Let number of bad eggs=x
Probability of getting a bad egg=\(x\over800\)
A.T.Q.   \({x\over800}=0.125\)
x=0.125 x 800=100

Total number of cards = 52
Number of aces and kings = 4 + 4 = 8
Number of cards which are neither ace nor king = 44
Probability that the card drawn is neither an ace nor a king = \(\frac{44}{52}=\frac{11}{13}\)

Total number of ways = 6
Number of ways to get a number less than 3 = 2
\(\therefore\)  Required probability = \(\frac{2}{6}=\frac{1}{3}\)

Area of restangle = 3 \(\times\) 2 = 6 m²
and area of circle of radius \(\frac{1}{2} \mathrm{~m}=\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4} \mathrm{~m}^2\)
\(\left[\because \text { diameter }=1 \mathrm{~m} \Rightarrow \text { radius }=\frac{1}{2} \mathrm{~m}\right]\)
Now, probanility that the die land inside the circle
\(=\frac{\text { Area of circle }}{\text { Area of rectangle }}=\frac{\pi / 4}{6}=\frac{\pi}{24}\)

(i) Let E be an event of getting an orange flavoured candy.  All possible outcomes are against the event E because in a bag, all candies are lemon flavoured. So, P(E) = 0
(ii) Let F be an event of getting a lemon flavoured candy. All possible outcomes are favourable to event F because all the candies in the bag are lemon flavoured.
So. P(F) =1

Card removed = king, queen and jack of clubs = 3 
Cards left = 52- 3 = 49
(i) Number. of hearts = 13
Probability of drawing a heart = \(\frac {13}{49}\) 
(ii) Number of king = 4
Number of kings left = 4 - 1 = 3
Probability of drawing a king = \(\frac {3}{49}\) 
(iii) 3 clubs are removed
Number of clubs left=13-3=10
Probability of drawing a club = \(\frac {10}{49}\)
(iv) There is only one '10' of hearts.
\(\therefore\)Probability of drawing a '10' of hearts =\(\frac {1}{49}\)

As black jacks, black kings and black aces are removed
Then remaining cards are 52 - 6 = 46.
\(\therefore \) Total outcomes of drawing the card are 46.
(a) Favourable outcomes of a red card are 26.
 \(\therefore \)   Probability of a red card = \(\frac { 26 }{ 46 } =\frac { 13 }{ 23 } \)
(b) Favourable outcomes for not a diamond are 46-13 = 33
\(\therefore \) Probability of not a diamond card = \(\frac { 33 }{ 46 } \).

From the total playing 52 cards. red coloured jacks, queen. kings and aces are removed (i.e.. 2 jacks. 2 queens, 2 kings, 2 aces) \(\therefore\) Remaining cards = 52-8 = 44
(i) Favourable cases for a black queen are 2 (i.e., queen club or spade)
\(\therefore\)Probability of drawing a black queen
=\(\frac { \ Favourable \ cases \ for \ black \ queen } {Total \ possible \ cases}\) =\(\frac {2}{44}=\frac {1}{22}\)
(ii) Favourable cases for red cards are 26 -8 = 18 (as 8 cards have been removed) (i.e. 9 diamonds + 9 hearts) 
\(\therefore\)Probability of drawing a red card =  \(\frac { \ Favourable \ cases \ for \ \ a red \ card } {Total \ possible \ cases}\) =\(\frac {18}{44}=\frac {9}{22}\)
(iii) Favourable cases for a black jack are 2 (i.e. jacks of club or spade)
\(\therefore\)Probability of drawing a black jack =\(\frac { \ Favourable \ cases \ for \ a \ black \ jack } {Total \ possible \ cases}\)=\(\frac {2}{44}=\frac {1}{22}\)
(iv)Favourable cases for a picture card 6 (i.e. 2 black jacks, queens and kings each)
\(\therefore\)Probability of drawing a picture card
=\(\frac { \ Favourable \ cases \ for \ a \ picture \ card } {Total \ possible \ cases}\)=\(\frac {6}{44}=\frac {3}{22}\)

Total number of balls = 19
(i) Prime numbers from 1 to 19 are 2, 3, 5, 7, 9, 11, 13, 17, 19 = Total 8 prime numbers
\(\therefore\)  Probability of drawing a prime number = \(\frac{8}{19}\)
(ii) Numbers divisible by 3 or 5 are 3, 6, 9, 15, 18, 10, 5, 12 = Total 8 numbers
\(\therefore\) Probability of drawing a number divisible by 3 or 5 = \(\frac{8}{19}\)
(iii) Number divisible by 5 and 10 are 5, 10, 15 = Total 3
\(\therefore\)  Numbers which are neither divisible by 5 nor 10 are 19 - 3 = 16
 \(\therefore\) Probability of drawing a number which is neither divisible by 5 nor by 10 = \(\frac{16}{19}\)
(iv) Even numbers from 1- 19 are 2, 4, 6, 8, 10, 12, 14, 16, 18 [Total 9 even numbers]
\(\therefore\)  Probability of drawing an even number = \(\frac{9} {19}\)

