BOARD OF SECONDARY EDUCATION, TELANGANA
CLASS 10 MATHEMATICS - CHAPTER 13: PROBABILITY
Complete Exercise-Wise Questions & Step-by-Step Solutions (Ex 13.1 - 13.2) | Muqeem Tutorial
P(E) = (Number of outcomes favorable to E) / (Total number of possible outcomes)
0 ≤ P(E) ≤ 1 | P(E) + P(not E) = 1
0 ≤ P(E) ≤ 1 | P(E) + P(not E) = 1
EXERCISE 13.1 — Theoretical Probability & Elementary Events (Q1 to Q25)
Q1. Complete the following statements:
• (i) Probability of an event E + Probability of the event 'not E' = __________. 👉 1
• (ii) The probability of an event that cannot happen is __________. Such an event is called an __________. 👉 0, Impossible event
• (iii) The probability of an event that is certain to happen is __________. Such an event is called a __________. 👉 1, Sure or Certain event
• (iv) The sum of the probabilities of all the elementary events of an experiment is __________. 👉 1
• (v) The probability of an event is greater than or equal to __________ and less than or equal to __________. 👉 0, 1
• (ii) The probability of an event that cannot happen is __________. Such an event is called an __________. 👉 0, Impossible event
• (iii) The probability of an event that is certain to happen is __________. Such an event is called a __________. 👉 1, Sure or Certain event
• (iv) The sum of the probabilities of all the elementary events of an experiment is __________. 👉 1
• (v) The probability of an event is greater than or equal to __________ and less than or equal to __________. 👉 0, 1
Q2. 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:
👉 NOT equally likely (Depends on mechanical condition, battery, and fuel).
• (ii) A player attempts to shoot a basketball. She/he shoots or misses the shot:
👉 NOT equally likely (Depends on skill, distance, and defense).
• (iii) A trial is made to answer a true-false question. The answer is right or wrong:
👉 Equally likely (Exactly two outcomes with equal probability 1/2).
• (iv) A baby is born. It is a boy or a girl:
👉 Equally likely (Two natural outcomes with equal probability 1/2).
👉 NOT equally likely (Depends on mechanical condition, battery, and fuel).
• (ii) A player attempts to shoot a basketball. She/he shoots or misses the shot:
👉 NOT equally likely (Depends on skill, distance, and defense).
• (iii) A trial is made to answer a true-false question. The answer is right or wrong:
👉 Equally likely (Exactly two outcomes with equal probability 1/2).
• (iv) A baby is born. It is a boy or a girl:
👉 Equally likely (Two natural outcomes with equal probability 1/2).
Q3. 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?
Solution:
Tossing a coin has only two mutually exclusive and equally likely outcomes (Head or Tail).
P(Head) = 1/2 and P(Tail) = 1/2. Neither team can influence or predict the result.
👉 Coin toss is a fair and completely unbiased decision method.
Tossing a coin has only two mutually exclusive and equally likely outcomes (Head or Tail).
P(Head) = 1/2 and P(Tail) = 1/2. Neither team can influence or predict the result.
👉 Coin toss is a fair and completely unbiased decision method.
Q4. Which of the following cannot be the probability of an event? (A) 2/3 (B) -1.5 (C) 15% (D) 0.7
Solution:
Since 0 ≤ P(E) ≤ 1, a probability value can never be negative.
👉 Answer: (B) -1.5
Since 0 ≤ P(E) ≤ 1, a probability value can never be negative.
👉 Answer: (B) -1.5
Q5. If P(E) = 0.05, what is the probability of 'not E'?
Solution:
P(not E) = 1 - P(E) = 1 - 0.05 = 0.95.
👉 P(not E) = 0.95
P(not E) = 1 - P(E) = 1 - 0.05 = 0.95.
👉 P(not E) = 0.95
Q6. 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) Orange candy: P(orange) = 0 (Impossible event)
• (ii) Lemon candy: P(lemon) = 1 (Sure event)
• (ii) Lemon candy: P(lemon) = 1 (Sure event)
Q7. 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?
