Possible outcomes are 4, 9, 16, 25, 36, 49, i.e. 6.
\(\Rightarrow\) P(perfect square number) =  \(\frac{6}{48}​​\)or \(\frac{1}{8}​​\)

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Total number of cards in a deck of card = 52
Number of clubs = 13
\(\therefore\) Probability of drawing a club = \(\frac{13}{52}\)
= \(\frac{1}{4}\)

Total number of cards = 18
Even numbers from 3 to 20 are 4, 6, 8, 10, 12, 14, 16, 18, 20 = 9 numbers
Probability that the number on the card taken out is an even number
 \(=\frac{Number \ \ of \ \ favourable \ \ outcomes}{Total \ \ number \ \ of \ \ outcomes}=\frac{9}{18}=\frac{1}{2}\)

When we toss a coin,we get either head or tail which are equally likely outcomes. Hence, the result of the toss of a coin is completely unpredictable or unbiased. So, tossing a coin is a fair way of deciding.

(i)Total number of cards=52
Number of red coloured kings=2
p(a king of red colour)=\({2\over52}={1\over26}\)
(ii)Number of face cards=12
p(a face card)\(={12\over52}={3\over13}\)
(iii)Number of queen of diamonds=1
p(the queen of diamonds)\(={1\over52}\)

Number of cards left in pack=52-6
=46
(i)let A=black king is drawn
Number of black king in the pack=2
\(\therefore\ P(A)={2\over 46}={1\over 23}\)
(ii)B=a card of red colour is drawn
No.of red cards left in the pack = 26-6=20
\(\therefore\ P(B)={20\over 46}={10\over 23}\)
(iii)Let C = A card of black colour is drawn
No. of black card in the pack =26
\(\therefore\ P(C)={26\over46}={13\over 23}\)

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 =\(\frac {1}{49}\)

Let the number of blue balls be x.
Total number of balls in the bag = 5 + x
\(\therefore\) Probability of drawing a red ball = \(\frac{5}{5+x}\) and probability of drawing a blue ball = \(\frac{x}{5+x}\)
Given probability of drawing a blue ball = 3 X probability of drawing a red ball
\(3\times\frac{5}{5+x}=\frac{x}{5+x}\Rightarrow15=x\)
\(\Rightarrow\) Number of blue balls = 15

Total number of playing cards = 52
(i) Favourable cases for a king or a jack are 8 (4 kings + 4 jacks)
\(\therefore\)  Probability of drawing a king or a jack = \(\frac{8}{52}=\frac{2}{13}\)
(ii) Favourable cases for a non-ace are 48 (52 cards - 4 aces)
\(\therefore\)  Probability of drawing a non-ace = \(\frac{48}{52}=\frac{12}{13}\)
(iii) Favourable cases for a red cards are 26 (13 hearts + 13 diamonds)
\(\therefore\)  Probability of drawing a red card = \(\frac{26}{52}=\frac{1}{2}\)
(iv) Favourable cases for neither a king nor a queen are 44 (52 cards - 4 kings - 4 queen)
\(\therefore\)  Probability of drawing neither a king nor a queen = \(\frac{44}{52}=\frac{11}{13}\)

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,
\(P(\bar{E})=1-P(E)=1-\frac{11}{36}=\frac{25}{36}\)

Total number of ball pens = 144
\(\therefore\) Number of all possible outcomes = 144
(i) Let E = event of getting a good ball pen
Then, number of outcomes favourable to E
= Total number of ball pens - Defective ball pens
= 144 - 20 = 124
She will buy a pen, if it is a good pen.
\(\therefore\) Required probability = \(P(E)=\frac{124}{144}=\frac{31}{36}\)
(ii) She will not buy a pen, if it is not a good pen.
\(\therefore\) Required probability
= 1 - Probability of getting a good pen
\([\because P(E)+P(\bar{E})=1]\)
\(=1-P(E)=1-\frac{31}{36}=\frac{5}{36}\)

Total number of points on the circle = 8
(i) Let E₁ = Event of getting arrow at number 8
\(\therefore\) Number of outcomes favourable to E₁ = 1
Probability that arrow cones at number 8,
\(P\left(E_1\right)=\frac{1}{8}\)
(ii) Let E₂ = Event of getting arrow at an odd number
Here, odd numbers are 1, 3, 5 and 7.
\(\therefore\) Number of outcomes favourable to E₂ = 4
Probability that arrow comes at an odd number,
\(P\left(E_2\right)=\frac{4}{8}=\frac{1}{2}\)
(iii) Let E₃ = Event of getting arrow at a number greater than 2, i.e. at 3, 4, 5, 6, 7 or 8
\(\therefore\) Number of outcomes favourable to E₃ = 6
Probability that arrow comes at a number greater than 2,
\(P\left(E_3\right)=\frac{6}{8}=\frac{3}{4}\)
(iv) P(a number less than 9) = \(\frac{8}{8}\)=1

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)=\(\frac{13}{17}\)

Total number of pieces = 8 + 10 = 18
(i) No.of triangles = 8. Hence, P(triangle is lost) = \(\frac{8}{18}=\frac{4}{9}\)
(ii) No.of squares = 10. Hence, P(square is lost) = \(\frac{10}{18}=\frac{5}{9}\)
(iii) No.of squares of blue colour = 6. So, P(square of blue colour is lost) = \(\frac{6}{18}=\frac{1}{3}\)
(iv) No.of triangles of red colour = 8 - 3 = 5. So, P(triangle of red colour is lost) = \(\frac{5}{18}\)