Class 9 Mathematics Ganita Manjari Part 1 NCERT Solutions
Chapter-7: The Mathematics of Maybe: Introduction to Probability
End-Of-Chapter Exercise
1. Fill in the blanks.
(i) The probability of an impossible event is ____.
(ii) The set of all possible outcomes of a random experiment is called the __________.
(iii) The probability of an event that is certain to happen is _______.
(iv) Tossing a fair coin has a probability of ______ for getting heads.
Solutions:
(i) The probability of an impossible event is 0.
(ii) The set of all possible outcomes of a random experiment is called the sample space.
(iii) The probability of an event that is certain to happen is 1.
(iv) Tossing a fair coin has a probability of \underline{\bm{\frac{1}{2}}} for getting heads.
2. In a survey of 50 students, 15 students said they liked football. The number of students who like football is 15, and the ________ (frequency/relative frequency) is __________ (fill in the fraction or decimal).
Solution:
The number of students who like football is 15, which is the frequency.
Relative frequency = \frac{\text{frequency}}{\text{Total number of students}}
= \frac{15}{50}=\frac{3}{10}=0.3
Answer:
In a survey of 50 students, 15 students said they liked football. The number of students who like football is 15, and the relative frequency is 0.3
3. 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) Tossing a fair coin once.
(iii) Rolling a fair 6-sided die.
(iv) Choosing a marble randomly from a bag that contains 3 red marbles and 7 blue marbles.
(v) A baby is born. It is a boy or a girl.
Solution:
An experiment has equally likely outcomes if every possible result has the exact same probability or chance of occurring.
(i) Not equally likely.
Explanation: The outcome depends entirely on external factors like the mechanical condition of the car, fuel levels, and battery health. Under normal circumstances, a functional car is much more likely to start than not start.
(ii) Equally likely
Explanation: Because the coin is specified as “fair,” it is perfectly symmetrical. A fair coin has two outcomes, Head and Tail, each with a probability of \frac{1}{2}.
(iii) Equally likely
Explanation: A fair die has six faces. Each of the six faces has an equal probability of \frac{1}{6}.
(iv) Not equally likely.
Explanation: The probability depends on the quantity of each color. The probability of picking a blue marble is \frac{7}{10}, while picking a red marble is only \frac{3}{10}. Blue is much more likely to be chosen.
(v) Equally likely
Explanation: The birth of a baby is treated as a random biological event where the two outcomes(boy or girl) are independent and have an equal chance \frac{1}{2} of occurring.
4. Write the sample space and calculate the probability based on the given information.
(i) Two coins are tossed at the same time. What is the probability of getting at least one head?
Solution:
Sample Space: S={HH, HT, TH, TT}
n(S)= 4
Event: getting at least one head
E={HH, HT, TH}
n(E)= 3
P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}=\frac{n(E)}{n(S)}
= \frac{3}{4}
(ii) Ten identical cards numbered 1 to 10 are placed in a box. One card is drawn at random. What is the probability of drawing a card with an even number?
Solution:
Sample space: S={1,2,3,4,5,6,7,8,9,10}
n(S)= 10
Event: getting an even number card
E= {2,4,6,8,10}
n(E)= 5
P(E) = \frac{n(E)}{n(S)}
= \frac{5}{10}=\frac{1}{2}
(iii) A die is rolled once. What is the probability of getting a number greater than 4?
Solution:
Sample space: S={1,2,3,4,5,6}
n(S)= 6
Event: getting a number greater than 4
E= {5,6}
n(E)= 2
P(E) = \frac{n(E)}{n(S)}
= \frac{2}{6}=\frac{1}{3}
(iv) A bag contains 3 red balls, 2 blue balls, and 1 green ball. One ball is picked at random. What is the probability that it is not red?
Solution:
Sample space: S={Red, Red, Red, Blue, Blue, Green}
n(S)= 6
Event: ball drawn is not red
E= {Blue, Blue, Green}
n(E)= 3
P(E) = \frac{n(E)}{n(S)}
= \frac{3}{6}=\frac{1}{2}
(v) Three coins are tossed simultaneously. What is the probability of getting exactly two heads?
Solution:
Sample space: S={HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S)= 8
Event: getting exactly two heads
E= {HHT, HTH, THH}
n(E)= 3
P(E) = \frac{n(E)}{n(S)}
= \frac{3}{8}
5. A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?
Solution:
Sample space: S={Strawberry, Lemon, Mint}
n(S)= 3
Event: getting a strawberry candy
E= {Strawberry}
n(E)= 1
P(E) = \frac{n(E)}{n(S)}
= \frac{1}{3}
6. A child has 2 shirts (one red and one blue) and 3 types of pants (jeans, khakis, and shorts). List all the possible combinations of outfits consisting of one shirt and one pair of pants. Display your answer in a table format.
