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Maharashtra SSC Class 10 Mathematics Chapter 5 Probability: textbook solutions
Solutions written by Shishya, checked, not the board's official answers
Step-by-step working for the practice-set questions of Chapter 5, “Probability”, in Mathematics Part I (Algebra), Standard X (Maharashtra SSC, English medium; the chapter is on pages 115-128 of the printed book). Each question is in our own short words and cited by its set and number; open the textbook for the exact wording and figures.
How these were checked: drafted with AI on 28 Oct 2026. A solution is shown only if a calculator check recomputed every final value from the question's own numbers and every arithmetic step in the working agreed. 25 of 32 questions in this chapter passed; the others are listed below as not solved here. These are not the board's official answers: the textbook and your teacher are the authority, and if something here disagrees with the book, follow the book and tell us.
Textbook PDF read on 28 Oct 2026 from the publisher's official site.
Practice Set 5.1
Not solved here yet: Q2, Q3, Q4 (see the textbook).
Practice Set 5.1, Q1
How many possible outcomes are there? (1) Picking one of 8 places to visit (2) Picking one day of the week at random (3) Drawing one card from a pack of 52 cards (4) Drawing one card at random from cards numbered 10 to 20, one number on each card
- (1) There are 8 distinct places to choose from, so there are 8 possible outcomes.
- (2) A week has 7 days, so picking one day at random gives 7 possible outcomes.
- (3) A standard pack contains 52 cards, so drawing one card gives 52 possible outcomes.
- (4) The cards are numbered from 10 to 20 inclusive, which is 20 − 10 + 1 = 11 cards, so there are 11 possible outcomes.
Answer: (1) 8; (2) 7; (3) 52; (4) 11
Practice Set 5.2
Not solved here yet: Q1 (see the textbook).
Practice Set 5.3
Not solved here yet: Q1 (see the textbook).
Practice Set 5.4
Practice Set 5.4, Q1
Two coins are tossed. Find the probability of (1) at least one head (2) no head.
- When two coins are tossed, the total number of possible outcomes is 4.
- The outcomes are HH, HT, TH, and TT.
- (1) The outcomes with at least one head are HH, HT, and TH.
- (1) The number of favorable outcomes for at least one head is 3.
- (1) Probability of at least one head = 3/4.
- (2) The outcome with no head is TT.
- (2) The number of favorable outcomes for no head is 1.
- (2) Probability of no head = 1/4.
Answer: (1) 3/4; (2) 1/4
Practice Set 5.4, Q2
Two dice are rolled. Find the probability that (1) the sum of the numbers is at least 10 (2) the sum is 33 (3) the number on the first die is greater than the number on the second.
- When two dice are rolled, the total number of possible outcomes is 6 × 6 = 36.
- (1) The outcomes where the sum is at least 10 are (4,6), (5,5), (5,6), (6,4), (6,5), and (6,6).
- (1) The number of favourable outcomes for sum at least 10 is 6.
- (1) Probability = 6/36 = 1/6.
- (2) The maximum possible sum of two dice is 6 + 6 = 12.
- (2) Since 33 is greater than 12, it is impossible to get a sum of 33.
- (2) The number of favourable outcomes is 0, so the probability is 0.
- (3) The outcomes where the first die is greater than the second are (2,1), (3,1), (3,2), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3), (6,4), and (6,5).
- (3) The number of favourable outcomes is 15.
- (3) Probability = 15/36 = 5/12.
Answer: (1) 1/6; (2) 0; (3) 5/12
Practice Set 5.4, Q3
15 tickets numbered 1 to 15 are in a box; one is drawn at random. Find the probability that it shows (1) an even number (2) a multiple of 5.
- Total number of tickets in the box is 15.
- (1) The even numbers from 1 to 15 are 2, 4, 6, 8, 10, 12 and 14.
- (1) The number of favourable outcomes for an even number is 7.
- (1) Probability of drawing an even number = 7/15.
- (2) The multiples of 5 from 1 to 15 are 5, 10 and 15.
- (2) The number of favourable outcomes for a multiple of 5 is 3.
- (2) Probability of drawing a multiple of 5 = 3/15.
- (2) Simplifying 3/15 gives 1/5.
Answer: (1) 7/15; (2) 1/5
Practice Set 5.4, Q4
A two-digit number is formed from the digits 2, 3, 5, 7, 9 without repetition. Find the probability that it is (1) odd (2) a multiple of 5.
