CBSE Class 10 Mathematics Probability Assignment Set 01

CBSE Class 10 Mathematics Probability Assignment Set A. Students are advised to refer to the attached assignments and practise them regularly. This will help them to identify their weak areas and will help them to score better in examination. Parents should download and give the assignments to their children for practice.

Question. The Ace, number 10 and jack of clubs are removed from a deck of 52 playing cards and remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of
(a) heart
(b) Ace
(c) clubs
(d) either 10 or jack
Answer : 13/49 , 3/49 , 104/9 , 6/49

Question. All the black Ace cards are removed from a pack of 52 playing cards. The remaining cards are well shuffled and then a card is drawn at random. Find the probability of getting
i) a Ace card ii) a red card iii) a black card iv) a Jack
Answer : 1/25 , 13/25 , 12/25, 2/25

Question. Cards marked with the number 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from the box. Find the probability that the number on the card is:
(i) An even number (ii) A number less than 14 (iii) A number is perfect square (iv) A prime number less than 20
Answer : 1/2 , 3/25 , 9/100 , 2/25

Question. A bag contains white, black and red balls only. A ball is drawn at random from the bag. The probability of getting a white ball is 310 and that of black is 25 . Find the probability of getting a red ball. If the bag contains 20 black balls, then find the total number of balls in the bag.
Answer : 3/10 , 50

Question. A bag contains 8 red balls & some blue balls. If the probability of drawing a blue ball is 3 times of a red ball, find the number of blue balls in the bag.
Answer : 24

Question. Cards, marked with numbers 5 to 50, are placed in a box and mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the taken card is (i) a prime number less than 10. (ii) a number which is a perfect square.
Answer : Total no. of cards = 46
Total no. of ways to select a card = 46
(i) Prime no. less than 10 in these cards are 5, 7
∴ No. of ways to select a prime no. less than 10 = 2.
∴ Probability that the number on the card is prime = 2/46 = 1/23
(ii) No. which is a perfect square, i.e. 9, 16, 25, 36, 49.
No. of ways to select a card with perfect square = 5.
∴ Probability = 5/46

Question. A card is drawn at random from a well-shuffled deck of playing cards. Find the probability of drawing a (i) face card (ii) card which is neither a king nor a red card.
Answer : Total number of outcomes = 52
(i) Number of face cards = 12
∴ Probability of drawing a face card = 12/52 = 3/13
(ii) Number of cards which are neither king nor red = 24
∴ Probability of drawing a card which is neither a king nor a red card = 24/52 = 6/13

Question. Find the probability of getting 53 Fridays in a leap year.
Answer : Leap year contains 366 days.
52 weeks + 2 days left
Now, 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.
∴ P (53 Fridays) = 2/7

Question. 15 cards, numbered 1, 2, 3, ..., 15 are put in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the card drawn bears (i) an even number (ii) a number divisible by 2 or 3.
Answer : Total number of ways to select one card out of 15 cards = 15
(i) Number of cards bears even numbers i.e. 2, 4, 6, 8, 10, 12 or 14 = 7
∴ Required Probability = 7/15
(ii) Number of Cards bears a number divisible by 2 or 3 i.e. 2, 3, 4, 6, 8, 9, 10, 12, 14 or 15 = 10
∴ Required Probability = 10/15 = 2/3

Question. Cards marked with numbers 3, 4, 5, ...., 50 are placed in a box and mixed thoroughly. One cardis drawn at random from the box. Find the probability that number on the drawn card is (i) divisible by 7 (ii) a number which is a perfect square.
Answer : 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 = 7/48
(ii) Perfect squares from 3 to 50 are = 9, 16, 25, 36 and 49 [Total 5 numbers]
∴ Probability of drawing a perfect square = 5/48