Solution:
P(same birthday) = 1 - 0.992 = 0.008.
👉 Probability = 0.008
P(same birthday) = 1 - 0.992 = 0.008.
👉 Probability = 0.008
Q8. 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?
Total balls = 3 + 5 = 8.
• (i) P(red) = 3 / 8.
• (ii) P(not red) = 1 - 3/8 = 5 / 8.
👉 (i) P(red) = 3/8 | (ii) P(not red) = 5/8
• (i) P(red) = 3 / 8.
• (ii) P(not red) = 1 - 3/8 = 5 / 8.
👉 (i) P(red) = 3/8 | (ii) P(not red) = 5/8
Q9. 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?
Total marbles = 5 + 8 + 4 = 17.
• (i) P(red) = 5 / 17.
• (ii) P(white) = 8 / 17.
• (iii) P(green) = 4 / 17 ⇒ P(not green) = 1 - 4/17 = 13 / 17.
👉 (i) 5/17 | (ii) 8/17 | (iii) 13/17
• (i) P(red) = 5 / 17.
• (ii) P(white) = 8 / 17.
• (iii) P(green) = 4 / 17 ⇒ P(not green) = 1 - 4/17 = 13 / 17.
👉 (i) 5/17 | (ii) 8/17 | (iii) 13/17
Q10. 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: (i) will be a 50p coin? (ii) will not be a ₹5 coin?
Total coins = 100 + 50 + 20 + 10 = 180 coins.
• (i) P(50p) = 100 / 180 = 5 / 9.
• (ii) Non-₹5 coins = 180 - 10 = 170 ⇒ P(not ₹5) = 170 / 180 = 17 / 18.
👉 (i) 5/9 | (ii) 17/18
• (i) P(50p) = 100 / 180 = 5 / 9.
• (ii) Non-₹5 coins = 180 - 10 = 170 ⇒ P(not ₹5) = 170 / 180 = 17 / 18.
👉 (i) 5/9 | (ii) 17/18
Q11. 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. What is the probability that the fish taken out is a male fish?
Total fish = 5 + 8 = 13. Male fish = 5.
👉 P(male fish) = 5/13
👉 P(male fish) = 5/13
Q12. 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. 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?
Total outcomes = 8.
• (i) P(8) = 1 / 8.
• (ii) Odd numbers = {1, 3, 5, 7} (4) ⇒ P(odd) = 4 / 8 = 1 / 2.
• (iii) Number > 2 = {3, 4, 5, 6, 7, 8} (6) ⇒ P(>2) = 6 / 8 = 3 / 4.
• (iv) Number < 9 = All 8 numbers ⇒ P(<9) = 8 / 8 = 1.
👉 (i) 1/8 | (ii) 1/2 | (iii) 3/4 | (iv) 1
• (i) P(8) = 1 / 8.
• (ii) Odd numbers = {1, 3, 5, 7} (4) ⇒ P(odd) = 4 / 8 = 1 / 2.
• (iii) Number > 2 = {3, 4, 5, 6, 7, 8} (6) ⇒ P(>2) = 6 / 8 = 3 / 4.
• (iv) Number < 9 = All 8 numbers ⇒ P(<9) = 8 / 8 = 1.
👉 (i) 1/8 | (ii) 1/2 | (iii) 3/4 | (iv) 1
Q13. 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.
Sample space S = {1, 2, 3, 4, 5, 6} (6 outcomes).
• (i) Prime numbers = {2, 3, 5} (3) ⇒ P = 3/6 = 1/2.
• (ii) Between 2 and 6 = {3, 4, 5} (3) ⇒ P = 3/6 = 1/2.
• (iii) Odd numbers = {1, 3, 5} (3) ⇒ P = 3/6 = 1/2.
👉 All probabilities = 1/2
• (i) Prime numbers = {2, 3, 5} (3) ⇒ P = 3/6 = 1/2.
• (ii) Between 2 and 6 = {3, 4, 5} (3) ⇒ P = 3/6 = 1/2.
