Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be:
(i) an even number
(ii) a multiple of 3
(iii) an even number and a multiple of 3
(iv) an even number or a multiple of 3
There are 9 cards from which one card is drawn.
Total number of elementary events = n(S) = 9
(i) From numbers 2 to 10, there are 5 even numbers i.e. 2, 4, 6, 8, 10
Favorable number of events = n(E) = 5
Probability of selecting a card with an even number = 
(ii) From numbers 2 to 10, there are 3 numbers which are multiples of 3 i.e. 3, 6, 9
Favorable number of events = n(E) = 3
Probability of selecting a card with a multiple of 3=
(iii) From numbers 2 to 10, there is one number which is an even number as well as multiple of 3 i.e. 6
Favorable number of events = n(E) = 1
Probability of selecting a card with a number which is an even number as well as multiple of 3 = 
(iv) From numbers 2 to 10, there are 7 numbers which are even numbers or a multiple of 3 i.e. 2, 3, 4, 6, 8, 9, 10
Favorable number of events = n(E) = 7
Probability of selecting a card with a number which is an even number or a multiple of 3 = 
Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is:
(i) a multiple of 5
(ii) a multiple of 6
(iii) between 40 and 60
(iv) greater than 85
(v) less than 48
There are 100 cards from which one card is drawn.
Total number of elementary events = n(S) = 100
(i) From numbers 1 to 100, there are 20 numbers which are multiple of 5 i.e. {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100} Favorable number of events = n(E) = 20
Probability of selecting a card with a multiple of 5 =
(ii) From numbers 1 to 100, there are 16 numbers which are multiple of 6 i.e. {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96} Favorable number of events = n(E) = 16
Probability of selecting a card with a multiple of 6 =
(iii) From numbers 1 to 100, there are 19 numbers which are between 40 and 60 i.e. {41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59} Favorable number of events = n(E) = 19
Probability of selecting a card between 40 and 60=
(iv) From numbers 1 to 100, there are 15 numbers which are greater than 85 i.e. {86, 87, …., 98, 99, 100} Favorable number of events = n(E) = 15
Probability of selecting a card with a number greater than 85 =
(v) From numbers 1 to 100, there are 47 numbers which are less than 48 i.e. {1, 2, ……….., 46, 47} Favorable number of events = n(E) = 47
Probability of selecting a card with a number less than 48 =
From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:
(i) 3
(ii) 5
(iii) 3 and 5
(iv) 3 or 5
There are 25 cards from which one card is drawn.
Total number of elementary events = n(S) = 25
(i) From numbers 1 to 25, there are 8 numbers which are multiple of 3 i.e. {3, 6, 9, 12, 15, 18, 21, 24} Favorable number of events = n(E) = 8
Probability of selecting a card with a multiple of 3 =
(ii) From numbers 1 to 25, there are 5 numbers which are multiple of 5 i.e. {5, 10, 15, 20, 25} Favorable number of events = n(E) = 5
Probability of selecting a card with a multiple of 5 =
(iii) From numbers 1 to 25, there is only one number which is multiple of 3 and 5 i.e. {15} Favorable number of events = n(E) = 1
Probability of selecting a card with a multiple of 3 and 5 =
(iv) From numbers 1 to 25, there are 12 numbers which are multiple of 3 or 5 i.e. {3, 5, 6, 9, 10, 12, 15, 18, 20,21, 24, 25} Favorable number of events = n(E) = 12
Probability of selecting a card with a multiple of 3 or 5 =
A die is thrown once. Find the probability of getting a number:
(i) less than 3
(ii) greater than or equal to 4
(iii) less than 8
(iv) greater than 6
In throwing a dice, total possible outcomes = {1, 2, 3, 4, 5, 6}
n(S) = 6
(i) On a dice, numbers less than 3 = {1, 2}
n(E) = 2
Probability of getting a number less than 3 = 
(ii) On a dice, numbers greater than or equal to 4 = {4, 5, 6}
n(E) = 3
Probability of getting a number greater than or equal to 4 =
(iii) On a dice, numbers less than 8 = {1, 2, 3, 4, 5, 6}
n(E) = 6
Probability of getting a number less than 8 = 
(iv) On a dice, numbers greater than 6 = 0
n(E) = 0
Probability of getting a number greater than 6 = 
A book contains 85 pages. A page is chosen at random. What is the probability that the sum of the digits on the page is 8?
Number of pages in the book = 85
Number of possible outcomes = n(S) = 85
Out of 85 pages, pages that sum up to 8 = {8, 17, 26, 35, 44, 53, 62, 71, 80}
pages that sum up to 8 = n(E) = 9
Probability of choosing a page with the sum of digits on the page equals 8 = 
A pair of dice is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first die.