Question. A box contains 19 balls bearing numbers 1, 2, 3, ....,19. A ball is drawn at random from the box. What is the probability that the number on the ball is (i) a prime number (ii) divisible by 3 or 5 (iii) neither divisible by 5 nor by 10 (iv) an even number.
Answer : Total number of balls = 19
(i) Prime numbers from 1 to 19 are 2, 3, 5, 7, 11, 13, 17, 19 = Total 8 prime numbers
∴ Probability of drawing a prime number = 8/19
(ii) Numbers divisible by 3 or 5 are 3, 6, 9, 15, 18, 10, 5, 12 = Total 8 numbers
∴ Probability of drawing a number divisible by 3 or 5 = 8/19
(iii) Number divisible by 5 and 10 are 5, 10, 15 = Total 3
∴ Numbers which are neither divisible by 5 nor 10 are 19 – 3 = 16
∴ Probability of drawing a number which is neither divisible by 5 nor by 10 = 16/19
(iv) Even numbers from 1 – 19 are 2, 4, 6, 8, 10, 12, 14, 16, 18 [Total 9 even numbers]
∴ Probability of drawing an even number = 9/19

Question. Three cards of spades are lost from a pack of 52 playing cards. The remaining cards were well shuffled and then a card was drawn at random from them. Find the probability that the drawn cards are of black colour.
Answer : 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
Total number of ways to draw a card = 49
Number of ways to draw a black card = 23
Required probability = 23/49

Question. A bag contains 5 red, 4 blue and 3 green balls. A ball is taken out of the bag at random. Find the probability that the selected ball is (i) of red colour (ii) not of green colour.
Answer : Total number of balls in the bag = 5 + 4 + 3 = 12
(i) Number of red balls = 5
Probability of drawing a red ball = 5/12
(ii) Number of ball not green in colour = Red + blue = 5 + 4 = 9
∴ Probability of drawing a ball which is not of green colour = 9/12 = 3/4

Question. Two different dice are tossed together. Find the probability (i) that the number on each dice is even (ii) that the sum of numbers appearing on two dice is 5.
Answer : 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)
Probability of getting even number on both dice = 9/36 = 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.
Probability of getting sum of numbers appearing on two dice is 5 = 4/36 = 1/9

Question. A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is (i) a king or a jack (ii) a non-ace (iii) a red card (iv) neither a king nor a queen.
Answer : Total number of playing cards = 52
(i) Favourable cases for a king or a jack are 8 (4 kings + 4 jacks)
∴ Probability of drawing a king or a jack = 8/52 = 2/13
(ii) Favourable cases for a non-ace are 48 (52 cards – 4 aces)
∴ Probability of drawing a non-ace = 48/52 = 12/13
(iii) Favourable cases for a red cards are 26 (13 hearts + 13 diamonds)
∴ Probability of drawing a red card = 26/52 = 1/2
(iv) Favourable cases for neither a king nor a queen are 44(52 cards – 4 kings – 4 queen)
∴ Probability of drawing neither a king nor a queen = 44/52 = 11/13

Question. Two dice are rolled once. Find the probability of getting such numbers on the two ice, whose product is 12.
Answer : Total number of elementary events = 36
Favourable events are (4, 3), (3, 4), (6, 2), (2, 6);
Required probability = 4/36 = 1/9

Question. A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither an ace nor a king.
Answer : Total number of cards = 52
Number of aces and kings = 4 + 4 = 8
Number of cards which are neither ace nor king = 52 - 8 = 44
Probability that the card drawn is neither an ace nor a king = 44/52 = 11/13

Question. Two dice are thrown simultaneously. What is the probability that (a) 5 will not come up on either of them? (b) 5 will come up on at least one? (c) 5 will come up at both dice?
Answer : Total no. of outcomes = 36
(a) Number of outcomes in which 5 will not come up on either of them = 25.
∴ Required Probability = 25/36
(b) Number of outcomes in which 5 will come up at least one die = 11.
∴ Required Probability = 11/36
(c) Number of outcomes in which 5 will come up at both die = 1.
∴ Required Probability = 1/36

