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NCERT Class 10 Mathematics Exercise 14.1 Solutions: Probability

AI-written, checked on 23 Oct 2026

Solutions written by Shishya, checked — not the board's official answers

Step-by-step working for every question of Exercise 14.1 in NCERT's Class 10 Mathematics book, chapter “Probability”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

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Ex 14.1 Q1
Complete: (i) P(E) + P(not E) = … (ii) an event that cannot happen has probability … and is called … (iii) an event certain to happen has probability … and is called … (iv) the probabilities of all elementary events add up to … (v) a probability is at least … and at most …
1. (i) The sum of the probability of an event and its complement is always 1.
  why?
  An event either happens or does not, so P(E) + P(not E) = 1, and P(not E) = 1 − P(E).
2. (ii) An event that cannot happen has a probability of 0 and is called an impossible event.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
3. (iii) An event that is certain to happen has a probability of 1 and is called a sure event.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
4. (iv) The sum of the probabilities of all elementary events in a sample space is 1.
5. (v) The probability of any event lies between 0 and 1 inclusive.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
Answer: (i) 1; (ii) 0, impossible event; (iii) 1, sure event; (iv) 1; (v) 0, 1
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Ex 14.1 Q2
Which have equally likely outcomes? (i) a car starts or does not start (ii) a player shoots or misses a basketball shot (iii) an answer to a true-false question is right or wrong (iv) a baby born is a boy or a girl.
1. (i) A car starting or not depends on mechanical condition, so the outcomes are not equally likely.
2. (ii) A player shooting or missing depends on skill and conditions, so the outcomes are not equally likely.
3. (iii) For a true-false question, the probability of guessing right or wrong is 1/2 each, so the outcomes are equally likely.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
4. (iv) The probability of a baby being born a boy or a girl is approximately 1/2 each, so the outcomes are equally likely.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
Answer: (i) Not equally likely; (ii) Not equally likely; (iii) Equally likely; (iv) Equally likely
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Ex 14.1 Q3
Why is tossing a coin a fair way to decide which team gets the ball at the start of a football game?
1. A coin has two distinct faces: heads and tails.
2. When a coin is tossed, it is equally likely to land on either face.
3. This means the probability of getting heads is 1/2 and the probability of getting tails is 1/2.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
4. Since both outcomes have the same probability, neither team has an advantage over the other.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
5. Therefore, tossing a coin is a fair method for making the decision.
Answer: Tossing a coin is fair because the probabilities of heads and tails are equal (1/2 each).
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Ex 14.1 Q4
Which cannot be the probability of an event: (A) 2/3 (B) −1.5 (C) 15% (D) 0.7?
1. The probability of any event must be a number between 0 and 1, inclusive.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
2. Option (A) is 2/3, which is approximately 0.67 and lies within the range 0 to 1.
3. Option (C) is 15%, which equals 0.15 and lies within the range 0 to 1.
  why?
  Per cent means 'out of 100', so x% of a quantity = (x / 100) × the quantity.
4. Option (D) is 0.7, which lies within the range 0 to 1.
5. Option (B) is −1.5, which is less than 0 and therefore cannot be a probability.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
Answer: (B)
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Ex 14.1 Q5
P(E) = 0.05. Find P(not E).
1. We know that the sum of the probability of an event and the probability of its complement is 1.
  why?
  An event either happens or does not, so P(E) + P(not E) = 1, and P(not E) = 1 − P(E).
2. So, P(E) + P(not E) = 1.
  why?
  An event either happens or does not, so P(E) + P(not E) = 1, and P(not E) = 1 − P(E).
3. Substitute the given value P(E) = 0.05 into the equation.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
4. 0.05 + P(not E) = 1.
  why?
  An event either happens or does not, so P(E) + P(not E) = 1, and P(not E) = 1 − P(E).