• (iii) Odd numbers = {1, 3, 5} (3) ⇒ P = 3/6 = 1/2.
👉 All probabilities = 1/2
Q14. 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.
Total cards = 52.
• (i) Red Kings = 2 ⇒ P = 2/52 = 1/26.
• (ii) Face cards = 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.
👉 (i) 1/26 | (ii) 3/13 | (iii) 3/26 | (iv) 1/52 | (v) 1/4 | (vi) 1/52
• (i) Red Kings = 2 ⇒ P = 2/52 = 1/26.
• (ii) Face cards = 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.
👉 (i) 1/26 | (ii) 3/13 | (iii) 3/26 | (iv) 1/52 | (v) 1/4 | (vi) 1/52
Q15. Five cards — the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is 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) P(Queen) = 1 / 5.
• (ii) Queen put aside ⇒ 4 cards left {10, J, K, A}:
(a) P(Ace) = 1 / 4.
(b) P(Queen) = 0 / 4 = 0.
👉 (i) 1/5 | (ii)(a) 1/4 | (ii)(b) 0
• (ii) Queen put aside ⇒ 4 cards left {10, J, K, A}:
(a) P(Ace) = 1 / 4.
(b) P(Queen) = 0 / 4 = 0.
👉 (i) 1/5 | (ii)(a) 1/4 | (ii)(b) 0
Q16. 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.
Solution:
Good pens = 132, Defective pens = 12.
Total pens = 132 + 12 = 144.
P(good pen) = 132 / 144 = 11 / 12.
👉 P(good pen) = 11/12
Good pens = 132, Defective pens = 12.
Total pens = 132 + 12 = 144.
P(good pen) = 132 / 144 = 11 / 12.
👉 P(good pen) = 11/12
Q17. (i) 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? (ii) Suppose the bulb drawn in (i) 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?
Solution:
• (i) Total = 20, Defective = 4 ⇒ P(defective) = 4 / 20 = 1 / 5.
• (ii) Non-defective drawn and not replaced ⇒ Total remaining = 19 bulbs.
Non-defective remaining = 16 - 1 = 15.
P(not defective) = 15 / 19.
👉 (i) 1/5 | (ii) 15/19
• (i) Total = 20, Defective = 4 ⇒ P(defective) = 4 / 20 = 1 / 5.
• (ii) Non-defective drawn and not replaced ⇒ Total remaining = 19 bulbs.
Non-defective remaining = 16 - 1 = 15.
P(not defective) = 15 / 19.
👉 (i) 1/5 | (ii) 15/19
Q18. 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.
Solution:
Total discs = 90.
• (i) Two-digit numbers = 10 to 90 (81 discs) ⇒ P = 81 / 90 = 9 / 10.
• (ii) Perfect squares = {1, 4, 9, 16, 25, 36, 49, 64, 81} (9 discs) ⇒ P = 9 / 90 = 1 / 10.
• (iii) Divisible by 5 = {5, 10, 15, ..., 90} (18 discs) ⇒ P = 18 / 90 = 1 / 5.
👉 (i) 9/10 | (ii) 1/10 | (iii) 1/5
Total discs = 90.
• (i) Two-digit numbers = 10 to 90 (81 discs) ⇒ P = 81 / 90 = 9 / 10.
• (ii) Perfect squares = {1, 4, 9, 16, 25, 36, 49, 64, 81} (9 discs) ⇒ P = 9 / 90 = 1 / 10.
• (iii) Divisible by 5 = {5, 10, 15, ..., 90} (18 discs) ⇒ P = 18 / 90 = 1 / 5.
👉 (i) 9/10 | (ii) 1/10 | (iii) 1/5
Q19. 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?
Solution:
Total faces = 6.
• (i) Faces with 'A' = 2 ⇒ P(A) = 2 / 6 = 1 / 3.
• (ii) Faces with 'D' = 1 ⇒ P(D) = 1 / 6.
👉 (i) P(A) = 1/3 | (ii) P(D) = 1/6
Total faces = 6.