Question. From a pack of 52 playing cards, jacks, queens, kings and aces of red colour are removed. From the remaining a card is drawn at random. Find the probability that the card drawn is (i) a black queen (ii) a red card (iii) a black jack (iv) a face card.
Answer : From the total playing 52 cards, red coloured jacks, queen, kings and aces are removed(i.e., 2 jacks, 2 queens, 2 kings, 2 aces) ∴ Remaining cards = 52 – 8 = 44
(i) Favourable cases for a black queen are 2 (i.e., queen of club or spade)
∴ Probability of drawing a black queen = 2/44 = 1/22
(ii) Favourable cases for red cards are 26 – 8 = 18 (as 8 cards have been removed)
(i.e.9 diamonds + 9 hearts)
∴ Probability of drawing a red card = 18/44 = 9/22
(iii) Favourable cases for a black jack are 2 (i.e. jacks of club or spade)
∴ Probability of drawing a black jack = 2/44 = 1/22
(iv) Favourable cases for a picture card are 6 (i.e. 2 black jacks, queens and kings each)
∴ Probability of drawing a picture card = 6/44 = 3/22

Question. A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is (i) a card of spade or an ace (ii) a red king (iii) neither a king nor a queen (iv) either a king or queen.
Answer : (i) Total number of spade = 13
Number of ace = 3
[There are 4 ace but 1 ace is of spade which has been included in spades]
Total number of ace or spades = 16
Probability of drawing a spade or ace = 16/52 = 4/13
(ii) There are 2 red kings.
Therefore, probability of drawing a red king = 2/52 = 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
Probability of drawing neither a king nor a queen = 44/52 = 11/13
(iv) Number of kings and queens = 4 kings + 4 queens = 8
Probability of drawing a king or queen = 8/52 = 2/13

Question. Cards bearing numbers 1, 3, 5, ..........., 35 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing (a) a prime number less than 15. (b) a number divisible by 3 and 5.
Answer : Total number of cards = 18
(a) Prime numbers less than 15 = 3, 5, 7, 11, 13
P(a prime number less than 15) = 5/18
(b) P(a number divisible by 3 and 5) = 1/18

Question. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of a red ball, find the number of blue balls in the bag.
Answer : Let the number of blue balls be x. Total number of balls in the bag = 5 + x
∴ Probability of drawing a red ball = 5 / 5 + x
and probability of drawing a blue ball = 5/x + x
Given probability of drawing a blue ball = 3 × probability of drawing a red ball
⇒ 3 x 5 + x = 5 + x ⇒ = 15
Number of blue balls = 15

Question. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing: (i) an ace (ii) ‘2’ of spades (iii) ‘10’ of a black suit.
Answer : Total number of cards = 52
(i) Number of ace = 4
∴ Probability of drawing an ace = 4/52 = 1/13
(ii) There is only one ‘2’ of spades
∴ Probability of drawing a ‘2’ of spade = 2/52 = 1/26
(iii) ‘10’ of a black suit there are two cards
∴ Probability of drawing 10 of black suit = 2/52 = 1/26

Question. A bag contains 18 balls out of which x balls are red. (i) If one ball is drawn at random from the bag, what is the probability that it is not red? (ii) If 2 more red balls are put in the bag, the probability of drawing a red ball will be 9/8 times the probability of drawing a red ball in the first case. Find the value of x.
Answer : Number of non red balls = 18 – x
Probability that ball drawn is not red = 18 - x / 18
Probability that ball drawn is red = x/18
When 2 more red balls are put in the bag then number of balls in the bag = 20
Number of red balls = x + 2
Now, Probability that ball drawn is red = x + 2'/20
According to the question, x + 2 / 20 = 9/8 x x/18 ⇒ x + 2 / 20 = x/6
⇒ 16x + 32 = 20x ⇒ 4x = 32 ⇒ x = 8

Question. All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting (i) a black face card, (ii) a queen, (iii) a black card.
Answer : Cards removed = 3 face cards of spade
Number of cards remaining = 52 – 3 = 49
(i) Number of black face cards left = 3
∴ Probability of drawing a black face card = 3/49
(ii) Number of queens = 4 – 1 = 3 [Since a queen of spade has been removed]
∴ Probability of drawing a queen = 3/49
(iii) Total number of black cards = 26 – 3 = 23 [as 3 black cards have been removed]
∴ Probability of drawing a black card = 23/49