Total number of outcomes = 36
(i) Let E = Event of getting 5 on atleast one die
Then, E would consist of 11 outcomes, namely (1,5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 6), (5, 1), (5, 2), (5, 3), (5, 4) and (5, 6).
\(\therefore\) Number of outcomes favourable to E = 11
Hence, probability that 5 will come up atleast once,
\(P(E)=\frac{11}{36}\)
(ii) Probability that 5 will not come up either time,
\(P(\bar{E})=1-P(E)=1-\frac{11}{36}=\frac{25}{36}\)

On a die, there are six numbers 1, 2, 3, 4, 5 and 6.
\(\therefore\) Total number of possible outcomes =6
(i) Let E₁ = Event of getting a prime number
Then, E₁ would consist of three outcomes namely 2, 3 and 5.
\(\therefore\)Number of outcomes favourable to E₁ = 3
Probability of getting a prime number,
\(P\left(E_1\right)=\frac{3}{6}=\frac{1}{2}\)
(ii) Let E₂ = Event of getting a number lying between 2 and 6
Then, E₂ would consist of three outcomes, namely 3, 4 and 5.
\(\therefore\) Number of outcomes favourable to E₂ = 3
Probability of getting a number lying between 2 and 6,
\(P\left(E_2\right)=\frac{3}{6}=\frac{1}{2}\)
(iii) Odd numbers = 1, 3, 5
\(\therefore\)  P(an odd number) = \(\frac{3}{6}=\frac{1}{2}\)

Total number of marbles = 5 + 8 + 4 = 17
(i) P(red marble) = \(\frac{5}{17}\)
(ii) P(white marble) = \(\frac{8}{17}\)
(iii) Let E₃ = Event of drawing a marble which is not green = Event of drawing either red or white marble
\(\therefore\) Number of outcomes favourable to E₃ = 5 + 8 = 13
So, probability of taken out not a green marble,
P(E₃)=\(\frac{13}{17}\)

Number of outcomes in a throw of a pair of dice 6 X 6 = 36
(i) Number of cases getting the sum 3 = 2 i.e., [(1, 2), (2,1)]
\(\therefore\)Probability of getting sum 3 =\(\frac {2}{36}=\frac {1}{18}\) 
Number of cases getting the sum 4 = 3 i.e., [(1,3)(2,2)(3,1)]
\(\therefore\) Probability of getting sum 4= \(\frac {3}{36}=\frac {1}{12}\) 
Number of cases of getting the sum 5 = 4 i.e., [(1,4),(2,3),(3,2),(4,1)]
 \(\therefore\)Probability of getting sum 5= 36 9 Number of cases of getting the sum 6 = 5 i.e.,[(1,5),(2,4),(3,3),(4,2),(5,1)]
\(\therefore\)Probability of getting sum 6 = \(\frac{5}{36}\)
 Number of cases of getting the sum 7 = 6 i.e.,[ (1,6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)]
\(\therefore\) Probability of getting sum 7=\(\frac{6}{36}=\frac{1}{6}\)
Number of cases of getting the sum 9 = 4 i.e., [(3,6),(4,5),(5,4),(6,3)]
\(\therefore\)Probability of getting sum 9 = \(\frac{4}{36}=\frac{1}{9}\)
Number of cases of getting the sum 10= 3 i.e., [(5,5),(4,6),(6,4)]
\(\therefore\)Probability of getting sum 10 =\(\frac{3}{36}=\frac{1}{12}\)
Number of cases of getting the sum 11 = 2  i.e.,[(5,6),(6,5)]
\(\therefore\)Probability of getting sum 11 = \(\frac{2}{36}=\frac{1}{18}\)

(ii) No, the outcomes as 11 different sum are not equally likely. 

Possible outcomes on tossing a coin 3 times, are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
\(\therefore\)  Total number of outcomes = 8
(i) Let E₁ be the event that Sweta losses the entry fee
i.e. shen tosses tail three times i.e. TTT.
\(\therefore\) Number of outcomes favourable to E₁ = 1
Hence, required probability = P(E₁) = \(\frac{1}{8}\)
(ii) Let E₂, be the event that Sweta gets double entry fee
i.e. she tosses heads three times
i.e. HHH
\(\therefore\) Number of outcomes favourable to E₂ = 1
Hence, required probability \(=P\left(E_2\right)=\frac{1}{8}\)
(iii) Let E₃ be the event that Sweta gets her entry fee back
i.e. Sweta gets heads one or two times
i.e. event of getting
HTT, THT, TTH, HHT, HTH or THH
\(\therefore\) Number of outcomes favourable to E₃ = 6
Hence, required probability = P(E₃) = \(\frac{6}{8}=\frac{3}{4}\)