- The available digits are 2, 3, 5, 7, 9.
- Two-digit numbers without repetition: 5 choices for the tens digit and 4 for the units digit, so n(S) = 5 × 4 = 20.
- (1) Odd numbers must end in 3, 5, 7, or 9, giving 4 choices for the units digit.
- (1) For each odd units digit, the tens digit can be any of the other 4 digits, so favourable cases = 4 × 4 = 16.
- (1) Probability of an odd number = 16/20 = 4/5.
- (2) A multiple of 5 must end in 5, giving 1 choice for the units digit.
- (2) With units digit 5, the tens digit can be any of the other 4 digits, so favourable cases = 1 × 4 = 4.
- (2) Probability of a multiple of 5 = 4/20 = 1/5.
Answer: (1) 4/5; (2) 1/5
Practice Set 5.4, Q5
A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that it is (1) an ace (2) a spade.
- Total number of cards in a pack = 52.
- (1) Number of aces in a pack = 4.
- (1) Probability of drawing an ace = 4/52 = 1/13.
- (2) Number of spades in a pack = 13.
- (2) Probability of drawing a spade = 13/52 = 1/4.
Answer: (1) 1/13; (2) 1/4
Problem Set 5
Not solved here yet: Q15, Q17 (see the textbook).
Problem Set 5, Q1(1)
Choose the correct option: which number cannot be a probability? (A) 2/3 (B) 1.5 (C) 15% (D) 0.7
- A probability value must always lie between 0 and 1, inclusive.
- Option (A) is 2/3, which equals 0.666... and is within the range.
- Option (C) is 15%, which equals 0.15 and is within the range.
- Option (D) is 0.7, which is within the range.
- Option (B) is 1.5, which is greater than 1.
- Since 1.5 exceeds the maximum possible probability of 1, it cannot be a probability.
- The correct option is (B).
Answer: (B)
Problem Set 5, Q1(2)
Choose the correct option: a die is rolled; the probability of a number less than 3 is (A) 1/6 (B) 1/3 (C) 1/2 (D) 0
- Total possible outcomes when a die is rolled = 6
- Numbers less than 3 are 1 and 2
- Number of favourable outcomes = 2
- Probability = Favourable outcomes ÷ Total outcomes
- Probability = 2 ÷ 6
- Probability = 1/3
- This matches option (B)
Answer: (B) 1/3
Problem Set 5, Q1(3)
Choose the correct option: the probability that a number chosen from 1 to 100 is prime is (A) 1/5 (B) 6/25 (C) 1/4 (D) 13/50
- The total number of integers from 1 to 100 is 100.
- The prime numbers from 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97 (1 is not prime).
- Counting these primes, there are 25 prime numbers in the range.
- The probability of choosing a prime number is the ratio of prime numbers to total numbers, which is 25/100.
- Simplifying the fraction 25/100 gives 1/4.
- This corresponds to option (C).
Answer: (C) 1/4
Problem Set 5, Q1(4)
Choose the correct option: 40 cards are numbered 1 to 40; the probability that a card drawn at random shows a multiple of 5 is (A) 1/5 (B) 3/5 (C) 4/5 (D) 1/3
- Total number of cards = 40.
- Multiples of 5 from 1 to 40 are 5, 10, 15, 20, 25, 30, 35, 40.
- Number of favourable outcomes = 8.
- Probability = Number of favourable outcomes / Total number of outcomes.
- Probability = 8/40.
- Probability = 1/5.
- This matches option (A).
Answer: (A) 1/5
Problem Set 5, Q1(5)
Choose the correct option: n(A) = 2 and P(A) = 1/5, so n(S) = (A) 10 (B) 5/2 (C) 2/5 (D) 1/3
- Given n(A) = 2 and P(A) = 1/5.
- The formula for probability is P(A) = n(A) / n(S).
- Substitute the values into the formula: 1/5 = 2 / n(S).
- Cross-multiply to solve for n(S): n(S) = 2 × 5.
- Calculate the value: n(S) = 10.
- This matches option (A).
Answer: (A) n(S) = 10
Problem Set 5, Q2
John, Vasim and Akash succeed at a basket drop with probabilities 4/5, 0.83 and 58%. Who has the greatest probability of success?
