Chapter 14 of NCERT Maths Class 10 delves into the concepts of probability, which measures the likelihood of an event occurring.
Key Concepts:
- Experiment: An activity with a well-defined outcome.
- Trial: Performing an experiment once.
- Outcome: A possible result of an experiment.
- Event: A set of outcomes.
- Probability: The measure of the likelihood of an event occurring.
Types of Probability:
- Theoretical Probability: Based on mathematical reasoning.
- Experimental Probability: Based on observations or experiments.
Probability of an Event:
- The probability of an event E, denoted by P(E), is the ratio of favorable outcomes to the total possible outcomes.
- 0 ≤ P(E) ≤ 1
Properties of Probability:
- The probability of a certain event is 1.
- The probability of an impossible event is 0.
Conditional Probability:
- The probability of event B occurring given that event A has already occurred is denoted by P(B|A).
Independent and Dependent Events:
- Two events A and B are dependent if the occurrence of one affects the occurrence of the other.
Applications:
- Real-world problems involving probability (e.g., games of chance, weather forecasting, quality control)
- Decision-making based on probability assessments
In essence, the chapter on probability provides a foundational understanding of the mathematical concept of chance, enabling students to analyze and make predictions in various situations.
Exercise 14.1
1. 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 ………..
Answer :
(i) 1
(ii) impossible event.
(iii) 1, Sure event.
(iv) 1.
(v) 0 , 1.
2.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.
Ans :
3. Why is tossing a coin considered to be a fair way of deciding which team should get the bail at the beginning of a football game?
Ans :
Tossing a coin is considered a fair way of deciding which team should get the ball at the beginning of a football game because it ensures an equal probability of outcome for both teams.
4. Which of the following cannot be the probability of an event?
(A) 2/3
(B) -1.5
(C) 15%
(D) 0.7
Ans :
- (A) 2/3: This is a valid probability as it lies between 0 and 1.
- (B) -1.5: This cannot be a probability because it is negative. Probability values are always non-negative.
- (C) 15%: This is a valid probability. 15% is equivalent to 0.15, which lies between 0 and 1.
- (D) 0.7: This is a valid probability as it lies between 0 and 1.
Therefore, -1.5 cannot be the probability of an event.
5. If P (E) = 0.05, what is the probability of ‘not E’?
Ans :
P(E) + P(not E) = 1
We know from the problem that:
P(E) = 0.05
P(not E) = 1 – P(E)
Now, we calculate:
P(not E) = 1 – 0.05 = 0.95
Therefore, the probability of “not E” is 0.95
6. 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?
Ans :
(i)Since the bag only contains lemon flavored candies, it’s impossible for Malini to take out an orange flavored one. event is 0.
(ii) Since all the candies in the bag are lemon flavored, it’s certain that Malini will take out a lemon flavored candy. event is 1.
7. 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?
Ans :
Let E be the event that 2 students have the same birthday.
We are given that P(not E) = 0.992.
Since the probability of an event and its complement always sum up to 1, we have:
P(E) + P(not E) = 1
Therefore, the probability that the 2 students have the same birthday (P(E)) is:
P(E) = 1 – P(not E) P(E) = 1 – 0.992 P(E) = 0.008
So, the probability that the 2 students have the same birthday is 0.008.
8. 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?
Ans :
(i) Probability of drawing a red ball:
There are 3 red balls and 5 black balls, making a total of 8 balls.
Number of red balls / Total number of balls = 3 / 8
(ii) Probability of drawing a ball that is not red:
Since all the balls that are not red are black, this is the same as the probability of drawing a black ball.
Therefore, the probability of drawing a black ball (or a ball that is not red) is:
Number of black balls / Total number of balls = 5 / 8
9. 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?
Ans :
Total marbles: 5 red + 8 white + 4 green = 17 marbles
Probability of drawing a red marble:
- Number of red marbles = 5
- Total number of marbles = 17
- Probability = Number of red marbles / Total number of marbles = 5 / 17
Probability of drawing a white marble:
- Number of white marbles = 8
- Total number of marbles = 17
- Probability = Number of white marbles / Total number of marbles = 8 / 17
Marble that is not green:
- This is the same as the probability of drawing a red or white marble.
- We can add the probabilities of drawing red and white marbles.
