April 10, 2026
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Chapter 14: Probability
Complete NCERT Solutions | CBSE Class 10 Mathematics (2025-26)
Lato 14px · Classical Probability · Exercises 14.1 Solutions
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Exercise 14.1 (Page 283)
12 marks
Complete the following statements:
(i) Probability of an event E + Probability of the event 'not E' = ______.
(ii) The probability of an event that cannot happen is ______. Such an event is called ______.
(iii) The probability of an event that is certain to happen is ______. Such an event is called ______.
(iv) The sum of the probabilities of all the elementary events of an experiment is ______.
(v) The probability of an event is greater than or equal to ______ and less than or equal to ______.
(i) Probability of an event E + Probability of the event 'not E' = ______.
(ii) The probability of an event that cannot happen is ______. Such an event is called ______.
(iii) The probability of an event that is certain to happen is ______. Such an event is called ______.
(iv) The sum of the probabilities of all the elementary events of an experiment is ______.
(v) The probability of an event is greater than or equal to ______ and less than or equal to ______.
Answer:
(i) 1
(ii) 0, impossible event
(iii) 1, sure event
(iv) 1
(v) 0, 1
(i) 1
(ii) 0, impossible event
(iii) 1, sure event
(iv) 1
(v) 0, 1
22 marks
Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Answer:
(i) Not equally likely (car may have mechanical issues).
(ii) Not equally likely (depends on skill).
(iii) Equally likely (true/false have equal chance if guessing).
(iv) Equally likely (boy/girl probability approximately equal).
(i) Not equally likely (car may have mechanical issues).
(ii) Not equally likely (depends on skill).
(iii) Equally likely (true/false have equal chance if guessing).
(iv) Equally likely (boy/girl probability approximately equal).
32 marks
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Answer: Tossing a coin has two equally likely outcomes (head/tail), so both teams have equal chance of winning the toss. Hence it is fair.
43 marks
Which of the following cannot be the probability of an event?
(A) 2/3 (B) -1.5 (C) 15% (D) 0.7
(A) 2/3 (B) -1.5 (C) 15% (D) 0.7
Answer: (B) -1.5, because probability cannot be negative.
53 marks
If P(E) = 0.05, what is the probability of 'not E'?
Answer: P(not E) = 1 - P(E) = 1 - 0.05 = 0.95.
63 marks
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Answer:
(i) Probability of orange candy = 0 (impossible event).
(ii) Probability of lemon candy = 1 (sure event).
(i) Probability of orange candy = 0 (impossible event).
(ii) Probability of lemon candy = 1 (sure event).
73 marks
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Answer: P(same birthday) = 1 - P(not same birthday) = 1 - 0.992 = 0.008.
83 marks
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?
Answer:
Total balls = 8.
(i) P(red) = 3/8.
(ii) P(not red) = 1 - 3/8 = 5/8.
Total balls = 8.
(i) P(red) = 3/8.
(ii) P(not red) = 1 - 3/8 = 5/8.
93 marks
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red? (ii) white? (iii) not green?
(i) red? (ii) white? (iii) not green?
Answer:
Total marbles = 5+8+4 = 17.
(i) P(red) = 5/17.
(ii) P(white) = 8/17.
(iii) P(not green) = 1 - P(green) = 1 - 4/17 = 13/17.
Total marbles = 5+8+4 = 17.
(i) P(red) = 5/17.
(ii) P(white) = 8/17.
(iii) P(not green) = 1 - P(green) = 1 - 4/17 = 13/17.
103 marks
A piggy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin which falls out will be (i) a 50p coin? (ii) a ₹5 coin?
Answer:
Total coins = 100+50+20+10 = 180.
(i) P(50p) = 100/180 = 5/9.
(ii) P(₹5) = 10/180 = 1/18.
Total coins = 100+50+20+10 = 180.
(i) P(50p) = 100/180 = 5/9.
(ii) P(₹5) = 10/180 = 1/18.
113 marks
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see Fig. 14.4). What is the probability that the fish taken out is a male fish?
Answer:
Total fish = 5+8 = 13. P(male) = 5/13.
Total fish = 5+8 = 13. P(male) = 5/13.
123 marks
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 14.5), and these are equally likely outcomes. What is the probability that it will point at
(i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?
(i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?
Answer:
Total outcomes = 8.
(i) P(8) = 1/8.
(ii) Odd numbers: 1,3,5,7 → 4 outcomes → P = 4/8 = 1/2.
(iii) Numbers >2: 3,4,5,6,7,8 → 6 outcomes → P = 6/8 = 3/4.
(iv) Numbers <9: all 8 numbers → P = 1.
Total outcomes = 8.
(i) P(8) = 1/8.