5. Subtract 0.05 from both sides to find P(not E).
  why?
  Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
6. P(not E) = 1 - 0.05.
  why?
  An event either happens or does not, so P(E) + P(not E) = 1, and P(not E) = 1 − P(E).
7. P(not E) = 0.95.
  why?
  An event either happens or does not, so P(E) + P(not E) = 1, and P(not E) = 1 − P(E).
Answer: P(not E) = 0.95
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Ex 14.1 Q6
A bag has only lemon candies. Malini takes one without looking. Find the probability it is (i) orange flavoured (ii) lemon flavoured.
1. (i) The bag contains only lemon candies, so there are 0 orange flavoured candies.
2. (i) The probability of an impossible event is 0.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
3. (i) Therefore, the probability of drawing an orange flavoured candy is 0.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
4. (ii) Since the bag has only lemon candies, every candy in the bag is lemon flavoured.
5. (ii) The probability of a sure event is 1.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
6. (ii) Therefore, the probability of drawing a lemon flavoured candy is 1.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
Answer: (i) 0; (ii) 1
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Ex 14.1 Q7
The probability that 2 students of a group do not share a birthday is 0.992. Find the probability that they do.
1. The sum of the probability of an event and the probability of its complement is 1.
  why?
  An event either happens or does not, so P(E) + P(not E) = 1, and P(not E) = 1 − P(E).
2. Let P(not sharing a birthday) = 0.992.
  why?
  An event either happens or does not, so P(E) + P(not E) = 1, and P(not E) = 1 − P(E).
3. The probability that they do share a birthday is 1 − 0.992.
  why?
  An event either happens or does not, so P(E) + P(not E) = 1, and P(not E) = 1 − P(E).
4. 1 − 0.992 = 0.008.
Answer: 0.008
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Ex 14.1 Q8
A bag has 3 red and 5 black balls; one is drawn at random. Find P(red) and P(not red).
1. Total number of balls in the bag is 3 + 5 = 8.
2. Number of red balls is 3.
3. Probability of drawing a red ball is number of red balls divided by total balls.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
4. P(red) = 3/8.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
5. Number of black balls is 5, which represents the event of not drawing a red ball.
6. Probability of drawing a ball that is not red is number of black balls divided by total balls.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
7. P(not red) = 5/8.
  why?
  An event either happens or does not, so P(E) + P(not E) = 1, and P(not E) = 1 − P(E).
Answer: (i) 3/8; (ii) 5/8
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Ex 14.1 Q9
A box has 5 red, 8 white and 4 green marbles; one is drawn at random. Find P(red), P(white) and P(not green).
1. Total number of marbles = 5 + 8 + 4 = 17
2. Number of red marbles = 5
3. P(red) = Number of red marbles / Total marbles = 5/17
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
4. Number of white marbles = 8
5. P(white) = Number of white marbles / Total marbles = 8/17
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
6. Number of green marbles = 4
7. P(green) = Number of green marbles / Total marbles = 4/17
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
8. P(not green) = 1 - P(green) = 1 - 4/17 = 13/17
  why?
  An event either happens or does not, so P(E) + P(not E) = 1, and P(not E) = 1 − P(E).
Answer: (i) 5/17; (ii) 8/17; (iii) 13/17
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Ex 14.1 Q10
A piggy bank has hundred 50p coins, fifty ₹1, twenty ₹2 and ten ₹5 coins; one falls out at random. Find P(50p coin) and P(not a ₹5 coin).
1. Total number of coins = 100 + 50 + 20 + 10 = 180.
2. Number of 50p coins = 100.
3. P(50p coin) = 100/180 = 5/9.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
4. Number of ₹5 coins = 10.
5. Number of coins that are not ₹5 = 180 − 10 = 170.
6. P(not a ₹5 coin) = 170/180 = 17/18.
  why?
  An event either happens or does not, so P(E) + P(not E) = 1, and P(not E) = 1 − P(E).
Answer: (i) 5/9; (ii) 17/18
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Ex 14.1 Q11
A tank has 5 male and 8 female fish; one is taken out at random. Find the probability it is male.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. The total number of fish in the tank is the sum of male and female fish, which is 5 + 8 = 13.
2. The number of favourable outcomes for selecting a male fish is 5.