In throwing a dice, total possible outcomes = {1, 2, 3, 4, 5, 6}
n(S) = 6
For two dice, n(S) = 6 
6 = 36
Favorable cases where the sum is 10 or more with 5 on 1st die = {(5, 5), (5, 6)}
Event of getting the sum is 10 or more with 5 on 1st die =n(E) = 2
Probability of getting a sum of 10 or more with 5 on 1st die =
If two coins are tossed once, what is the probability of getting:
(i) both heads.
(ii) at least one head.
(iii) both heads or both tails.
When two coins are tossed together possible number of outcomes = {HH, TH, HT, TT}
n(S) = 4
(i) E = event of getting both heads = {HH}
n(E) = 1
Probability of getting both heads = 
(ii) E = event of getting at least one head = {HH, TH, HT}
n(E) = 3
Probability of getting at least one head = 
(iii) E = event of getting both heads or both tails = {HH, TT}
n(E) = 2
Probability of getting both heads or both tails = 
Two dice are rolled together. Find the probability of getting:
(i) a total of at least 10.
(ii) a multiple of 2 on one die and an odd number on the other die.
In throwing a dice, total possible outcomes = {1, 2, 3, 4, 5, 6}
n(S) = 6
For two dice, n(S) = 6 
6 = 36
(i) E = event of getting a total of at least 10 = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
n(E) = 6
Probability of getting a total of at least 10 = 
(ii) E = event of getting a multiple of 2 on one die and an odd number on the other = {(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (6, 5), (1, 2), (3, 2), (5, 2), (1, 4), (3, 4), (5, 4), (1, 6), (3, 6), (5, 6)}
n(E) = 18
Probability of getting a multiple of 2 on one die and an odd number on the other = 
A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card drawn is:
(i) a spade(v) Jack or queen
(ii) a red card(vi) ace and king
(iii) a face card(vii) a red and a king
(iv) 5 of heart or diamond(viii) a red or a king
Number of possible outcomes when card is drawn from pack of 52 cards = 52
n(S) = 52
(i) Number of spade cards = 13 = E = event of drawing a spade
n(E) = 13
Probability of drawing a spade = 
(ii) Number of red cards(hearts + diamonds) = 26 = E = event of drawing a red card
n(E) = 26
Probability of drawing a red card = 
(iii) Number of face cards(4 kings + 4 queens + 4 Jacks)= 12 = E = Event of drawing a face card
n(E) = 12
Probability of drawing a face card = 
(iv) E = event of drawing a 5 of heart or of diamond = {5H, 5D}
n(E) = 2
Probability of drawing a 5 of heart or of diamond =
(v) E = event of drawing a jack or a queen = {JH, JS, JD, JC, QH, QS, QD, QC}
n(E) = 8
Probability of drawing a jack or a queen = 
(vi) A card cannot be both an ace as well as a king.
E =event of drawing an ace and a king = 0
n(E) = 0
Probability of drawing an ace and a king = 
(vii) E = event of drawing a red and a king = {KH, KD }
n(E) = 2
Probability of drawing a red and a king =
(viii) E = event of drawing a red or a king = 26 red cards (13 h + 13D) + 2 black kings[since 26 red cards contain 2 red kings]
n(E) = 28
Probability of drawing a red or a king =
A bag contains 16 colored balls. Six are green, 7 are red and 3 are white. A ball is chosen, without looking into the bag. Find the probability that the ball chosen is:
(i) red(v) green or red
(ii) not red(vi) white or green
(iii) white(vii) green or red or white
(iv) not white
Balls in the bag = 16 = Number of balls that could be drawn
n(S) = 16
(i) E = Event of drawing a red ball = number of red balls = 7
n(E) = 7
Probability of drawing a red ball = 
(ii) Not a red ball = 16 - number of red balls = 16 -7 = 9 = E
n(E) = 9
Probability of not drawing a red ball = 
(iii) E = Event of drawing a white ball = number of white balls = 3
n(E) = 3
Probability of drawing a white ball = 
(iv) Not a white ball = 16 - number of white balls = 16-3 = 13=E
n(E) = 13
Probability of not drawing a white ball = 
(v) E = Event of drawing a green or a red ball = number of green balls + number of red balls = 6 + 7 = 13
n(E) = 13
Probability of drawing a green or a red ball =
(vi) E = Event of drawing a green or a white ball = number of green balls + number of white balls = 6 + 3 = 9
n(E) = 9
Probability of drawing a green or a white ball =
(vii) E = Event of drawing a green or a white or a red ball = number of green balls + number of white balls + number of red balls = 6 + 3 + 7 = 16
n(E) = 16
Probability of drawing a green or a white or a red ball =
A ball is drawn at random from a box containing 12 white, 16 red and 20 green balls. Determine the probability that the ball drawn is:
(i) white(iii) not green
(ii) red(iv) red or white
Total number of balls in the box = 48
Total possible outcomes on drawing a ball = 48
