac{41}{90} \end{equation}\)

Since, every student get one chocolate. So, number of chocolates Rohit has is equal to the number of students in the class.
(i) (a): Let number of milk chocolates Rohit has = x
Probability of distributing milk chocolates = \(\begin{equation} \frac{1}{3} \end{equation}\)
\(\begin{equation} \Rightarrow \frac{x}{54}=\frac{1}{3} \Rightarrow x=\frac{54}{3}=18 \end{equation}\)
(ii) (c): Let number of dark chocolates Rohit has = y
Probability of distributing dark chocolates = \(\begin{equation} \frac{4}{9} \end{equation}\)
\(\begin{equation} \Rightarrow \frac{y}{54}=\frac{4}{9} \Rightarrow y=\frac{4 \times 54}{9}=24 \end{equation}\)
(iii) (d) : Number of white chocolates Rohit has = 54 -(18 + 24) = 12
Required probability = \(\begin{equation} \frac{12}{54}=\frac{2}{9} \end{equation}\)
(iv) (b) : Total number of milk and white chocolates = 18 + 12 = 30
Required probability = \(\begin{equation} \frac{30}{54}=\frac{5}{9} \end{equation}\)
(v) (b): Since all students gets one chocolate. So, total number of chocolates distributed = 54
\(\begin{equation} \text { Required probability }=\frac{54}{54}=1 \end{equation}\)

Total number of puppets in claw crane = 58 + 42 + 36 + 64 = 200
(i) (b): P(picking a tiger) = \(\begin{equation} \frac{36}{200}=\frac{9}{50} \end{equation}\)
(ii) (a): P(picking a monkey) = \(\begin{equation} \frac{64}{200}=\frac{8}{25} \end{equation}\)
(iii) (c) : P(picking a teddy bear) = \(\begin{equation} \frac{58}{200}=\frac{29}{100} \end{equation}\)
(iv) (d) : P(not picking a monkey) = 1 - P(picking a monkey)
\(\begin{equation} =1-\frac{8}{25}=\frac{17}{25} \end{equation}\)
(v) (d): P(picking a pokemon) = \(\begin{equation} \frac{42}{200}=\frac{21}{100} \end{equation}\)
P(not picking a pokemon) = 1 - P(picking a pokemon)
\(\begin{equation} =1-\frac{21}{100}=\frac{79}{100} \end{equation}\)