3. The probability of an event is calculated as the number of favourable outcomes divided by the total number of possible outcomes.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
4. Substituting the values, the probability is 5/13.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
Answer: 5/13
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Ex 14.1 Q12
A spinner stops at one of 1 to 8, equally likely. Find the probability it points at (i) 8 (ii) an odd number (iii) a number greater than 2 (iv) a number less than 9.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. The spinner has 8 equally likely outcomes: 1, 2, 3, 4, 5, 6, 7, 8.
2. Total number of possible outcomes is 8.
3. (i) The event of pointing at 8 has only 1 favorable outcome.
4. (i) Probability = 1/8.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
5. (ii) The odd numbers are 1, 3, 5, 7, so there are 4 favorable outcomes.
6. (ii) Probability = 4/8 = 1/2.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
7. (iii) The numbers greater than 2 are 3, 4, 5, 6, 7, 8, so there are 6 favorable outcomes.
8. (iii) Probability = 6/8 = 3/4.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
9. (iv) All numbers from 1 to 8 are less than 9, so there are 8 favorable outcomes.
10. (iv) Probability = 8/8 = 1.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
Answer: (i) 1/8; (ii) 1/2; (iii) 3/4; (iv) 1
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Ex 14.1 Q13
A die is thrown once. Find the probability of (i) a prime number (ii) a number between 2 and 6 (iii) an odd number.
1. (i) The possible outcomes when a die is thrown are 1, 2, 3, 4, 5, 6, so total outcomes = 6.
2. (i) The prime numbers on a die are 2, 3, and 5, so favorable outcomes = 3.
3. (i) Probability of a prime number = 3/6 = 1/2.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
4. (ii) The numbers strictly between 2 and 6 are 3, 4, and 5, so favorable outcomes = 3.
5. (ii) Probability of a number between 2 and 6 = 3/6 = 1/2.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
6. (iii) The odd numbers on a die are 1, 3, and 5, so favorable outcomes = 3.
7. (iii) Probability of an odd number = 3/6 = 1/2.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
Answer: (i) 1/2; (ii) 1/2; (iii) 1/2
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Ex 14.1 Q14
One card is drawn from a well-shuffled deck of 52. Find the probability of (i) a red king (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds.
1. A deck has 52 equally likely cards: 4 suits (hearts and diamonds are red; clubs and spades are black) of 13 cards each.
2. (i) Red kings: the king of hearts and the king of diamonds, 2 cards. P = 2/52 = 1/26.
3. (ii) Face cards are the jack, queen and king of each suit: 3 × 4 = 12 cards. P = 12/52 = 3/13.
4. (iii) Red face cards: 3 in hearts + 3 in diamonds = 6 cards. P = 6/52 = 3/26.
5. (iv) There is exactly one jack of hearts. P = 1/52.
6. (v) There are 13 spades. P = 13/52 = 1/4.
7. (vi) There is exactly one queen of diamonds. P = 1/52.
Answer: (i) 1/26; (ii) 3/13; (iii) 3/26; (iv) 1/52; (v) 1/4; (vi) 1/52
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Ex 14.1 Q15
The 10, J, Q, K and A of diamonds are shuffled face down and one is picked. (i) Find P(queen). (ii) If the queen is put aside, find the probability that the next card is (a) an ace (b) a queen.
1. (i) There are 5 cards in total: 10, J, Q, K, A of diamonds.
2. (i) Only one of these cards is a queen.
3. (i) The probability of picking a queen is 1/5.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
4. (ii) The queen is removed, so 4 cards remain: 10, J, K, A.
5. (a) Only one of the remaining cards is an ace.
6. (a) The probability of picking an ace is 1/4.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
7. (b) Since the queen was already removed, there are no queens left.
8. (b) The probability of picking a queen is 0.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
Answer: (i) 1/5; (a) 1/4; (b) 0
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Ex 14.1 Q16
12 defective pens are mixed with 132 good ones; one is taken at random. Find the probability it is good.
1. Total number of pens = 12 defective + 132 good = 144
2. Number of good pens = 132
3. Probability of drawing a good pen = (Number of good pens) / (Total number of pens)
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
4. Probability = 132 / 144
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
5. Simplifying the fraction by dividing numerator and denominator by 12 gives 11/12