n(S) = 48
(i) Event of drawing a white ball = E = 12
n(E) = 12
Probability of drawing a white ball = 
(ii) Event of drawing a red ball = E = 16
n(E) = 16
Probability of drawing a red ball = 
(iii) Event of drawing a green ball = E = 20
n(E) = 20
Probability of drawing a green ball = 
Probability of not drawing a green ball = 1 - 
(iv) red or a white ball = 12 + 16 = 28 balls
Event of drawing a red or white ball = E = 28
n(E) = 28
Probability of not drawing a green ball = 
A card is drawn from a pack of 52 cards. Find the probability that the card drawn is:
(i) a red card(v) a black ace
(ii) a black card(vi) ace of diamonds
(iii) a spade(vii) not a club
(iv) an ace(viii) a queen or a jack
Number of possible outcomes when card is drawn from pack of 52 cards = 52
n(S) = 52
(i) Number of red cards(hearts + diamonds) = 26 = E
n(E) = 26
Probability of drawing a red card = 
(ii) Number of black cards(spade + clubs) = 26 = E
n(E) = 26
Probability of drawing a black card =
(iii) Number of spade cards = 13 = E = event of drawing a spade
n(E) = 13
Probability of drawing a spade = 
(iv) Number of ace cards = 4 = E = event of drawing an ace
n(E) = 4
Probability of drawing an ace = 
(v) Number of black ace cards = 2 = E = event of drawing an ace
n(E) = 2
Probability of drawing a black ace = 
(vi) There is only one ace of diamonds.
E = event of drawing an ace of diamonds
n(E) = 1
Probability of drawing an ace of diamonds = 
(vii) Number of club cards = 13 = E = event of drawing a club card
n(E) = 13
Probability of drawing a club card =
Probability of not drawing a club card = 1 - 
(viii) E = event of drawing a jack or a queen = {JH, JS, JD, JC, QH, QS, QD, QC}
n(E) = 8
Probability of drawing a jack or a queen = 
Thirty identical cards are marked with numbers 1 to 30. If one card is drawn at random, find the probability that it is:
(i) a multiple of 4 or 6
(ii) a multiple of 3 and 5
(iii) a multiple of 3 or 5
There are 30 cards from which one card is drawn.
Total number of elementary events = n(S) = 30
(i) From numbers 1 to 30, there are 10 numbers which are multiple of 4 or 6 i.e. {4, 6, 8, 12, 16, 18, 20, 24, 28, 30} Favorable number of events = n(E) = 10
Probability of selecting a card with a multiple of 4 or 6 =
(ii) From numbers 1 to 30, there are 2 numbers which are multiple of 3 and 5 i.e. {15, 30} Favorable number of events = n(E) = 2
Probability of selecting a card with a multiple of 3 and 5 =
(iii) From numbers 1 to 30, there are 14 numbers which is multiple of 3 or 5 i.e. {3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30} Favorable number of events = n(E) = 14
Probability of selecting a card with a multiple of 3 or 5 =
In a single throw of two dice, find the probability of:
(i) a doublet
(ii) a number less than 3 on each dice
(iii) an odd number as a sum
(iv) a total of at most 10
(v) an odd number on one dice and a number less than or equal to 4 on the other dice.
The number of possible outcomes is 6 × 6 = 36. We write them as given below :
{(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)
(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2)
(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3)
(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4)
(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)
(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)}
n(S) = 36
(i) E = Event of getting a doublet = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
n(E) = 6
Probability of getting a doublet = 
(ii) E = Event of getting a number less than 3 on each dice
= {(1, 1), (1, 2), (2, 1), (2, 2)}
n(E) = 4
Probability of getting a number less than 3 on each dice = 
(iii) E = Event of getting an odd number as a sum = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)}
n(E) = 18
Probability of getting an odd number as sum = 
(iv) E = Event of getting a total of at most 10 =
{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5) ,(3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2),(5,3), (5,4), (5,5)
(6,1), (6,2), (6,3), (6,4)}
Therefore total number of favorable ways = 33 = n(E)
Probability of getting a total of at most 10 = 
(v) E = Event of getting an odd number on dice 1 and a number less than or equal to 4 on dice 2 =
{(1,1), (1,2), (1,3), (1,4), (1, 5)
(2, 1), (2, 3), (2, 5)
(3,1), (3,2), (3,3), (3,4), (3, 5)
(4, 1), (4, 3), (4, 5)
(5,1), (5,2),(5,3), (5,4)}
Therefore total number of favorable ways = 20 = n(E)
Probability of getting an odd number on dice 1 and a number less than or equal to 4 on dice 2 = 
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