The Mathematics of Maybe: Introduction to Probability Class 9 Maths Ganita Manjari Part 1 Chapter 7 NCERT Solutions

The Mathematics of Maybe: Introduction to Probability Class 9 Maths Ganita Manjari Part 1 Chapter 7 NCERT Solutions Looking for the The Mathematics of Maybe: Introduction to Probability Class 9 Maths Ganita Manjari Part 1 Chapter 7 NCERT Solutions? You are in the right place. This chapter introduces students to the basics of probability, helping them understand how to calculate the chances of different events using simple and practical examples.

Exercise Set 7.1 (Page 159)

1. Rank the following events on a scale from 0 (Impossible) to 1 (Certain). Label each event: Impossible, less likely, equally likely (even chance), more likely, certain. Give reasons why you gave each event its ranking.
(i) The next Monday will come after Sunday.
(ii) It will snow in Mumbai in July.
(iii) An elephant will walk through your classroom today.
(iv) You will greet at least one friend at school tomorrow.
Solution:

Exercise Set 7.2 (Page 165 – 166)

1. A teacher mixes a large bag of sweets of different colours and randomly selects a sample of 30 sweets. She counts the number of sweets of each colour:
10 red sweets | 8 green sweets | 7 yellow sweets | 5 blue sweets
(i) Calculate the probability that a randomly picked sweet from the sample is green.
(ii) If there are 600 sweets in total in the large bag, estimate how many are likely to be yellow, based on the sample results.
Solution:
(i) Number of green sweets
Total sweets in sample = 30
P (green sweet) = $\frac{N u m b e r o f g r e e n s w e e t s}{T o t a l n u m b e r o f s w e e t s}$ = $\frac{8}{30}$ = $\frac{4}{15}$ = 0.2666… ≈ 0.267
So, the probability is 0.267 or 26.7%.

(ii) Number of yellow sweets = 7
Total sweets in sample = 30
P (yellow sweet) = $\frac{N u m b e r o f y e l l o w s w e e t s}{T o t a l n u m b e r o f s w e e t s}$ = $\frac{7}{30}$ = 0.2333…
Total sweets in the large bag = 600
Estimated number of yellow sweets in 600 sweets = $\frac{7}{30}$ × 600 = 7 × 20 = 140.
Therefore, approximately 140 yellow sweets are likely to be in the large bag.

2. A survey is conducted at a school where a random sample of 40 students is asked about their favourite club. The responses are: 14 students: Science Club | 11 students: Arts Club | 9 students: Sports Club | 6 students: Debate Club Assume there are 800 students in the whole school.
(i) What is the probability that a randomly chosen student from the sample prefers the Arts Club?
(ii) Using the sample results, estimate how many students in the whole school are likely to prefer the Sports Club.
Solution:
Number of students preferring Science Club = 14
Number of students preferring Arts Club = 11
Number of students preferring Sports Club = 9
Number of students preferring Debate Club = 6
Total number of students surveyed = 14 + 11 + 9 + 6 = 40
(i) P(Arts Club) = $\frac{N u m b e r o f s t u d e n t s p r e f e r r i n g A r t s C l u b}{T o t a l n u m b e r o f s t u d e n t s s u r v e y e d}$ = $\frac{11}{40}$ = 0.275
Therefore, the probability that a randomly chosen student prefers the Arts Club is 0.275

(ii) P (Sports Club) = $\frac{N u m b e r o f s t u d e n t s p r e f e r r i n g S p o r t s C l u b}{T o t a l n u m b e r o f s t u d e n t s s u r v e y e d}$ = $\frac{9}{40}$ = 0.225
Total number of students in the school = 800
Estimated number of students preferring Sports Club = $\frac{9}{40}$ × 800 = 9 × 20 = 180.
Therefore, approximately 180 students are likely to prefer the Sports Club in the whole school.

3. Toss a coin 20 times and record the result each time (heads or tails).
(i) How many times did you get heads?
(ii) How many times did you get tails?
(iii) Calculate the experimental probability of getting heads.
(iv) If you toss the coin once more, what is the probability of getting tails?
Solution:
(i) Number of times heads appeared = 12
(ii) Number of times tails appeared = 8
(iii) P (Heads) = $\frac{N u m b e r o f t i m e s h e a d s a p p e a r e d}{T o t a l n u m b e r o f t o s s e s}$ = $\frac{12}{20}$ = $\frac{3}{5}$
Therefore, the experimental probability of getting heads is $\frac{3}{5}$ or 0.6.

(iv) Since a fair coin has two equally likely outcomes (Head or Tail),
P (Tails) = $\frac{1}{2}$
Therefore, the probability of getting tails is $\frac{1}{2}$ .

4. Toss a paper cup into the air 100 times. After each toss record whether the cup lands on its bottom, upside down on its top or on its side (See Fig. 7.5). Assign probabilities to the outcomes by using experimental probability.

Solution:
Let us assume the following results after tossing the paper cup 100 times:

Experimental probability = $\frac{N u m b e r o f t i m e s e v e n t o c c u r s}{T o t a l n u m b e r o f t r a i l s}$
(i) P (bottom) = $\frac{45}{100}$ = $\frac{9}{20}$
(ii) P (top) = $\frac{20}{100}$ = $\frac{1}{5}$
(iii) P (side) = $\frac{35}{100}$ = $\frac{7}{20}$

5. What is the probability of getting an even number when rolling a fair 6-sided die?
Solution:
A fair 6-sided die has the numbers: 1, 2, 3, 4, 5, 6
Even numbers on the die = 2, 4, 6
Number of favourable outcomes = 3
Total number of possible outcomes = 6
P (getting an even number) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{3}{6}$ = $\frac{1}{2}$ = 0.5.
Therefore, the probability of getting an even number is $\frac{1}{2}$ or 0.5.