- Probability = (Number of red marbles + Number of white marbles) / Total number of marbles
- Probability = (5 + 8) / 17 = 13 / 17
10. A piggy bank contains hundred 50 p 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 50 p coin?
(ii) will not be a ₹ 5 coin?
Ans :
Total coins:
- 100 * 50 paise coins = 50 rupees
- 50 * ₹1 coins = 50 rupees
- 20 * ₹2 coins = 40 rupees
- 10 * ₹5 coins = 50 rupees
- Total rupees = 50 + 50 + 40 + 50 = 190 rupees
Total number of coins: 100 + 50 + 20 + 10 = 180 coins
Probability of a 50 paise coin:
- Number of 50 paise coins = 100
- Total number of coins = 180
- Probability = 100/180 = 5/9
Probability of not a ₹5 coin:
- Number of coins that are not ₹5 = 100 + 50 + 20 = 170
- Total number of coins = 180
- Probability = 170/180 = 17/18
11. 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 figure). What is the probability that the fish taken out is a male fish?
Ans :
Total fish: 5 male + 8 female = 13 fish
Probability of male fish:
- Number of male fish = 5
- Total number of fish = 13
Probability = Number of male fish / Total number of fish = 5 / 13
Therefore, the probability that the fish taken out is a male fish is 5/13.
12. A game of chance consists of spinning an arro w which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see figure.), 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?
Ans :
- (i) 1/8
- (ii) 1/2
- (iii) 3/4
- (iv) 1
13. A die is thrown once. Find the probability of getting
(i) a prime number
(ii) a number lying between 2 and 6
(ill) an odd number
Ans :
Possible outcomes when a die is thrown once: 1, 2, 3, 4, 5, 6
Total number of outcomes: 6
(i)
- Number of favorable outcomes: 3
- Probability = 3/6 = 1/2
(ii)
- Number of favorable outcomes: 3
- Probability = 3/6 = 1/2
(iii)
- Odd numbers: 1, 3, 5
- Number of favorable outcomes: 3
- Probability = 3/6 = 1/2
14. 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
Ans :
Total cards: 52
(i) Probability of getting a king of red color:
- Kings of red color: King of Hearts, King of Diamonds
- Number of favorable outcomes: 2
- Probability = 2/52 = 1/26
(ii) Probability of getting a face card:
- Face cards: King, Queen, Jack
- Number of face cards: 12 (4 kings, 4 queens, 4 jacks)
- Probability = 12/52 = 3/13
(iii) red face card:
- Red face cards: King of Hearts, Queen of Hearts, Jack of Hearts, King of Diamonds, Queen of Diamonds, Jack of Diamonds
- Number of red face cards: 6
- Probability = 6/52 = 3/26
(iv) Probability of getting the jack of hearts:
- There is only one Jack of Hearts.
- Probability = 1/52
(v) Probability of getting a spade:
- There are 13 spades in a deck.
- Probability = 13/52 = 1/4
(vi) Probability of getting the queen of diamonds:
- There is only one Queen of Diamonds.
- Probability = 1/52
15. 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?
Ans :
(i) Probability of picking the queen:
- Total cards: 5
- Number of queens: 1
- Probability = Number of queens / Total cards = 1/5
(ii) If the queen is drawn and put aside, there are now 4 cards left: 10, jack, king, and ace of diamonds.
(a) Probability of picking an ace:
- Number of aces: 1
- Total cards remaining: 4
- Probability = 1/4
(b) Probability of picking a queen:
- Since the queen was already drawn and put aside, there are no queens left.
- Probability = 0/4 = 0
16. 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.
Ans :
Total pens = 12 defective + 132 good = 144 pens
Probability of a good pen = Number of good pens / Total pens
= 132 / 144 = 11/12
17. (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?
Ans :
(i) Probability of drawing a defective bulb:
- Total bulbs = 20
- Number of defective bulbs = 4
- Probability = Number of defective bulbs / Total bulbs = 4/20 = 1/5
(ii) Probability of drawing a non-defective bulb (second draw):
- Since one non-defective bulb was drawn in the first draw, there are now 19 remaining bulbs and 3 defective bulbs.
- Probability = Number of non-defective bulbs / Total bulbs = 16/19
18. 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.
Ans :
(i) Probability of a two-digit number:
- Two-digit numbers range from 10 to 90.
