Get complete, step-by-step solutions for NCERT Class 10 Maths Chapter 14 Exercise 14.1 on Introduction to Probability. Solutions cover fundamental concepts, including definitions (Q.1, Q.4), mutually exclusive events, complementary events ($P(E) + P(\text{not } E) = 1$) (Q.5, Q.7), and equally likely outcomes (Q.2, Q.3). Practice calculations involving coins (Q.23, Q.25), dice (Q.13, Q.22, Q.24), cards (Q.14, Q.15), marbles/balls (Q.8, Q.9), and real-world scenarios like defective items (Q.16, Q.17, Q.21) and geometric probability (Q.20). Essential for mastering the basics of theoretical probability.
This exercise covers the basic concepts and formulas of theoretical probability.
1. Completing Statements (Basic Definitions)
(i) Probability of an event E + Probability of the event ‘not E’ =1. (or P(E)+P(not E)=1) (ii) The probability of an event that cannot happen is 0. Such an event is called an impossible event. (iii) The probability of an event that is certain to happen is 1. Such an event is called a 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 0 and less than or equal to 1.
2. Equally Likely Outcomes
Equally likely outcomes mean each outcome has the same chance of occurring.
(i) A driver attempts to start a car. The car starts or does not start. * Not Equally Likely. The likelihood depends on the condition of the car, the battery, etc. A new car is far more likely to start than a very old, faulty one. (ii) A player attempts to shoot a basketball. She/he shoots or misses the shot. * Not Equally Likely. The likelihood depends on the player’s skill, which varies widely (e.g., a professional player is far more likely to shoot than a beginner). (iii) A trial is made to answer a true-false question. The answer is right or wrong. * Equally Likely. Assuming the person has no knowledge of the subject, there is an equal chance (1/2) of guessing True or False. (iv) A baby is born. It is a boy or a girl. * Equally Likely. (In real life, the probability is very close to 1/2 for each, and for basic probability problems, we assume it is exactly 1/2).
3. Fairness of Tossing a Coin
Tossing a coin is considered a fair way because the two outcomes (Head and Tail) are equally likely . There is no factor that influences the outcome to favor one side or the other. Thus, the decision is made completely by chance.
4. Which cannot be the Probability?
The probability of an event E, P(E), must satisfy 0≤P(E)≤1.
(A) 2/3≈0.67 (Valid) (B) –1.5 (Invalid). Probability cannot be negative. (C) 15%=0.15 (Valid) (D) 0.7 (Valid)
The answer is (B) –1.5.
5. Probability of ‘not E’
- Given: P(E)=0.05.
- Formula: P(not E)=1−P(E).
- Calculation: P(not E)=1−0.05=0.95.
6. Lemon Flavored Candies
The bag contains only lemon flavored candies.
(i) Probability of an orange flavoured candy? * Number of orange flavored candies = 0. * P(orange)=Total Candies0=0. (Impossible event) (ii) Probability of a lemon flavoured candy? * Number of lemon flavored candies = Total number of candies. * P(lemon)=Total CandiesTotal Candies=1. (Sure event)
7. Probability of Same Birthday
- Given: P(not having same birthday)=P(not E)=0.992.
- Find: P(having same birthday)=P(E).
- Calculation: P(E)=1−P(not E)=1−0.992=0.008.
8. Balls in a Bag
- Total Balls: 3 (Red)+5 (Black)=8.
(i) Probability that the ball drawn is red? * P(Red)=Total BallsNumber of Red Balls=3/8. (ii) Probability that the ball drawn is not red? * ‘Not Red’ means Black. * P(Not Red)=P(Black)=85. * Alternatively: P(Not Red)=1−P(Red)=1−3/8=5/8.
9. Marbles in a Box
- Total Marbles: 5 (Red)+8 (White)+4 (Green)=17.
(i) Probability that the marble taken out will be red? * P(Red)=175. (ii) Probability that the marble taken out will be white? * P(White)=178. (iii) Probability that the marble taken out will be not green? * ‘Not Green’ means Red or White (5+8=13 marbles). * P(Not Green)=1713.