Question. A bag contains 12 balls out of which x are white. (i) If one ball is drawn at random, what is the probability that it will be a white ball? (ii) If 6 more white balls are put in the bag, the probability of drawing a white ball will be double than that in (i). Find x.
Answer : Total number of balls = 12
(i) Number of white balls = x
∴ Probability of getting white ball = x/12
(ii) Number of white balls = x + 6
Total number of balls = 12 + 6 = 18.
∴ Probability of getting white ball = x + 6/18
According to the question, x+6/18 = 2 x x/12 ⇒ 6x + 36 = 18x ⇒ 12x = 36 ⇒ x = 3

Question. Two dice are rolled once. Find the probability of getting such numbers on two dice, whose product is a perfect square.
Answer : Total no. of outcomes = 36
Number of outcomes in which product is a perfect square = 8.
i.e. (1,1), (1, 4), (2,2), (3,3)(4,1), (4,4), (5,5) (6,6)
∴ Required Probability = 8/36 = 2/9

Question. A child’s game has 8 triangles of which 3 are blue and rest are red, and 10 squares of which 6are blue and rest are red. One piece is lost at random. Find the probability that it is a (i) triangle (ii) square (iii) square of blue colour (iv) triangle of red colour
Answer : Total number of pieces = 8 + 10 = 18
(i) No. of triangles = 8. Hence, P(triangle is lost) = 8/18 = 4/9
(ii) No. of squares = 10. Hence, P(square is lost) = 10/18 = 5/9
(iii) No. of squares of blue colour = 6. So, P(square of blue colour is lost) = 6/18 = 1/3
(iv) No. of triangles of red colour = 8 – 3 = 5. So, P(triangle of red colour is lost) = 5/18

Question. Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly.
One card is drawn from this box. Find the probability that the number on the card is
(i) an even number (ii) a number less than 14 (iii) a number which is a perfect square
(iv) a prime number less than 20.
Answer : Total number of cards = 100
(i) Number of cards bearing even number = 50
∴ Probability of drawing an even number = 50/100 = 1/2
(ii) Total cards with number less than 14 = 12
∴ Probability of drawing a number less than 14 = 12/100 = 3/25
(iii) Total perfect squares {4, 9, 16, 25, 36, 49, 64, 81, 100} = 9
∴ Probability of drawing a perfect square = 9/100
(iv) Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19
∴ Probability of drawing a prime number less than 20 is = 8/100 = 2/25

Question. The king, queen and jack of diamonds are removed from a pack of 52 cards and then the pack is well shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of (i)diamonds, (ii) a jack
Answer : 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 = 10/49
(ii) Number of jacks left = 4 – 1 = 3 [as jack of diamond has been removed]
Probability of drawing a jack = 3/49

Question. A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is (i) white or blue (ii) red or black (iii) not white (iv) neither white nor black.
Answer : Total number of balls = 5 + 7 + 4 + 2 = 18
(i) Number of white or blue balls = 5 + 2 = 7
∴ Probability of drawing a white or blue balls = 7/18
(ii) Number of red or black balls = 7 + 4 = 11
∴ Probability of drawing a red or black ball = 11/18
(iii) Number of balls that are not white = 13
∴ Probability of drawing a ball which is not white = 13/18
(iv) Number of balls that are not white or black = 7 + 2 = 9
∴ Probability of drawing neither white nor black ball = 9/18 = 1/2

Level 1

Question 1. A coin is tossed once. What is the probability of getting a head?
Answer: Total number of outcomes = 2 (Head and Tail)
Number of favourable outcomes for getting a head = 1
Probability \(P(H) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{1}{2}\)
In simple words: A coin has 2 sides, and getting a head is 1 of them, so the probability is \(\frac{1}{2}\).

Exam Tip: Always state the formula clearly before substituting values to secure full marks in board exams.

Question 2. Two coins are tossed once (or one coin is tossed twice). What is the probability of: (i) getting two heads (ii) at least one tail.
Answer: Possible outcomes of the experiment = {HH, HT, TH, TT}
Total number of events \(n(S) = 4\)
(i) Favourable outcomes for getting two heads = {HH}
Number of favourable outcomes = 1
\(P(\text{two heads}) = \frac{1}{4}\)
(ii) Favourable outcomes for at least one tail = {HT, TH, TT}
Number of favourable outcomes = 3
\(P(\text{at least one tail}) = \frac{3}{4}\)
In simple words: Write down all possible outcomes first, count the ones that match the condition, and divide by the total.