- Convert John's probability of success to a decimal: 4/5 = 0.80.
- Note Vasim's probability of success is given as 0.83.
- Convert Akash's probability of success from percentage to decimal: 58% = 58/100 = 0.58.
- Compare the three probabilities: 0.80 (John), 0.83 (Vasim), and 0.58 (Akash).
- Since 0.83 is greater than 0.80 and 0.58, Vasim has the greatest probability of success.
Answer: Vasim
Problem Set 5, Q3
A hockey squad has 6 defenders, 4 offenders and 1 goalie; a captain is chosen at random. Find the probability that the captain is (1) the goalie (2) a defender.
- Total players = 6 + 4 + 1 = 11
- (1) Number of goalies = 1
- (1) P(goalie) = 1/11
- (2) Number of defenders = 6
- (2) P(defender) = 6/11
Answer: (1) 1/11; (2) 6/11
Problem Set 5, Q4
26 cards each carry one letter of the English alphabet; one is drawn at random. Find the probability that it is a vowel.
- Total number of cards = 26.
- Vowels in the English alphabet are A, E, I, O, U.
- Number of vowel cards = 5.
- Probability = Number of favourable outcomes / Total number of outcomes.
- Probability = 5 / 26.
Answer: 5/26
Problem Set 5, Q5
A vendor has 2 red, 3 blue and 4 green balloons and picks one at random. Find the probability that it is (1) red (2) blue (3) green.
- Total number of balloons = 2 + 3 + 4 = 9, so n(S) = 9
- (1) Number of red balloons = 2
- (1) P(Red) = 2/9
- (2) Number of blue balloons = 3
- (2) P(Blue) = 3/9 = 1/3
- (3) Number of green balloons = 4
- (3) P(Green) = 4/9
Answer: (1) 2/9; (2) 1/3; (3) 4/9
Problem Set 5, Q6
A box has 5 red, 8 blue and 3 green pens; one is picked at random. Find the probability that it is blue.
- Total number of pens = 5 + 8 + 3
- Total number of pens = 16
- Number of blue pens = 8
- Probability of picking a blue pen = Number of blue pens / Total number of pens
- Probability = 8/16
- Probability = 1/2
Answer: 1/2
Problem Set 5, Q7
A die's six faces show A, B, C, D, E, A. It is rolled once. Find the probability that (1) A shows (2) D shows.
- The die has six faces with the letters A, B, C, D, E, A.
- Total number of possible outcomes = 6.
- (1) The letter A appears on two faces.
- Number of favourable outcomes for A = 2.
- P(A) = 2/6 = 1/3.
- (2) The letter D appears on one face.
- Number of favourable outcomes for D = 1.
- P(D) = 1/6.
Answer: (1) 1/3; (2) 1/6
Problem Set 5, Q8
30 tickets are numbered 1 to 30; one is drawn at random. Find the probability that it shows (1) an odd number (2) a perfect square.
- Total number of tickets = 30
- (1) Odd numbers from 1 to 30 are 1, 3, 5, ..., 29
- Number of odd numbers = 15
- Probability of drawing an odd number = 15/30
- Simplifying 15/30 gives 1/2
- (2) Perfect squares from 1 to 30 are 1, 4, 9, 16, 25
- Number of perfect squares = 5
- Probability of drawing a perfect square = 5/30
- Simplifying 5/30 gives 1/6
Answer: (1) 1/2; (2) 1/6
Problem Set 5, Q9
A rectangular garden is 77 m by 50 m and has a circular lake of diameter 14 m. A towel falls at random into the garden. Find the probability that it falls in the lake (take π = 22/7).
- Area of rectangular garden = 77 × 50
- Area of rectangular garden = 3850 m²
- Radius of circular lake = 14 ÷ 2
- Radius of circular lake = 7 m
- Area of circular lake = (22/7) × 7 × 7
- Area of circular lake = 154 m²
- Probability = Area of lake ÷ Area of garden
- Probability = 154 ÷ 3850
- Probability = 1/25
Answer: 1/25
Problem Set 5, Q10
An arrow stops at random on one of the numbers 1 to 8. Find the probability that it stops at (1) 8 (2) an odd number (3) a number greater than 2 (4) a number less than 9.
- Total possible outcomes are the numbers 1, 2, 3, 4, 5, 6, 7, 8.