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
Answer: 11/12
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Ex 14.1 Q17
(i) 4 of 20 bulbs are defective; find P(defective) for one drawn at random. (ii) If that bulb was good and not replaced, find the probability the next bulb drawn is not defective.
1. (i) Total number of bulbs = 20.
2. (i) Number of defective bulbs = 4.
3. (i) Probability of drawing a defective bulb = Number of defective bulbs / Total number of bulbs.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
4. (i) P(defective) = 4/20 = 1/5.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
5. (ii) The first bulb drawn was good and not replaced.
6. (ii) Remaining total bulbs = 20 - 1 = 19.
7. (ii) Number of defective bulbs remaining = 4.
8. (ii) Number of good bulbs remaining = 19 - 4 = 15.
9. (ii) Probability of drawing a non-defective bulb = Number of good bulbs remaining / Remaining total bulbs.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
10. (ii) P(not defective) = 15/19.
  why?
  An event either happens or does not, so P(E) + P(not E) = 1, and P(not E) = 1 − P(E).
Answer: (i) 1/5; (ii) 15/19
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Ex 14.1 Q18
Discs numbered 1 to 90; one is drawn at random. Find the probability of (i) a two-digit number (ii) a perfect square (iii) a number divisible by 5.
1. Total number of discs is 90, so total outcomes n(S) = 90.
2. (i) Two-digit numbers range from 10 to 90.
3. Number of two-digit discs = 90 − 9 = 81.
4. Probability of a two-digit number = 81/90 = 9/10.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
5. (ii) Perfect squares between 1 and 90 are 1, 4, 9, 16, 25, 36, 49, 64, and 81.
6. There are 9 perfect squares.
7. Probability of a perfect square = 9/90 = 1/10.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
8. (iii) Numbers divisible by 5 are 5, 10, 15, ..., 90.
9. Number of such discs = 90 ÷ 5 = 18.
10. Probability of a number divisible by 5 = 18/90 = 1/5.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
Answer: (i) 9/10; (ii) 1/10; (iii) 1/5
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Ex 14.1 Q19
A child's die has faces A, B, C, D, E, A. Find P(A) and P(D) in one throw.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. The die has 6 faces in total.
2. The faces are A, B, C, D, E, and A.
3. The face A appears on 2 faces of the die.
4. The probability of getting A is the number of faces with A divided by the total number of faces.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
5. P(A) = 2 ÷ 6 = 1/3.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
6. The face D appears on 1 face of the die.
7. The probability of getting D is the number of faces with D divided by the total number of faces.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
8. P(D) = 1 ÷ 6 = 1/6.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
Answer: (i) 1/3; (ii) 1/6
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Ex 14.1 Q20
A die is dropped at random on a 3 m × 2 m rectangle containing a circle of diameter 1 m. Find the probability it lands inside the circle.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. The total area of the rectangular region is 3 m × 2 m = 6 m².
2. The diameter of the circle is 1 m, so its radius is 1/2 m.
3. The area of the circle is π × (1/2)² = π/4 m².
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
4. The probability of the die landing inside the circle is the ratio of the circle's area to the rectangle's area.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
5. Probability = (π/4) / 6 = π/24.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
Answer: π/24
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Ex 14.1 Q21
Of 144 ball pens, 20 are defective. Nuri buys the pen drawn at random only if it is good. Find P(she buys it) and P(she does not).
1. Total number of ball pens = 144
2. Number of defective pens = 20
3. Number of good pens = 144 − 20 = 124
4. P(Nuri buys the pen) = Number of good pens ÷ Total pens = 124 ÷ 144
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
5. Simplifying 124/144 by dividing numerator and denominator by 4 gives 31/36
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
6. P(Nuri does not buy the pen) = Number of defective pens ÷ Total pens = 20 ÷ 144
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
7. Simplifying 20/144 by dividing numerator and denominator by 4 gives 5/36
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
Answer: (i) 31/36; (ii) 5/36
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Ex 14.1 Q22