• (i) Faces with 'A' = 2 ⇒ P(A) = 2 / 6 = 1 / 3.
• (ii) Faces with 'D' = 1 ⇒ P(D) = 1 / 6.
👉 (i) P(A) = 1/3 | (ii) P(D) = 1/6
Q20. Suppose you drop a die at random on the rectangular region of dimensions 3 m × 2 m. What is the probability that it will land inside the circle with diameter 1 m?
Solution:
Area of rectangle = 3 × 2 = 6 m2.
Circle diameter = 1 m ⇒ Radius r = 0.5 m = 1/2 m.
Area of circle = π r2 = π (1/2)2 = π / 4 m2.
P(landing inside circle) = (Area of circle) / (Area of rectangle) = (π / 4) / 6 = π / 24.
👉 Probability = π / 24 ≈ 0.1309
Area of rectangle = 3 × 2 = 6 m2.
Circle diameter = 1 m ⇒ Radius r = 0.5 m = 1/2 m.
Area of circle = π r2 = π (1/2)2 = π / 4 m2.
P(landing inside circle) = (Area of circle) / (Area of rectangle) = (π / 4) / 6 = π / 24.
👉 Probability = π / 24 ≈ 0.1309
Q21. 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?
Solution:
Total pens = 144. Defective = 20 ⇒ Good pens = 124.
• (i) P(buys) = P(good) = 124 / 144 = 31 / 36.
• (ii) P(does not buy) = P(defective) = 20 / 144 = 5 / 36.
👉 (i) 31/36 | (ii) 5/36
Total pens = 144. Defective = 20 ⇒ Good pens = 124.
• (i) P(buys) = P(good) = 124 / 144 = 31 / 36.
• (ii) P(does not buy) = P(defective) = 20 / 144 = 5 / 36.
👉 (i) 31/36 | (ii) 5/36
Q22. Two dice, one blue and one grey, are thrown at the same time. (i) Complete the table for sum on two dice. (ii) A student argues that there are 11 possible sums (2 to 12) and therefore each of them has a probability 1/11. Do you agree with this argument?
Solution:
Total outcomes = 36.
• (ii) NO, the argument is INCORRECT because the 11 sums are NOT equally likely outcomes.
Total outcomes = 36.
| Sum | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Prob. | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |
Q23. 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.
Solution:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} (8 outcomes).
Winning = {HHH, TTT} (2 outcomes). Losing = 8 - 2 = 6 outcomes.
P(Hanif loses) = 6 / 8 = 3 / 4.
👉 Probability Hanif loses = 3/4 = 0.75
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} (8 outcomes).
Winning = {HHH, TTT} (2 outcomes). Losing = 8 - 2 = 6 outcomes.
P(Hanif loses) = 6 / 8 = 3 / 4.
👉 Probability Hanif loses = 3/4 = 0.75
Q24. 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?
Solution:
Total outcomes = 36.
Outcomes with 5 at least once = 11.
• (i) P(no 5) = (36 - 11) / 36 = 25 / 36.
• (ii) P(at least one 5) = 11 / 36.
👉 (i) 25/36 | (ii) 11/36
Total outcomes = 36.
Outcomes with 5 at least once = 11.
• (i) P(no 5) = (36 - 11) / 36 = 25 / 36.
• (ii) P(at least one 5) = 11 / 36.
👉 (i) 25/36 | (ii) 11/36
Q25. Which of the following arguments are correct and which are not? Give reasons. (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.
Solution:
• (i) INCORRECT: Sample space S = {HH, HT, TH, TT} (4 outcomes). "One of each" means {HT, TH} with probability 2/4 = 1/2, not 1/3.
• (ii) CORRECT: Odd = {1, 3, 5} and Even = {2, 4, 6}. Both have 3 outcomes each out of 6, so P(odd) = 3/6 = 1/2.
👉 (i) Incorrect | (ii) Correct
• (i) INCORRECT: Sample space S = {HH, HT, TH, TT} (4 outcomes). "One of each" means {HT, TH} with probability 2/4 = 1/2, not 1/3.
