(a) When three coins are tossed simultaneously, then the number of possible outcomes = 8,
(i.e. , HHH, HTH, THH, TTH, HHT, HTT, THT, TTT)
Number of favourable outcomes (three heads) = l, (i.e., HHH)
\(\therefore\)Required probability = \(\frac{1}{8}\) 
(b) Number of favourable outcomes (exactly 2 3, (i.e., HTH, THH, HHT)
\(\therefore\)Required probability =\(\frac{3}{8}\) 
(c) Number of favourable outcomes (at least two heads) = 4, (i.e., HTH, THH, HHT, HHH)
Required probability =\(\frac{4}{8}=\frac{1}{2}\) 

Number of blue marbles = 35
Number of white marbles = 25
Number of red marbles = 40
Total marbles in box = 35 + 25 + 40
=100
(i) Probability of a white marble = P (white) 
=\(\frac{25}{100}=\frac{1}{4}\)
(ii) Number of non-blue marbles = 100-35
 = 65
P (not blue) = \(\frac{65}{100}=\frac{13}{20}\) 
(iii) Number of white or blue marbles= 25 + 35
= 60
Number of neither white nor blue marbles = 100-60
= 40
P (neither white nor-blue marbles) =\(\frac{40}{100}=\frac{2}{5}\)

Total number of cards = 52
(i) Number of ace = 4
\(\therefore\)  Probability of drawing an ace = \(\frac{4}{52}=\frac{1}{13}\)
(ii) There is only one '2' of spades
\(\therefore\) Probability of drawing a '2' of spade = \(\frac{1}{52}\)
(iii) '10' of a black suit there are two cards
\(\therefore\)  Probability of drawing 10 of black suit = \(\frac{2}{52}=\frac{1}{26}\)

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 = \(\frac{9} {19}\)

(i) Total number of cards = 5
\(\therefore\) Number of all possible outcomes = 5
P (picking a queen card)= \(\frac{1}{5}\)
[\(\because\) as there is only one queen]
(ii) Suppose a queen is drawn and put a side. then, four cards are left namely, ten, jack, king and ace of diamonds.
Now, number of all possible outcomes = 4
(a) P(the second card picked up is an ace) = \(\frac{1}{4}\)
(b) P (the second card picked up is a queen)
\(=\frac{0}{4}=0\)   [\(\because\) queen is drawn before]

Since all the black face cards are removed, the total number of remaining cards = 46
(i) P(face card) = \(\frac{6}{46}\) = \(\frac{3}{23}\) 
(ii) P(red card) = \(\frac{26}{46}\) = \(\frac{13}{23}\)
(iii) P(black card) = -\(\frac{20}{46}\)= \(\frac{10}{23}\)
(iv) P(king card) = \(\frac{2}{46}\)= \(\frac{1}{23}\)

\((i){4\over13}(ii){1\over26}(iii){11\over13}(iv){2\over13}\)

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) = \(\frac{5}{18}\)

(i) Total number of discs in a box = 90
\(\therefore\) Number of all possible outcomes = 90
Let E₁ = Event of getting a disc bearing a two-digit number
Here, two-digit numbers are 10, 11, .., 90
\(\therefore\)Number of outcomes favourable to E₁ =81
Hence, probability of getting a disc bearing a two-digit number, \(P\left(E_1\right)=\frac{81}{90}=\frac{9}{10}\)
(ii) Let E₂ = Event of getting a disc bearing a perfect square number
Here, perfect square number are 1, 4, 9, 16, 25, 36, 49, 64 and 81.
\(\therefore\) Number of outcomes favourable to E₂ =9
Hence, probability of getting a disc bearing a perfect square number, \(P\left(E_2\right)=\frac{9}{90}=\frac{1}{10}\)
(iii) Let E₃ = Event of getting a disc bearing a number divisible by 5
Here, the numbers divisible by 5 are
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90
\(\therefore\) Number of outcomes favourable to E₃ = 18
Hence, required probability = \(P\left(E_3\right)=\frac{18}{90}=\frac{1}{5}\)

Possible outcomes are HH, HT, TH, TT
\(\Rightarrow\) Number of total outcomes = 4
Let A = Atleast one head
Favourable outcomes of event A are HH, HT or TH
\(\therefore\) Number of favourable outcomes = 3
 P(A) = \(\frac{3}{4}\)

Total number of cards in the box = 48
(i) Numbers divisible by 7 are = 7, 14, 21, 28, 35, 42, 49 [Total 7 numbers]
Thus the probability of drawing a number divisible by 7 = \(\frac { 7 }{ 48 } \)
(ii) Perfect squares from 3 to 50 are = 9, 16, 25,36 and 49 [Total 5 numbers]
\(\therefore \)Probability of drawing a perfect square = \(\frac { 5 }{ 48 } \)

Total number of tickets = 40 and multiple of 5 are 5, 10, 15, 20, 25, 30, 35, 40
P(a number which is a multiple of 5)
\(=\frac{8}{40}=\frac{1}{5}\)

Let S and R denote the events that Sangeeta wins the match and Reshma wins the match, respectively.
The probability of sangeeta's winning = P(S) = 0.62 (given)
The probability of Reshma's winning = P(R) = 1 - P(S)
[As the events R and S are complementary]
= 1 - 0.62 = 0.38

Leap year contains 366 days.  \(\Rightarrow \)  52 weeks + 2 days
52 weeks contain 52 Fridays
We will get 53 Fridays if one of the remaining two days is a Friday. Total possibilities for two days are:
(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)
There are 7 possibilities and out of these there are 2 favourable cases.
\(\therefore P(53 Fridays) = \frac{2}{7}\)

Total number of possible outcomes = 10
(i) (b): P(getting 100% discount) = \(\begin{equation} \frac{1}{10} \end{equation}\)
(ii) (c): P(getting no prize) = \(\begin{equation} \frac{3}{10} \end{equation}\)
(iii) (b) : P(getting a soccer ball) = \(\begin{equation} \frac{2}{10}=\frac{1}{5} \end{equation}\)
(iv) (b) : P(getting 15% discount) = \(\begin{equation} \frac{2}{10}=\frac{1}{5} \end{equation}\)
(v) (a): P(getting a free spin) = \(\begin{equation} \frac{1}{10} \end{equation}\)

(i) (b):Total number of possible outcomes on throwing two dice simultaneously = 6 x 6 = 36
(ii) (c): As we know, that maximum sum of numbers on two dice = 6 + 6 = 12
It is an impossible event.
So, required probability = 0
(iii) (c) : Let A be the event that both the numbers on dice are prime.
 A = {(2, 2), (2, 3), (2, 5), (3,2), (3, 3), (3, 5), (5, 2), (5,3), (5, 5)}
\(\Rightarrow\)n(A) = 9
\(\begin{equation} P(A)=\frac{9}{36}=\frac{1}{4} \end{equation}\)
(iv) (c) : Let B the event that product of two numbers is odd.
B = {(1, 1), (1, 3), (1, 5), (3,1), (3, 3), (3, 5), (5,1), (5,3), (5, 5)}
\(\Rightarrow\)n(B) = 9
\(\begin{equation} P(B)=\frac{9}{36}=\frac{1}{4} \end{equation}\)
(v) (c) : Let C be the event that difference of two numbers is zero.
C = {(1, 1), (2,2), (3, 3), (4,4), (5, 5), (6, 6)}
\(\Rightarrow\)n(C) = 6
\(\begin{equation} P(C)=\frac{6}{36}=\frac{1}{6} \end{equation}\)