- Total number of outcomes n(S) = 8.
- (1) The number 8 is a single outcome.
- Number of favourable outcomes for (1) = 1.
- P(stops at 8) = 1/8.
- (2) The odd numbers from 1 to 8 are 1, 3, 5, 7.
- Number of favourable outcomes for (2) = 4.
- P(stops at an odd number) = 4/8 = 1/2.
- (3) The numbers greater than 2 are 3, 4, 5, 6, 7, 8.
- Number of favourable outcomes for (3) = 6.
- P(stops at a number greater than 2) = 6/8 = 3/4.
- (4) All numbers from 1 to 8 are less than 9.
- Number of favourable outcomes for (4) = 8.
- P(stops at a number less than 9) = 8/8 = 1.
Answer: (1) 1/8; (2) 1/2; (3) 3/4; (4) 1
Problem Set 5, Q11
Six cards carry the numbers 0 to 5; one is drawn. Find the probability that it shows (1) a natural number (2) a number less than 1 (3) a whole number (4) a number greater than 5.
- The sample space S = {0, 1, 2, 3, 4, 5} and n(S) = 6.
- (1) Natural numbers are {1, 2, 3, 4, 5}, so n(A) = 5.
- (1) P(A) = n(A) / n(S) = 5/6.
- (2) The only number less than 1 is {0}, so n(B) = 1.
- (2) P(B) = n(B) / n(S) = 1/6.
- (3) Whole numbers are {0, 1, 2, 3, 4, 5}, so n(C) = 6.
- (3) P(C) = n(C) / n(S) = 6/6 = 1.
- (4) There are no numbers greater than 5 in S, so n(D) = 0.
- (4) P(D) = n(D) / n(S) = 0/6 = 0.
Answer: (1) 5/6; (2) 1/6; (3) 1; (4) 0
Problem Set 5, Q12
A bag has 3 red, 3 white and 3 green balls; one is drawn at random. Find the probability that it is (1) red (2) not red (3) red or white.
- Total number of balls = 3 + 3 + 3 = 9, so n(S) = 9
- (1) Number of red balls = 3
- (1) P(Red) = 3/9 = 1/3
- (2) Number of balls that are not red = 3 + 3 = 6
- (2) P(Not Red) = 6/9 = 2/3
- (3) Number of red or white balls = 3 + 3 = 6
- (3) P(Red or White) = 6/9 = 2/3
Answer: (1) 1/3; (2) 2/3; (3) 2/3
Problem Set 5, Q13
Each letter of the word "mathematics" is on a separate card; one card is drawn at random. Find the probability that it shows the letter m.
- Count the total number of letters in the word mathematics.
- Total letters = 11.
- Count the number of times the letter m appears in mathematics.
- Number of m's = 2.
- Probability = Number of favourable outcomes / Total number of outcomes.
- P(m) = 2/11.
Answer: 2/11
Problem Set 5, Q14
Of 200 students, 135 like kabaddi and the rest do not. One student is chosen at random. Find the probability that the student does not like kabaddi.
- Total number of students = 200
- Number of students who like kabaddi = 135
- Number of students who do not like kabaddi = 200 − 135
- Number of students who do not like kabaddi = 65
- Probability of not liking kabaddi = 65/200
- Simplifying the fraction by dividing numerator and denominator by 5
- Probability = 13/40
Answer: 13/40
Problem Set 5, Q16
A die whose faces show 0, 1, 2, 3, 4, 5 is rolled twice. Find the probability that the product of the two numbers is zero.
- The die has faces 0, 1, 2, 3, 4, 5, so there are 6 possible outcomes for each roll.
- The total number of outcomes when the die is rolled twice is 6 × 6 = 36.
- The product of the two numbers is zero if at least one of the numbers is 0.
- The favourable outcomes are (0,0), (0,1), (0,2), (0,3), (0,4), (0,5), (1,0), (2,0), (3,0), (4,0), (5,0).
- The number of favourable outcomes is 11.
- The probability is the number of favourable outcomes divided by the total number of outcomes.
- Probability = 11/36.
Answer: 11/36
Other chapters solved for this book
- Chapter 1: Linear Equations in Two Variables — 58 questions solved
- Chapter 2: Quadratic Equations — 73 questions solved
- Chapter 3: Arithmetic Progression — 47 questions solved