Exam Tip: 'At least one tail' means 1 tail or 2 tails, which includes all outcomes except both heads.

Question 3. An unbiased die is thrown. What is the probability of getting: (i) an even number (ii) a multiple of 3 (iii) a number 3 or 4 (iv) a number less than 5 (v) an odd number (vi) a number greater than 3.
Answer: Total number of possible outcomes when a die is thrown = 6 (1, 2, 3, 4, 5, 6)
(i) Even numbers = {2, 4, 6} (Total 3)
\(P(\text{even number}) = \frac{3}{6} = \frac{1}{2}\)
(ii) Multiples of 3 = {3, 6} (Total 2)
\(P(\text{multiple of 3}) = \frac{2}{6} = \frac{1}{3}\)
(iii) Number 3 or 4 = {3, 4} (Total 2)
\(P(\text{3 or 4}) = \frac{2}{6} = \frac{1}{3}\)
(iv) Numbers less than 5 = {1, 2, 3, 4} (Total 4)
\(P(\text{less than 5}) = \frac{4}{6} = \frac{2}{3}\)
(v) Odd numbers = {1, 3, 5} (Total 3)
\(P(\text{odd number}) = \frac{3}{6} = \frac{1}{2}\)
(vi) Numbers greater than 3 = {4, 5, 6} (Total 3)
\(P(\text{greater than 3}) = \frac{3}{6} = \frac{1}{2}\)
In simple words: Always simplify fractions to their lowest terms like \(\frac{1}{2}\) or \(\frac{1}{3}\).

Exam Tip: Double-check whether numbers like 3 are included or excluded based on terms like 'less than' versus 'less than or equal to'.

Question 4. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting the queen of diamonds.
Answer: Total number of cards in a deck = 52
Number of queen of diamonds in a deck = 1
\(P(\text{queen of diamonds}) = \frac{1}{52}\)
In simple words: There is only 1 queen of diamonds out of 52 cards.

Exam Tip: Remember standard deck composition: 52 cards, 4 suits (spades, hearts, diamonds, clubs), 13 cards per suit, and 4 cards of each rank.

Level 2

Question 1. A child has a block in the shape of a cube with one letter written on each face as shown below: A, B, C, D, E, A. The cube is thrown once. What is the probability of getting (i) A (ii) D?
Answer: Total number of faces on the cube = 6
(i) Faces bearing the letter A = 2
\(P(A) = \frac{2}{6} = \frac{1}{3}\)
(ii) Faces bearing the letter D = 1
\(P(D) = \frac{1}{6}\)
In simple words: Count how many times the specific letter appears and divide by 6.

Exam Tip: Watch out for repeated letters like 'A' appearing twice on the cube faces.

Question 2. A bag contains 3 red and 2 blue balls. A ball is drawn at random. What is the probability of drawing a blue ball?
Answer: Number of red balls = 3
Number of blue balls = 2
Total number of balls = 3 + 2 = 5
Number of favourable outcomes (blue balls) = 2
\(P(\text{blue ball}) = \frac{2}{5}\)
In simple words: Total balls in the bag are 5, and 2 of them are blue.

Exam Tip: Always calculate the total number of items in the sample space first by adding all individual quantities.

Question 3. It is given that in a group of 3 students, the probability of 2 students having not the same birthday is 0.992. What is the probability that 2 students have the same birthday?
Answer: We know that the sum of probabilities of complementary events is 1:
\(P(\text{Same Birthday}) + P(\text{not Same Birthday}) = 1\)
\(P(\text{same birthday}) + 0.992 = 1\)
\(P(\text{same birthday}) = 1 - 0.992 = 0.008\)
In simple words: Subtract the probability of an event not happening from 1 to find the probability of it happening.