Two dice are thrown. (i) Complete the table of probabilities of the sums 2 to 12. (ii) A student says each of the 11 sums has probability 1/11. Do you agree?
1. (i) Total outcomes when two dice are thrown = 6 × 6 = 36.
2. (i) Sum 2: (1,1) → 1 outcome, P = 1/36.
3. (i) Sum 3: (1,2),(2,1) → 2 outcomes, P = 2/36.
4. (i) Sum 4: (1,3),(2,2),(3,1) → 3 outcomes, P = 3/36.
5. (i) Sum 5: (1,4),(2,3),(3,2),(4,1) → 4 outcomes, P = 4/36.
6. (i) Sum 6: (1,5),(2,4),(3,3),(4,2),(5,1) → 5 outcomes, P = 5/36.
7. (i) Sum 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 outcomes, P = 6/36.
8. (i) Sum 8: (2,6),(3,5),(4,4),(5,3),(6,2) → 5 outcomes, P = 5/36.
9. (i) Sum 9: (3,6),(4,5),(5,4),(6,3) → 4 outcomes, P = 4/36.
10. (i) Sum 10: (4,6),(5,5),(6,4) → 3 outcomes, P = 3/36.
11. (i) Sum 11: (5,6),(6,5) → 2 outcomes, P = 2/36.
12. (i) Sum 12: (6,6) → 1 outcome, P = 1/36.
13. (ii) The student claims each sum has probability 1/11.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
14. (ii) From the table, probabilities are not equal (e.g., P(sum=7) = 6/36 ≠ 1/11).
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
15. (ii) Therefore, the sums are not equally likely.
Answer: (i) 1/36, 2/36, 3/36, 4/36, 5/36, 6/36, 5/36, 4/36, 3/36, 2/36, 1/36; (ii) do not agree
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Ex 14.1 Q23
A coin is tossed 3 times; Hanif wins if all three results are the same. Find the probability that he loses.
1. When a coin is tossed 3 times, the total number of possible outcomes is 2 × 2 × 2 = 8.
2. The outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
3. Hanif wins if all three results are the same, which means the outcome is either HHH or TTT.
4. So, the number of favourable outcomes for Hanif winning is 2.
5. The probability that Hanif wins is 2/8 = 1/4.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
6. The probability that Hanif loses is 1 − P(wins) = 1 − 1/4 = 3/4.
  why?
  An event either happens or does not, so P(E) + P(not E) = 1, and P(not E) = 1 − P(E).
Answer: 3/4
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Ex 14.1 Q24
A die is thrown twice. Find the probability that (i) 5 does not come up either time (ii) 5 comes up at least once.
1. When a die is thrown twice, the total number of possible outcomes is 6 × 6 = 36.
2. (i) For 5 not to come up either time, each throw can result in any number from 1 to 4 or 6, giving 5 options per throw.
3. The number of favourable outcomes for 5 not appearing is 5 × 5 = 25.
4. The probability that 5 does not come up either time is 25/36.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
5. (ii) The event '5 comes up at least once' is the complement of '5 does not come up either time'.
6. The number of favourable outcomes for 5 coming up at least once is 36 − 25 = 11.
7. The probability that 5 comes up at least once is 11/36.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
Answer: (i) 25/36; (ii) 11/36
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Ex 14.1 Q25
Correct or not? (i) Two coins tossed give three outcomes (two heads, two tails, one of each), so each has probability 1/3. (ii) A die gives an odd or an even number, so P(odd) = 1/2.
1. (i) When two coins are tossed, the possible outcomes are HH, HT, TH, and TT.
2. (i) There are 4 equally likely outcomes, not 3.
3. (i) The outcome 'one of each' includes both HT and TH, so it has 2 outcomes.
4. (i) Therefore, P(one of each) = 2/4 = 1/2, while P(two heads) = 1/4 and P(two tails) = 1/4.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
5. (i) The statement that each has probability 1/3 is incorrect.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
6. (ii) A die has 6 faces with numbers 1, 2, 3, 4, 5, 6.
7. (ii) The odd numbers are 1, 3, 5 (3 outcomes) and even numbers are 2, 4, 6 (3 outcomes).
8. (ii) Since there are 3 odd outcomes out of 6 total equally likely outcomes, P(odd) = 3/6 = 1/2.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
9. (ii) The statement that P(odd) = 1/2 is correct.
  why?
  Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).
Answer: (i) Incorrect; (ii) Correct
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NCERT Class 10 Mathematics Exercise 14.1 Solutions: Probability

Ex 14.1 Q1

Ex 14.1 Q2

Ex 14.1 Q3

Ex 14.1 Q4

Ex 14.1 Q5

Ex 14.1 Q6

Ex 14.1 Q7

Ex 14.1 Q8

Ex 14.1 Q9

Ex 14.1 Q10

Ex 14.1 Q11

Ex 14.1 Q12

Ex 14.1 Q13

Ex 14.1 Q14

Ex 14.1 Q15

Ex 14.1 Q16

Ex 14.1 Q17

Ex 14.1 Q18

Ex 14.1 Q19

Ex 14.1 Q20

Ex 14.1 Q21

Ex 14.1 Q22

Ex 14.1 Q23

Ex 14.1 Q24

Ex 14.1 Q25