- Number of two-digit numbers = 90 – 10 + 1 = 81
- Total number of discs = 90
- Probability = Number of two-digit numbers / Total number of discs = 81/90 = 9/10
(ii) Probability of a perfect square number:
- Perfect squares between 1 and 90: 1, 4, 9, 16, 25, 36, 49, 64, 81
- Number of perfect square numbers = 9
- Probability = Number of perfect square numbers / Total number of discs = 9/90 = 1/10
(iii)
- Numbers divisible by 5: 5, 10, 15, 20, …, 90
- Number of numbers divisible by 5 = 90 / 5 = 18
- Probability = Number of numbers divisible by 5 / Total number of discs = 18/90 = 1/5
19. A child has a die whose six faces show the letters as given below:
The die is thrown once. What is the probability of getting
(i) A?
(ii) D?
Ans :
(i) Probability of getting A:
- Number of faces with A: 2
- Total faces: 6
- Probability = Number of favorable outcomes / Total number of outcomes = 2/6 = 1/3
(ii) Probability of getting D:
- Number of faces with D: 1
- Total faces: 6
- Probability = Number of favorable outcomes / Total number of outcomes
- = ⅙
20. Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1 m?
Ans :
- Calculate the area of the rectangle:
- Area of rectangle = length * width = 3m * 2m = 6 m²
- Calculate the area of the circle:
- Radius of the circle = diameter / 2 = 1m / 2 = 0.5m
- Area of circle = π * radius² = π * (0.5m)² = π/4 m²
- Calculate the probability:
- The probability of the die landing inside the circle is the ratio of the area of the circle to the area of the rectangle.
- Probability = Area of circle / Area of rectangle = (π/4 m²) / (6 m²) = π/24
Therefore, the probability that the die will land inside the circle is π/24.
21. 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?
Ans :
Total pens = 144 Defective pens = 20 Good pens = 144 – 20 = 124
(i) Probability of buying (good pen):
- Probability = Number of good pens / Total pens = 124 / 144 = 31/36
(ii) Probability of not buying (defective pen):
- Probability = Number of defective pens / Total pens = 20 / 144 = 5/36
22. Two dice, one blue and one grey, are thrown at the same time. Now
(i) Complete the following table:
(ii) A student argues that-there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your answer.
Ans :
(i)
| Event: (Sum on 2 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) Student’s argument:
The student is incorrect. There are actually 36 possible outcomes when two dice are rolled. This is because each die has 6 possible outcomes, and the total number of outcomes is found by multiplying the number of outcomes for each die: 6 * 6 = 36.
Therefore, each outcome (sum of 2 dice) has a probability of 1/36, not 1/11.
23. 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.
Ans :
Probability of all heads:
- Each toss has a 1/2 probability of getting heads.
- For three tosses, the probability of all heads is (1/2) * (1/2) * (1/2) = 1/8
Probability of all tails:
- Similarly, the probability of all tails is 1/8.
Probability of losing (not winning):
- Losing means not getting all heads or all tails.
- So, the probability of losing is 1 – (Probability of all heads + Probability of all tails)
- Probability of losing = 1 – (1/8 + 1/8) = 1 – 2/8 = 6/8 = 3/4
Therefore, the probability that Hanif will lose the game is 3/4.
24. 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.]
Ans :
When a die is thrown once, there are 6 possible outcomes (1, 2, 3, 4, 5, 6). So, when a die is thrown twice, there are 6 * 6 = 36 possible outcomes.
Probabilities:
(i)
- For one throw, the probability of not getting 5 is 5/6 (since there are 5 other possible outcomes).
- For two throws, the probability of not getting 5 both times is (5/6) * (5/6) = 25/36
(ii)
- This is the complement of not getting 5 either time.
- So, P(5 comes up at least once) = 1 – P(5 does not come up either time)
- P(5 comes up at least once) = 1 – 25/36 = 11/36
25. 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.
Ans :
Argument (i):
- Incorrect.
- While there are three possible outcomes (two heads, two tails, or one of each), they are not equally likely.
- There are four possible combinations: HH, HT, TH, and TT.
- Only one of these combinations results in two heads and two tails.
- Therefore, the probability of each outcome is not 1/3.
Argument (ii):
- Correct.
- When a die is thrown, there are two possible outcomes: odd or even.
- Since these are the only two possibilities and they are mutually exclusive, the probability of getting an odd number is 1/2.
In conclusion, argument (i) is incorrect, while argument (ii) is correct.