10. Coins in a Piggy Bank
- Total Coins: 100 (50p)+50 (₹1)+20 (₹2)+10 (₹5)=180.
(i) Probability that the coin will be a 50 p coin? * P(50 p)=Total CoinsNumber of 50 p Coins=180100=5/9. (ii) Probability that the coin will not be a ₹ 5 coin? * ‘Not ₹ 5’ means 50 p, ₹1, or ₹2 (100+50+20=170 coins). * P(Not ₹ 5)=180170=17/18.
11. Fish in an Aquarium
- Total Fish: 5 (Male)+8 (Female)=13.
- Probability that the fish taken out is a male fish?
- P(Male)=Total FishNumber of Male Fish=5/13.
12. Spinning an Arrow
- Total Outcomes: {1,2,3,4,5,6,7,8}. Total number of outcomes =8.
(i) Probability that it will point at 8? * P(8)=81. (ii) Probability that it will point at an odd number? * Odd numbers: {1,3,5,7}. Number of favorable outcomes =4. * P(Odd)=84=1/2. (iii) Probability that it will point at a number greater than 2? * Numbers greater than 2: {3,4,5,6,7,8}. Number of favorable outcomes =6. * P(>2)=86=3/4. (iv) Probability that it will point at a number less than 9? * Numbers less than 9: {1,2,3,4,5,6,7,8}. Number of favorable outcomes =8. * P(<9)=88=1. (Sure event)
13. Throwing a Die Once
- Total Outcomes: {1,2,3,4,5,6}. Total number of outcomes =6.
(i) Probability of getting a prime number? * Prime numbers: {2,3,5}. Number of favorable outcomes =3. * P(Prime)=63=1/2. (ii) Probability of getting a number lying between 2 and 6? * Numbers between 2 and 6: {3,4,5}. Number of favorable outcomes =3. * P(Between 2 and 6)=63=1/2. (iii) Probability of getting an odd number? * Odd numbers: {1,3,5}. Number of favorable outcomes =3. * P(Odd)=63=1/2.
14. Drawing One Card
- Total Cards: 52.
(i) Probability of getting a king of red colour? * Red kings: King of Hearts and King of Diamonds. Number of favorable outcomes =2. * P(Red King)=522=1/26. (ii) Probability of getting a face card? * Face cards: Jack, Queen, King (3 per suit × 4 suits). Number of favorable outcomes =12. * P(Face Card)=5212=3/13. (iii) Probability of getting a red face card? * Red face cards: (3 in Hearts + 3 in Diamonds). Number of favorable outcomes =6. * P(Red Face Card)=526=3/26. (iv) Probability of getting the jack of hearts? * Jack of Hearts is one specific card. Number of favorable outcomes =1. * P(Jack of Hearts)=1/52. (v) Probability of getting a spade? * Spades are one suit. Number of favorable outcomes =13. * P(Spade)=5213=1/4. (vi) Probability of getting the queen of diamonds? * Queen of Diamonds is one specific card. Number of favorable outcomes =1. * P(Queen of Diamonds)=1/52.
15. Five Diamond Cards
- Initial Cards (Set S): {Ten, Jack, Queen, King, Ace} of Diamonds. Total cards =5.
(i) Probability that the card is the queen? * P(Queen)=51.
(ii) If the queen is drawn and put aside: * New Total Cards (Set S’): {Ten, Jack, King, Ace} of Diamonds. Total cards =4. * (a) Probability that the second card picked up is an ace? * Ace is 1 card in S’. P(Ace)=1/4. * (b) Probability that the second card picked up is a queen? * Queen is 0 cards in S’. P(Queen)=40=0.
16. Defective and Good Pens
- Good Pens: 132.
- Defective Pens: 12.
- Total Pens: 132+12=144.
- Probability that the pen taken out is a good one?
- P(Good)=Total PensNumber of Good Pens=144132=11/12.
17. Defective Bulbs
(i) Probability that the bulb is defective? * Initial Total Bulbs: 20. Defective: 4. * P(Defective)=204=1/5.
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. * New Total Bulbs: 20−1=19. * New Non-Defective Bulbs: 16−1=15. (Since the first one was NOT defective) * Probability that this bulb is not defective? * P(Not Defective)=New TotalNew Non-Defective=15/19.