Question 4. A bag contains 8 red, 6 white and 4 black marbles. A marble is drawn at random from the bag. Find the probability that the marble drawn is (i) Red or white (ii) not black (iii) neither white nor black.
Answer: Total number of marbles = 8 + 6 + 4 = 18
(i) Number of red or white marbles = 8 + 6 = 14
\(P(\text{red or white}) = \frac{14}{18} = \frac{7}{9}\)
(ii) Number of not black marbles (red or white) = 14
\(P(\text{not black marble}) = \frac{14}{18} = \frac{7}{9}\)
(iii) Number of neither white nor black marbles (which means red marbles) = 8
\(P(\text{neither white nor black}) = \frac{8}{18} = \frac{4}{9}\)
In simple words: Combine the counts of the allowed colors and divide by the total number of marbles.

Exam Tip: 'Neither white nor black' means the marble must be from the remaining color category, which is red.

Level 3

Question 1. Five cards - the ten, jack, queen, king and ace, are well shuffled with their face downwards. One card is then picked up at random. (i) What is the probability that the card is a queen? (ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace (b) a queen?
Answer: Total number of initial cards = 5 (10, J, Q, K, A)
(i) Number of queens = 1
\(P(\text{queen}) = \frac{1}{5}\)
(ii) When the queen is drawn and put aside, the remaining total number of cards = 4
(a) Number of aces remaining = 1
\(P(\text{ace}) = \frac{1}{4}\)
(b) Number of queens remaining = 0
\(P(\text{queen}) = \frac{0}{4} = 0\)
In simple words: When a card is removed without replacement, the total number of cards decreases by 1 for the second draw.

Exam Tip: Pay close attention to phrases like 'put aside' or 'without replacement', as they alter the sample space for subsequent events.

Question 2. A letter is chosen at random from the word 'ASSASSINATION'. Find the probability that the letter chosen is a (i) vowel (ii) consonant.
Answer: Total number of letters in the word 'ASSASSINATION' = 13
(i) Vowels in the word are A, A, I, A, I, O (Total 6 vowels)
\(P(\text{vowel}) = \frac{6}{13}\)
(ii) Consonants in the word = Total letters - Vowels = 13 - 6 = 7
\(P(\text{consonant}) = \frac{7}{13}\)
In simple words: Count all vowels and consonants individually, including repeated letters.

Exam Tip: Count every single letter instance in the given word, do not remove duplicates unless specified.

Question 3. If 65% of the population have black eyes, 25% have brown eyes and the remaining have blue eyes. What is the probability that a person selected at random has (i) Blue eyes (ii) Brown or black eyes (iii) Blue or black eyes?
Answer: Total percentage = 100%
Percentage with black eyes = 65%
Percentage with brown eyes = 25%
Percentage with blue eyes = 100% - (65% + 25%) = 100% - 90% = 10%
(i) \(P(\text{Blue eyes}) = \frac{10}{100} = \frac{1}{10}\)
(ii) \(P(\text{Brown or black eyes}) = \frac{25 + 65}{100} = \frac{90}{100} = \frac{9}{10}\)
(iii) \(P(\text{Blue or black eyes}) = \frac{10 + 65}{100} = \frac{75}{100} = \frac{3}{4}\)
In simple words: Percentages act directly out of 100 as favorable cases.

Exam Tip: Ensure percentages add up to 100% before answering sub-parts.

Question 4. Find the probability of getting 53 Fridays in a leap year.
Answer: Total number of days in a leap year = 366 days
366 days = 52 weeks and 2 days (\(52 \times 7 = 364\), leaving 2 extra days)
A leap year definitely has 52 Fridays. For the 53rd Friday, one of the 2 remaining days must be a Friday.
The possible combinations for the 2 consecutive remaining days are:
1. Sunday and Monday
2. Monday and Tuesday
3. Tuesday and Wednesday
4. Wednesday and Thursday
5. Thursday and Friday
6. Friday and Saturday
7. Saturday and Sunday
Total possible outcomes = 7 pairs
Favourable outcomes containing Friday = (Thursday and Friday), (Friday and Saturday) = 2 cases
\(P(53 \text{ Fridays}) = \frac{2}{7}\)
In simple words: A leap year has 2 leftover days which can form 7 possible pairs of consecutive days, out of which 2 pairs contain Friday.