Solution:
The possible outfit combinations are:
| Outfit | Shirt | Pants | Combination |
|---|---|---|---|
| 1 | Red | Jeans | (Red, Jeans) |
| 2 | Red | Khakis | (Red, Khakis) |
| 3 | Red | Shorts | (Red, Shorts) |
| 4 | Blue | Jeans | (Blue, Jeans) |
| 5 | Blue | Khakis | (Blue, Khakis) |
| 6 | Blue | Shorts | (Blue, Shorts) |
7. A tyre company records distances before replacement in 1000 cases.
| Distance (km) | Less than 4000 | 4001 to 9000 | 9001 to 14000 | More than 14000 |
|---|---|---|---|---|
| Number of cases | 20 | 210 | 325 | 445 |
Find the probability that a randomly chosen tyre lasts:
(i) Less than 4000 km.
(ii) Between 4000 and 14000 km.
(iii) More than 14000 km
Solution:
Total number of cases =20+210+325+445=1000
The probability of an event is:
P(E)= \frac{\text{Number of favourable cases}}{\text{Total number of cases}}
(i) Probability that a tyre lasts less than 4000 km
Number of favourable cases = 20
P(Less than 4000 km)= \frac{20}{1000}
= \frac{1}{50}
(ii) Probability that a tyre lasts between 4000 km and 14000 km
Number of cases between 4000km and 14000km= 210+325=535
P(Between 4000 km and 14000 km=\frac{535}{1000}
= \frac{107}{200}
(iii) Probability that a tyre lasts more than 14000 km
Number of favourable cases = 445
P(More than 14000 km)= \frac{445}{1000}
= \frac{89}{200}
8. The letters of the word ‘PEACE’ are placed on cards. Leela draws a card without looking.
(i) What is the probability that it is a P, E or C?
Solution:
P(P, E or C)= \frac{\text{Number of favourable cards}}{\text{Total number of cards}}
= \frac{4}{5}
(ii) What is the probability that it is not an E?
Solution:
The cards that are not E are: P, A, C
Number of favourable cards= 3
P(Not E)= \frac{\text{Number of favourable cards}}{\text{Total number of cards}}
= \frac{3}{5}
*8. A game of chance consists of spinning an arrow (see Fig. 7.7.) which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 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?
(v) A multiple of 3?
Solution:
Sample space: S={1,2,3,4,5,6,7,8}
Total number of outcomes: n(S)=8
(i) Event: the arrow points at 8
E= {8}
Number of favourable outcomes=1
P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total number of total outcomes}}
= \frac{1}{8}
(ii) Event: the arrow points at an odd number
E= {1,3,5,7}
Number of favourable outcomes=4
P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total number of total outcomes}}
= \frac{4}{8}=\frac{1}{2}
(iii) Event: the arrow points at a number greater than 2
E= {3,4,5,6,7,8}
Number of favourable outcomes=6
P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total number of total outcomes}}
= \frac{6}{8}=\frac{3}{4}
(iv) Event: the arrow points at a number less than 9
E= {1,2,3,4,5,6,7,8}
Number of favourable outcomes=8
P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total number of total outcomes}}
= \frac{8}{8}=1
(v) Event: the arrow points at a multiple of 3
E= {3,6}
Number of favourable outcomes=2
P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total number of total outcomes}}
= \frac{2}{8}=\frac{1}{4}
*10. A basket contains 4 red balls and 5 blue balls. One ball is drawn and laid aside, and a second ball is drawn. Draw a tree diagram to represent the possible outcomes and probabilities. Use the tree diagram to answer the following questions.
(i) What is the probability of drawing a red ball and then a blue ball?
(ii) What is the probability of drawing 2 blue balls?
*11. I throw a pair of 6-sided dice. Write down an event that has a probability of 0 and an outcome that has a probability of 1.
Solution:
Event with probability 0:
Getting a sum of 13.
(The maximum possible sum is 12 on two dice, so this can never happen.)
Event with probability 1:
Getting a sum between 2 and 12 (inclusive)
(Every possible throw of two dice gives a sum from 2 to 12, so this always happens.)
*12. Write the sample space and calculate the probability based on the given information.
(i) Two dice are rolled. What is the probability that the sum is a prime number greater than 5?
Solution:
Sample Space: S=\begin{Bmatrix}(1,1)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6)\\(2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\(3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6)\\(4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6)\\(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)\end{Bmatrix}
Total favourable outcomes = 8
P(sum is a prime number greater than 5)= \frac{\text{Number of favourable outcomes}}{\text{Total outcomes}}
=\frac{8}{36}=\frac{2}{9}
(ii) A bag contains 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours?
Solution:
Sample Space:
Total balls= 4 Red + 3 Green + 2 Blue = 9 balls
The first ball is drawn from the 9 balls and second ball is drawn from remaining 8 balls.
