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 \(\frac { 20 }{ 24 } =\frac { 5 }{ 6 } \)

No. of queens = 4, No. of kings = 4, No. of aces = 4
= 52 -12 = 40, After No. of cards left removing all kings, queens and aces
(i) No. of black face cards in the remaining cards (2 jacks) = 2
\(\therefore \)Probability of black face card \(=\frac { 2 }{ 40 } =\frac { 1 }{ 20 } \) 
(ii) Out of 12 cards removed, 6 are of red colour.
\(\therefore \)No. of red coloured cards left = 26-6 = 20
\(\therefore \)No. of ways to draw a red card = 20
\(\therefore \)Probability of getting a red card\(=\frac { 20 }{ 40 } =\frac { 1 }{ 2 } \)

Favourable events are (4,3), (3,4), (6,2), (2,6);
Total number of elementary events = 36
Required probability = \(\frac{4}{36}=\frac{1}{9}\)

Total number of cards = 18
(a) Prime numbers less than
15 = 3, 5, 7, 11, 13
P(a prime number less than 15) = \(\frac{5}{18}\)
(b) P(a number divisible by 3 and 5) = \(\frac{1}{18}\)

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 = \(\frac{10}{49}\)

Cards removed = 3 face cards of spade
Number ofcards remaining = 52- 3 = 49
(i) Number of black face cards left = 3
\(\therefore \)Probability of drawing a black face card = \(\frac { 3 }{ 49 } \)
(ii) Number of queens = 4-1=3 [Since a queen of spade has been removed]
\(\therefore \)Probability of drawing a queen =\(\frac { 3 }{ 49 } \)
(iii) Total number of black cards = 26-3 = 23 [as 3 black cards have been removed]
\(\therefore \)Probability of drawing a black  card = \(\frac { 23 }{ 49 } \)

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 = \(\frac { 5 }{ 48 } \)

(i)Total number of spade = 13
Number of ace = 3 [There are 4 ace but I ace is of spade which has been included in spades]
TotaI number of ace or spades = 16
\(\therefore \)Probability of drawing a spade or ace = \(\frac { 16 }{ 52 } =\frac { 4 }{ 13 } \)
 (ii)There are 2 red kings.
Therefore probability of drawing a red king = \(\frac { 2 }{ 52 } =\frac { 1 }{ 26 } \) 
(iii) Total number of kings and queens
 = 4 kings + 4 queens = 8
Number of cards which are neither kings nor queens = 52 —8 = 44
\(\therefore \)Probability of drawing neither a king nor a queen =\(\frac { 44 }{ 52 } =\frac { 11 }{ 13 } \)
(iv) Number of kings and queens = 4 kings + 4 queens = 8
\(\therefore \)Probability of drawing a king 2 8 or queen = \(\frac { 8 }{ 52 } =\frac { 2 }{ 13 } \)​​​​​​​

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= \(\frac { 5 }{ 7 } \)

Two different dice are tossed. Therefore, total outcomes are 36.
(i) Favourable outcomes for even number on both dice = 9, (2, 2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)
\(\therefore \)Probability of getting even number on both dice =\(\frac { 9 }{ 36 } =\frac { 1 }{ 4 } \)
(ii) Favourable outcomes that the sum of the numbers appearing in two dice is 5 are (1, 4), (2, 3), (3, 2), (4, 1), i.e. 4.
\(\therefore \)Probability of getting sum of numbers appearing on two dice is 5=\(\frac { 4 }{ 36 } =\frac { 1 }{ 9 } \)

Number of cards left = 52 - 3 = 49 and number of cards of spade left = 13 - 3 = 10
Number of black cards left = 13 + 10 = 23
( \(\because \) Spade is of black colour)
Total number of ways to draw a card = 49
Number of ways to draw a black card = 23
\(\therefore\) Required probability = \(\frac{23}{49}\).

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)
\(=\frac{8}{40}=\frac{1}{5}\)

Total number of cards in the deck=52
Number of cards removed = 3 [king, Queen & Jack of diamonds]
Number of cards remaining
= 52-3 = 49
(i) Number of diamonds left
13 - 3 = 10 [as 3 diamonds have been removed]
Probability of drawing a diamond = \(\frac { 10 }{ 49 } \)
(ii) Number of jacks left
4 - 1 = 3[asjack of diamond has been removed]
Probability of drawing a jack =\(\frac { 3 }{ 49 } \)  

Total outcomes = 8
Factors of 8 are 1, 2, 4, 8
\(\therefore\)  Required probability = \(\frac{4}{8}=\frac{1}{2}\)

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 = \(P(E)=\frac{21}{26}\)

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

Total number of cards = 52 and number of Jack of hearts  =1
\(\therefore\)  Probability of drawing a Jack of hearts = \(\frac{1}{52}\)

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 = \(\frac{2}{6}=\frac{1}{3}\)

Total outcomes = 52 and favourable outcomes = 6 (red face cards)
\(\therefore\)  Probability of getting a red face card
= \(\frac{6}{52}=\frac{3}{26}\)

Let number of tickets bought by Archana = x
Probability of winning 1st prize \(={x\over 12000}\)
\(\Rightarrow\ \ {x\over12000}=0.04\)
x = 0.04 x 12000 = 480

Let number of bad pens = x.
\(\therefore\)  Probability of getting a bad pen = \(\frac{x}{400}\)
A.T.Q. \(\frac{x}{400}\) =0.25
\(\Rightarrow\)   x = 0.25 x 400 = 100
\(\therefore\)  Number of good pens = 400 - 100 = 300

Here P(man will speak the truth) = \(\frac{5}{6}\)

when a face card is drawn from the pack of 52 playing cards then probability of man's reporting it is a face card is the probability of man's speaking the truth = \(\frac{5}{6}\).

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 = \(\frac { 2 }{ 7 } \) 

Here, P(A) + P(not A) = 1

When a fair dice is rolled then possible outcomes are 1, 2, 3, 4, 5 or 6.

\(\therefore\)  Probability of x = 1.

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 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 = \(\frac{44}{52}=\frac{11}{13}\)

(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

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) The car starts normally but when there is some defect, then car does not start. So, the outcomes are not equally likely.
(ii) The outcomes in this situation are not equally likely because the outcomes depends on many factors such as training ofthe player, quality of basketball, etc.
(iii) The outcomes in trial of true-false question is either true or false. Hence, the two outcomes are equally likely.
(iv) A new baby can be either a boy or a girl, so both the outcomes are equally likely.