Question 1. Three unbiased coins are tossed simultaneously. What is the probability of getting exactly two heads?
Answer: When three coins are tossed simultaneously, the sample space is:
\(S = \{\text{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}\}\)
Total number of outcomes \(n(S) = 8\)
Favourable outcomes for exactly two heads = {HHT, HTH, THH}
Number of favourable outcomes \(n(E) = 3\)
\(P(\text{exactly two heads}) = \frac{n(E)}{n(S)} = \frac{3}{8}\)
In simple words: List all 8 outcomes for 3 coins and pick the ones having exactly two H's.

Exam Tip: Use the formula \(2^n\) to verify the total number of outcomes for tossing \(n\) coins (\(2^3 = 8\)).

Question 2. A dice is thrown twice. Find the probability of getting (a) doublets (b) prime number on each die.
Answer: When a die is thrown twice, total number of outcomes \(n(S) = 6 \times 6 = 36\).
(a) Events getting doublets = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
Number of favourable outcomes = 6
\(P(\text{doublet}) = \frac{6}{36} = \frac{1}{6}\)
(b) Prime numbers on a die are {2, 3, 5}.
Events getting a prime number on each die = {(2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)}
Number of favourable outcomes \(n(E) = 9\)
\(P(\text{prime number on each die}) = \frac{9}{36} = \frac{1}{4}\)
In simple words: Doublets mean both numbers are identical. Prime numbers on a die are 2, 3, and 5.

Question 3. An integer is chosen at random from the first two hundred digits. What is the probability that the integer chosen is divisible by 6 or 8?
Answer: Total integers from 1 to 200 = 200
Multiples of 6 up to 200: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, 108, 114, 120, 126, 132, 138, 144, 150, 156, 162, 168, 174, 180, 186, 192, 198 (Total 33 multiples)
Multiples of 8 up to 200: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184, 192, 200 (Total 25 multiples)
Multiples of both 6 and 8 (i.e., multiples of LCM(6, 8) = 24) up to 200: 24, 48, 72, 96, 120, 144, 168, 192 (Total 8 multiples)
Using the inclusion-exclusion principle: Total multiples of 6 or 8 = \(33 + 25 - 8 = 50\)
\(P(\text{divisible by 6 or 8}) = \frac{50}{200} = \frac{1}{4}\)
In simple words: Find multiples of 6, multiples of 8, subtract common multiples (LCM), and divide by 200.

Exam Tip: Always subtract the intersection (LCM multiples) to avoid double counting when calculating 'A or B' probabilities.

Question 4. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is \(\frac{2}{3}\). Find the number of blue marbles in the jar.
Answer: Total number of marbles = 24
Let the number of green marbles be \(x\).
Given, \(P(\text{green marble}) = \frac{x}{24}\)
Also given that \(P(\text{green marble}) = \frac{2}{3}\)
Therefore, \(\frac{x}{24} = \frac{2}{3}\)
\(x = \frac{2 \times 24}{3} = 2 \times 8 = 16\)
Number of green marbles = 16
Number of blue marbles = Total marbles - Green marbles = 24 - 16 = 8
In simple words: Set up a proportion using the given probability to find the green marbles, then subtract from 24 to get the blue ones.

Exam Tip: Cross-multiplication is an efficient tool for solving linear equations involving fractions.

Question 5. A box contains 12 balls out of which \(x\) are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find \(x\).
Answer: Initial total number of balls = 12
Number of black balls = \(x\)
Initial probability of drawing a black ball = \(\frac{x}{12}\)
When 6 more black balls are added:
New total number of balls = \(12 + 6 = 18\)
New number of black balls = \(x + 6\)
New probability of drawing a black ball = \(\frac{x + 6}{18}\)
According to the question, the new probability is double the original probability:
\(\frac{x + 6}{18} = 2 \left( \frac{x}{12} \right)\)
\(\frac{x + 6}{18} = \frac{2x}{12} = \frac{x}{6}\)
Cross-multiplying:
\(6(x + 6) = 18x\)
\(6x + 36 = 18x\)
\(12x = 36\)
\(x = 3\)
In simple words: Form an equation based on the updated totals and solve for \(x\).

Exam Tip: Carefully update both the numerator (favourable outcomes) and the denominator (total outcomes) when new items are added to a container.