So, The total outcomes are: 9×8=72
Favourable Outcomes
The ordered pairs where the two balls are of different colours:
Red then green=4×3=12
Green then red=3×4=12
Red then blue=4×2=8
Blue then red=2×4=8
Green then blue=3×2=6
Blue then green=2×3=6
Total favourable outcomes= 12+12+8+8+6+6=52
P(different colours)= \frac{\text{Favourable outcomes}}{\text{Total outcomes}}
=\frac{52}{72}=\frac{13}{18}
(iii) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?
Solution:
Sample Space: S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}
Total outcomes = 8
Event: The first coin shows heads and exactly two heads occur in total
E= {HHT,HTH}
Favourable outcomes = 2
P(E)=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}
=\frac{2}{8}=\frac{1}{4}
(iv) A four-digit number is formed using the digits 1, 2, 3, and 4 with no repetition. What is the probability that the number is even?
Solution:
Sample space:
Total number of possible four-digit numbers= 4×3×2×1=24
Favourable outcomes:
For an even number, the last digit must be 2 or 4.
Case 1: Last digit = 2
Remaining digits are 1,3,4
Number of ways to arrange them = 3×2×1=6
Case 2: Last digit = 4
Remaining digits are 1,2,3
Number of ways to arrange them = 3×2×1=6
Favourable outcomes= 6+6=12
P(even number)=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}
=\frac{12}{24}=\frac{1}{2}
(v) A student takes a multiple-choice test with 3 questions, each having 4 options (A, B, C, D), with only one correct answer. What is the probability that the student guesses and gets exactly 2 answers correct?
Solution:
Sample space:
Total possible answer patterns: 4×4×4=64
Probability of guessing right(R) answer=\frac{1}{4}
Probability of guessing wrong(W) answer=\frac{3}{4}
Favourable outcomes:
Exactly two answers are correct: (R,R,W), (R,W,R), (W,R,R)
Probability of (R,R,W)=\frac{1}{4}×\frac{1}{4}×\frac{3}{4}=\frac{3}{64}
Probability of (R,W,R)=\frac{1}{4}×\frac{3}{4}×\frac{1}{4}=\frac{3}{64}
Probability of (W,R,R)=\frac{3}{4}×\frac{1}{4}×\frac{1}{4}=\frac{3}{64}
Required probability=Probability of (R,R,W), (R,W,R), (W,R,R)
=\frac{3}{64}×\frac{3}{64}×\frac{3}{64}
=\frac{9}{64}
*13. A box contains 4 balls numbered 1 to 4. Record a sample space using a tree diagram for the following experiments:
(i) A ball is drawn, and the number is recorded. Then the ball is returned, and a second ball is drawn and recorded.
(ii) A ball is drawn and recorded. Without replacing the first ball, the experimenter draws and records a second ball.
(iii) What are the sizes of these two sample spaces?
Solution:
*14. List the elements of a sample space for the simultaneous tossing of a coin and drawing of a card from a set of 6 cards numbered 1 through 6.
Solution:
Possible outcomes of tossing a coin= {H,T}
Possible outcomes of drawing a card= {1, 2, 3, 4, 5, 6}
Therefore, required sample space is:
S={(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}
*15. Three coins are tossed, and the number of heads is recorded. Which of the following lists is a sample space for this experiment? Why do the other lists fail to qualify as a sample space?
(i) {1, 2, 3}
(ii) {0, 1, 2}
(iii) {0, 1, 2, 3, 4}
(iv) {0, 1, 2, 3}
Solution:
When three coins are tossed, the possible numbers of heads are:
0 heads (TTT)
1 head (HTT, THT, TTH)
2 heads (HHT, HTH, THH)
3 heads (HHH)
So, the sample space: S={0,1,2,3}
Hence, option (iv) is the correct sample space.
(i) {1, 2, 3} It does not include the outcome 0 heads
(ii) {0, 1, 2} It does not include the outcome 3 heads
(iii) {0, 1, 2, 3, 4} 4 heads is impossible when only 3 coins are tossed.
(iv) {0, 1, 2, 3} This is the correct sample space because it includes all possible numbers of heads that can occur when three coins are tossed.
*16. Suppose you drop a dye at random on the rectangular region shown in Fig. 7.8. What is the probability that it will land inside the circle with a diameter of 1 m?
Solution:
Length of Rectangle= 3m
Breadth of rectangle= 2m
Area of rectangle= 3×2=6m^2
Diameter of circle= 1m
Radius of circle= \frac{1}{2}m
Area of circle= \pi r^2
=\pi \left(\frac{1}{2}\right)^2
=\frac{22}{7}×\frac{1}{4}=\frac{11}{14}m^2
P(inside the circle)=\frac{\text{Area of circle}}{\text{Area of rectangle}}
=\frac{\frac{11}{14}}{6}=\frac{11}{84}