(ii) Odd numbers: 1,3,5,7 → 4 outcomes → P = 4/8 = 1/2.
(iii) Numbers >2: 3,4,5,6,7,8 → 6 outcomes → P = 6/8 = 3/4.
(iv) Numbers <9: all 8 numbers → P = 1.
133 marks
A die is thrown once. Find the probability of getting
(i) a prime number (ii) a number lying between 2 and 6 (iii) an odd number.
(i) a prime number (ii) a number lying between 2 and 6 (iii) an odd number.
Answer:
Total outcomes = 6.
(i) Prime numbers: 2,3,5 → 3 outcomes → P = 3/6 = 1/2.
(ii) Numbers between 2 and 6: 3,4,5 → 3 outcomes → P = 3/6 = 1/2.
(iii) Odd numbers: 1,3,5 → 3 outcomes → P = 3/6 = 1/2.
Total outcomes = 6.
(i) Prime numbers: 2,3,5 → 3 outcomes → P = 3/6 = 1/2.
(ii) Numbers between 2 and 6: 3,4,5 → 3 outcomes → P = 3/6 = 1/2.
(iii) Odd numbers: 1,3,5 → 3 outcomes → P = 3/6 = 1/2.
143 marks
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds
(i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds
Answer:
Total cards = 52.
(i) Red kings: 2 (hearts, diamonds) → P = 2/52 = 1/26.
(ii) Face cards (J,Q,K): 12 → P = 12/52 = 3/13.
(iii) Red face cards: 6 → P = 6/52 = 3/26.
(iv) Jack of hearts: 1 → P = 1/52.
(v) Spades: 13 → P = 13/52 = 1/4.
(vi) Queen of diamonds: 1 → P = 1/52.
Total cards = 52.
(i) Red kings: 2 (hearts, diamonds) → P = 2/52 = 1/26.
(ii) Face cards (J,Q,K): 12 → P = 12/52 = 3/13.
(iii) Red face cards: 6 → P = 6/52 = 3/26.
(iv) Jack of hearts: 1 → P = 1/52.
(v) Spades: 13 → P = 13/52 = 1/4.
(vi) Queen of diamonds: 1 → P = 1/52.
153 marks
Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the 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?
(i) What is the probability that the card is the 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 cards = 5.
(i) P(queen) = 1/5.
(ii) After removing queen, remaining cards = 4.
(a) P(ace) = 1/4.
(b) P(queen) = 0 (queen already removed).
Total cards = 5.
(i) P(queen) = 1/5.
(ii) After removing queen, remaining cards = 4.
(a) P(ace) = 1/4.
(b) P(queen) = 0 (queen already removed).
163 marks
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Answer:
Total pens = 12+132 = 144. Good pens = 132.
P(good) = 132/144 = 11/12.
Total pens = 12+132 = 144. Good pens = 132.
P(good) = 132/144 = 11/12.
173 marks
A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? Suppose the bulb drawn in previous case is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Answer:
(i) P(defective) = 4/20 = 1/5.
(ii) After removing one non-defective, remaining bulbs = 19, defective still 4, good = 15.
P(not defective) = 15/19.
(i) P(defective) = 4/20 = 1/5.
(ii) After removing one non-defective, remaining bulbs = 19, defective still 4, good = 15.
P(not defective) = 15/19.
183 marks
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.
(i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.
Answer:
Total discs = 90.
(i) Two-digit numbers: 10 to 90 → 81 numbers → P = 81/90 = 9/10.
(ii) Perfect squares: 1,4,9,16,25,36,49,64,81 → 9 numbers → P = 9/90 = 1/10.
(iii) Divisible by 5: 5,10,...,90 → 18 numbers → P = 18/90 = 1/5.
Total discs = 90.
(i) Two-digit numbers: 10 to 90 → 81 numbers → P = 81/90 = 9/10.
(ii) Perfect squares: 1,4,9,16,25,36,49,64,81 → 9 numbers → P = 9/90 = 1/10.
(iii) Divisible by 5: 5,10,...,90 → 18 numbers → P = 18/90 = 1/5.
193 marks
A child has a die whose six faces show the letters as given below:
A, B, C, D, E, A
The die is thrown once. What is the probability of getting (i) A? (ii) D?
A, B, C, D, E, A
The die is thrown once. What is the probability of getting (i) A? (ii) D?
Answer:
Total faces = 6.
(i) A appears twice → P(A) = 2/6 = 1/3.
(ii) D appears once → P(D) = 1/6.
Total faces = 6.
(i) A appears twice → P(A) = 2/6 = 1/3.
(ii) D appears once → P(D) = 1/6.