• (ii) CORRECT: Odd = {1, 3, 5} and Even = {2, 4, 6}. Both have 3 outcomes each out of 6, so P(odd) = 3/6 = 1/2.
👉 (i) Incorrect | (ii) Correct
EXERCISE 13.2 — Advanced Experiments & Multi-Stage Events (Optional/Additional)
Q1. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on: (i) the same day? (ii) consecutive days? (iii) different days?
Solution:
Days = {Tue, Wed, Thu, Fri, Sat} (5 days). Total outcomes = 5 × 5 = 25.
• (i) Same day: {(T,T), (W,W), (Th,Th), (F,F), (S,S)} (5 outcomes) ⇒ P = 5/25 = 1/5.
• (ii) Consecutive days: {(T,W),(W,T),(W,Th),(Th,W),(Th,F),(F,Th),(F,S),(S,F)} (8 outcomes) ⇒ P = 8/25.
• (iii) Different days: 1 - P(same day) = 1 - 1/5 = 4/5.
👉 (i) 1/5 | (ii) 8/25 | (iii) 4/5
Days = {Tue, Wed, Thu, Fri, Sat} (5 days). Total outcomes = 5 × 5 = 25.
• (i) Same day: {(T,T), (W,W), (Th,Th), (F,F), (S,S)} (5 outcomes) ⇒ P = 5/25 = 1/5.
• (ii) Consecutive days: {(T,W),(W,T),(W,Th),(Th,W),(Th,F),(F,Th),(F,S),(S,F)} (8 outcomes) ⇒ P = 8/25.
• (iii) Different days: 1 - P(same day) = 1 - 1/5 = 4/5.
👉 (i) 1/5 | (ii) 8/25 | (iii) 4/5
Q2. A die whose faces are numbered 1, 2, 2, 3, 3, 6 is thrown twice and the total score in two throws is noted. (i) Complete the score table. (ii) What is the probability that the total score is: (a) even? (b) 6? (c) at least 6?
Solution:
Total outcomes = 6 × 6 = 36.
• (a) P(even score) = 18 / 36 = 1 / 2.
• (b) P(score = 6) = 4 / 36 = 1 / 9.
• (c) P(score ≥ 6) = 15 / 36 = 5 / 12.
👉 (a) 1/2 | (b) 1/9 | (c) 5/12
Total outcomes = 6 × 6 = 36.
• (a) P(even score) = 18 / 36 = 1 / 2.
• (b) P(score = 6) = 4 / 36 = 1 / 9.
• (c) P(score ≥ 6) = 15 / 36 = 5 / 12.
👉 (a) 1/2 | (b) 1/9 | (c) 5/12
Q3. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Solution:
Let blue balls = x. Total = 5 + x.
x / (5 + x) = 2 × [5 / (5 + x)] ⇒ x = 10.
👉 Number of Blue Balls = 10
Let blue balls = x. Total = 5 + x.
x / (5 + x) = 2 × [5 / (5 + x)] ⇒ x = 10.
👉 Number of Blue Balls = 10
Q4. A box contains 12 balls out of which x are black. If one ball is drawn at random, 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.
Solution:
Initial P(black) = x/12. New P(black) = (x+6)/18.
(x+6)/18 = 2 × (x/12) ⇒ x+6 = 3x ⇒ 2x = 6 ⇒ x = 3.
👉 Initial Black Balls x = 3 | P(black) = 1/4
Initial P(black) = x/12. New P(black) = (x+6)/18.
(x+6)/18 = 2 × (x/12) ⇒ x+6 = 3x ⇒ 2x = 6 ⇒ x = 3.
👉 Initial Black Balls x = 3 | P(black) = 1/4
Q5. 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 2/3. Find the number of blue marbles in the jar.
Solution:
Green marbles = (2/3) × 24 = 16.
Blue marbles = 24 - 16 = 8.
👉 Number of Blue Marbles = 8
Green marbles = (2/3) × 24 = 16.
Blue marbles = 24 - 16 = 8.
👉 Number of Blue Marbles = 8
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