18. Numbered Discs
- Total Discs: 90 (numbered 1 to 90).
(i) Probability that it bears a two-digit number? * Two-digit numbers: 10 to 90. Number of favorable outcomes =90−10+1=81. * P(Two-digit)=9081=9/10. (ii) Probability that it bears a perfect square number? * Perfect squares ≤90: {1,4,9,16,25,36,49,64,81}. Number of favorable outcomes =9. * P(Perfect Square)=909=1/10. (iii) Probability that it bears a number divisible by 5? * Numbers divisible by 5: {5,10,15,…,90}. Number of favorable outcomes =90/5=18. * P(Divisible by 5)=9018=1/5.
19. Child’s Die
- Total Outcomes (Faces): {A, B, C, D, E, A}. Total number of outcomes =6.
(i) Probability of getting A? * Number of A’s =2. * P(A)=62=1/3. (ii) Probability of getting D? * Number of D’s =1. * P(D)=61.
20. Dropping a Die (Geometric Probability)
- Area of Rectangular Region: 3 m×2 m=6 m2. (Total area)
- Area of Circular Region: Diameter 1 m⟹ Radius r=0.5 m. Acircle=πr2=π(0.5)2=0.25π m2
- Probability (that it will land inside the circle): P=Area of RectangleArea of Circle=60.25π=π/24
21. Ball Pens
- Total Pens: 144.
- Defective Pens: 20.
- Good Pens: 144−20=124.
(i) Probability that she will buy it? (She buys if it is good). * P(Good)=144124=31/36. (ii) Probability that she will not buy it? (She doesn’t buy if it is defective). * P(Defective)=14420=5/36.
22. Sum on Two Dice
- Total Outcomes: 6×6=36 equally likely outcomes.
(i) Complete the table:
| Event: ‘Sum on 2 dice’ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| Favorable Outcomes | 1 | 2 | 3 | 4 | 5 | 6 | 5 | 4 | 3 | 2 | 1 |
| Probability | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |
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(ii) A student argues that… Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your answer. * No, I do not agree. * Justification: The outcomes (sums 2, 3, 4, …, 12) are not equally likely. As shown in the table above, the sum of 7 has the highest frequency (6/36), while the sum of 2 and 12 have the lowest (1/36). The formula P(E)=Total OutcomesFavorable Outcomes only applies when all outcomes are equally likely.
23. Tossing a One Rupee Coin 3 Times
- Total Outcomes (Sample Space S): {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. Total number of outcomes =8.
- Hanif wins if: Three heads (HHH) or three tails (TTT).
- Hanif loses if: Any other outcome (6 outcomes).
- Probability that Hanif will lose the game:
- P(Lose)=Total OutcomesNumber of Losing Outcomes=86=3/4.
24. A Die is Thrown Twice
- Total Outcomes: 6×6=36.
(i) Probability that 5 will not come up either time? * Total Outcomes: 36. * Favorable Outcomes (5 does NOT come up): In each throw, there are 5 possibilities ({1,2,3,4,6}). * Number of favorable outcomes =5×5=25. * P(No 5)=3625.
(ii) Probability that 5 will come up at least once? * ‘At least once’ is the complement of ‘No 5’. * P(At least one 5)=1−P(No 5) * P(At least one 5)=1−3625=3636−25=11/36.
25. Correctness of Arguments
(i) If two coins are tossed simultaneously there are three possible outcomes… Therefore, for each of these outcomes, the probability is 1/3. * Not Correct. * Reason: The actual outcomes are {HH, TT, HT, TH}. There are 4 possible outcomes. The three listed events (two heads, two tails, one of each) are not elementary events and are not equally likely. The event ‘one of each’ (HT or TH) has a probability of 2/4=1/2, while the others have 1/4.
(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. * Correct. * Reason: The elementary outcomes are {1,2,3,4,5,6}. Odd outcomes are {1,3,5} (3 outcomes). Even outcomes are {2,4,6} (3 outcomes). Since the number of outcomes for ‘Odd’ equals the number of outcomes for ‘Even’, both are equally likely with a probability of 3/6=1/2.