203 marks
Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is the probability that it will land inside the circle with diameter 1 m? (Figure: rectangle 3 m × 2 m, circle of radius 0.5 m)
Answer:
Area of rectangle = 3×2 = 6 m².
Area of circle = πr² = π×(0.5)² = 0.25π m².
P(land inside circle) = (0.25π)/6 = π/24.
Area of rectangle = 3×2 = 6 m².
Area of circle = πr² = π×(0.5)² = 0.25π m².
P(land inside circle) = (0.25π)/6 = π/24.
213 marks
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it? (ii) She will not buy it?
(i) She will buy it? (ii) She will not buy it?
Answer:
Total pens = 144, good = 124, defective = 20.
(i) P(buy) = P(good) = 124/144 = 31/36.
(ii) P(not buy) = 1 - 31/36 = 5/36.
Total pens = 144, good = 124, defective = 20.
(i) P(buy) = P(good) = 124/144 = 31/36.
(ii) P(not buy) = 1 - 31/36 = 5/36.
223 marks
Two dice, one blue and one grey, are thrown at the same time. (i) Complete the following table:
Event 'Sum on two dice' → 2,3,4,5,6,7,8,9,10,11,12
Probability → 1/36, 2/36, 3/36, 4/36, 5/36, 6/36, 5/36, 4/36, 3/36, 2/36, 1/36
(ii) A student argues that 'there are 11 possible outcomes 2,3,4,5,6,7,8,9,10,11,12. Therefore, each of them has a probability 1/11. Do you agree? Why or why not?
Event 'Sum on two dice' → 2,3,4,5,6,7,8,9,10,11,12
Probability → 1/36, 2/36, 3/36, 4/36, 5/36, 6/36, 5/36, 4/36, 3/36, 2/36, 1/36
(ii) A student argues that 'there are 11 possible outcomes 2,3,4,5,6,7,8,9,10,11,12. Therefore, each of them has a probability 1/11. Do you agree? Why or why not?
Answer:
(ii) Disagree. The outcomes are not equally likely because different sums have different numbers of favourable outcomes (e.g., sum 7 has 6 ways, sum 2 has 1 way).
(ii) Disagree. The outcomes are not equally likely because different sums have different numbers of favourable outcomes (e.g., sum 7 has 6 ways, sum 2 has 1 way).
233 marks
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result, i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Answer:
Total outcomes = 2³ = 8. Favorable outcomes for win: HHH, TTT → 2.
P(win) = 2/8 = 1/4. P(lose) = 1 - 1/4 = 3/4.
Total outcomes = 2³ = 8. Favorable outcomes for win: HHH, TTT → 2.
P(win) = 2/8 = 1/4. P(lose) = 1 - 1/4 = 3/4.
244 marks
A die is thrown twice. What is the probability that
(i) 5 will not come up either time? (ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]
(i) 5 will not come up either time? (ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]
Answer:
Total outcomes = 6×6 = 36.
(i) Number of outcomes without 5: 5×5 = 25 → P = 25/36.
(ii) 5 comes up at least once = 1 - P(no 5) = 1 - 25/36 = 11/36.
Total outcomes = 6×6 = 36.
(i) Number of outcomes without 5: 5×5 = 25 → P = 25/36.
(ii) 5 comes up at least once = 1 - P(no 5) = 1 - 25/36 = 11/36.
254 marks
Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails, or one of each. Therefore, for each of these outcomes, the probability is 1/3.
(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.
(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails, or one of each. Therefore, for each of these outcomes, the probability is 1/3.
(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.
Answer:
(i) Incorrect. The outcomes are not equally likely. Possible outcomes: HH, HT, TH, TT (4 outcomes). P(two heads)=1/4, P(two tails)=1/4, P(one of each)=2/4=1/2.
(ii) Correct. Odd numbers: 1,3,5 (3 outcomes), even numbers: 2,4,6 (3 outcomes). So P(odd)=3/6=1/2.
(i) Incorrect. The outcomes are not equally likely. Possible outcomes: HH, HT, TH, TT (4 outcomes). P(two heads)=1/4, P(two tails)=1/4, P(one of each)=2/4=1/2.
(ii) Correct. Odd numbers: 1,3,5 (3 outcomes), even numbers: 2,4,6 (3 outcomes). So P(odd)=3/6=1/2.
Key Concepts Summary
Formula 1Probability of an Event
What is the formula for probability of an event E?
Answer: P(E) = Number of outcomes favourable to E / Total number of possible outcomes.
Formula 2Complementary Event
What is the relation between P(E) and P(not E)?
Answer: P(E) + P(not E) = 1.
NoteRange of Probability
What is the range of probability of an event?
Answer: 0 ≤ P(E) ≤ 1.
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