Total number of tickets 1000, Number tickets with prizes = 5
\(\therefore\) Probability of winning a prize = \(\frac{5}{1000}\)
  = 0.005

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) = \(\frac{3}{4}\)

Number of ways to draw a card by 1st player = 1000
perfect square greater than 500 are 529, 576, 625, 676, 729, 784, 841, 900, 961
\(\therefore \)Probability of 1st player winning the prize = \(\frac { 9 }{ 1000 } \)  
when 1st has won the prize then cards left = 999 (because card is not replaced)
Cards with no. which a perfect square greater than 500 = 8 (l winning number removed)
\(\therefore \)Probability of second player wins a prize when the first player has won =\(\frac { 8 }{ 999 } \)

Total balls in the bag = 5 + 8 + 4 + 7 = 24
(i) Number of black balls = 7. So Probability drawing a black ball =\(\frac { 7 }{ 24 } \) 
(ii) Number of red balls = 5 So Probability of drawing a red ball =\(\frac { 5 }{ 24 } \)
(iii) Number of balls which are not green = Number of red balls + Number of white + Number of black balls = 5+8+7=20
\(\therefore \)Probability of drawing a ball  which is not green is \(\frac { 20 }{ 24 } =\frac { 5 }{ 6 } \)

Given that king, queen and jack of club are removed from a pack then number of remaining cards 49
(i) then probability of getting a heart = \(\frac{13}{49}\)
(ii) probability of getting a queen = \(\frac{3}{49}\)
(iii) probability of getting a club = \(\frac{10}{49}\)

Number of days in a leap years = 366, Number of weeks = 52
\(\therefore \) Number of tuesdays in 52 weeks = 52
Number of days left after 52 weeks = 366-52 x 7 = 2.
Now, exactly 52 tuesday mean there should not be a tuesday in the remaining 2 days 
Possible outcome of remaing two days (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday) or (Sunday, Monday)
Total possible outcome = 7
Probability of not getting a Tuesday = \(\frac { 5 }{ 7 } \)
\(\therefore \)Probability of getting exactly 52 Tuesday= \(\frac { 5 }{ 7 } \)

We know that in English alphabet, there are 26 letters (5 vowels + 21 consonants).
So, total number of outcomes = 26
Let E be the event of choosing a consonant.
\(\therefore\)  Number of outcomes favourable to E = 21
Hence, required probability = \(P(E)=\frac{21}{26}\)

Total possibilities = 6 and favourable possibilities for number greater than 4 are 2 (i.e. 5, 6).
\(\therefore\)  Probability of getting a number greater than 4 = \(\frac{2}{6}=\frac{1}{3}\)

P(man will speak the truth) = \(\frac { 5 }{ 7 } \)
P(man will not speak the truth)
 = \(1-\frac { 5 }{ 7 } =\frac { 2 }{ 7 } \)
 when a number other than 6 comes up the probability of man's reporting it is a six is the probability of man's not speaking the truth = \(\frac { 2 }{ 7 } \) 

Total number of ways to throw a pair of dice = 36
Same number on each dice, i.e. (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)
\(\therefore\)  Number of ways of getting the same number on each dice = 6
Required probability = \(\frac{6}{36}=\frac{1}{6}\)

We have P(E) + P(not E) = 1
 P(E) + 0.992 = 1
 P(E) = 1 - 0.992 = 0.008

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.

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) = \(\frac{3}{6}=\frac{1}{2}\)

(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 = \(P\left(E_3\right)=\frac{18}{90}=\frac{1}{5}\)