6. Suppose you roll a 6-sided die 12 times and get a ‘3’ three times.
(i) What is the experimental probability of rolling a ‘3’?
(ii) What is the theoretical probability of rolling a ‘3’?
(iii) Why might these probabilities be different? What would you expect to happen if you roll the die 60, 600, or 6000 times?
Solution:
Number of times the die was rolled = 12
Number of times ‘3’ appeared = 3
(i) Experimental probability = $\frac{N u m b e r o f t i m e s e v e n t o c c u r s}{T o t a l n u m b e r o f t r i a l s}$
P (rolling a 3) = $\frac{3}{12}$ = $\frac{1}{4}$
Therefore, the experimental probability of rolling a ‘3’ is $\frac{1}{4}$ .

(ii) Outcomes of a fair 6-sided die = 1, 2, 3, 4, 5, 6
Number of favourable outcomes = 1 (only one face has 3)
Total number of possible outcomes = 6
Therotical probability = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$
P (rolling a 3) = $\frac{1}{6}$
Therefore, the theoretical probability of rolling a ‘3’ is $\frac{1}{6}$ .

(iii) The experimental probability may differ from the theoretical probability because the die was rolled only a small number of times (12 rolls). Random variation can cause different results in a small sample.
If the die is rolled 60, 600, or 6000 times, the experimental probability is expected to get closer to the theoretical probability $(\frac{1}{2})$ because a larger number of trials gives more reliable results.

Exercise Set 7.3 (Page 167 – 168)

1. When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?
Solution:
Experiment: Rolling a single 6-sided die once.
Sample Space: S = {1, 2, 3, 4, 5, 6}
Sample Size: n(S) = 6
Therefore, the the total number of outcomes in the sample space is 6.

2. For the following experiments write down the sample space S.
(i) Rolling a die and tossing a coin together.
(ii) Choosing a random integer between – 5 and + 5.
(iii) A box containing 5 green and 7 red balls. One ball is drawn at random.
Solution:
(i) Outcomes of a die: {1, 2, 3, 4, 5, 6}
Outcomes of a coin: {H, T}
Experiment: Rolling a die and tossing a coin together.
Sample Space: S = {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T), (5, T), (6, T)}.

(ii) Experiment: Choosing a random integer between – 5 and + 5.
Sample Space: S = {−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5}.

(iii) Experiment: Drawing randomly a ball from a box containing 5 green and 7 red balls.
Sample Space: S = {Green, Red}.

3. In a village fair, there are 3 popular snacks available: Samosa, Pakora, and Bhaji. For drinks, villagers can choose either Chai or Lassi.
(i) List the sample space of all possible snack and drink combinations a person could choose at the fair.
(ii) List the event ‘Selecting Samosa as a snack.’
Solution:
Snacks: Samosa, Pakora, Bhaji
Drinks: Chai, Lassi
(i) Experiment: Choosing all possible combinations of snack and drink at the fair.
Sample Space: S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}.

(ii) Event: ‘Selecting Samosa as a snack.’
E = {(Samosa, Chai), (Samosa, Lassi)}

Exercise Set 7.4 (Page 169)

1. There are two fruit baskets A and B. Basket A has one apple and two oranges. Basket B has one banana and one mango. You randomly pick one fruit from each basket.
(i) Draw a tree diagram showing all possible pairs of fruits.
(ii) List the sample space.
(iii) What is the probability of picking one apple and one banana?
Solution:
(i) Tree Diagram:

(ii) Sample Space: S = { (Apple, Banana), (Apple, Mango), (Orange₁, Banana), (Orange₁, Mango), (Orange₂, Banana), (Orange₂, Mango) }

(iii) Favourable outcome (Apple, Banana) = 1 outcome
P (Apple and Banana) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{1}{6}$ .

2. Let us say that you have a box containing 3 red pens, 4 black pens and 2 green pens. You pick a pen (without looking) from the box and put it back. Then your friend does the same.
(i) What are the possible outcomes of the pen colours? Can you draw a tree diagram representing the possible outcomes?
(ii) Can you use the tree diagram to guess the probability that both you and your friend pick pens of the same colour?
Solution:
(i) Since both you and your friend can pick a red (R), black (B), or green (G) pen, the possible outcomes are:
Sample space: S = {(R, R), (R, B), (R, G), (B, R), (B, B), (B, G), (G, R), (G, B), (G, G)}

(ii) Event: ‘Both you and your friend pick pens of the same color’.
E = {(R, R), (B, B), (G, G)}
P(E) = P(R, R) + P(B, B) + P(G, G)
= $(\frac{3}{9} \times \frac{3}{9}) + (\frac{4}{9} \times \frac{4}{9}) + (\frac{2}{9} \times \frac{2}{9})$
= $\frac{9}{81}$ + $\frac{16}{81}$ + $\frac{4}{81}$
= $\frac{9 + 16 + 4}{81}$ = $\frac{29}{81}$ ≈ 0.358 or 35.8%
So there is roughly a 36% chance that both of you pick a pen of the same colour.

End Of Chapter Exercises (Page 169 – 173)

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.
Solution:
(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 12 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, and the relative frequency is:
$\frac{15}{30}$ = $\frac{3}{10}$ = 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:
(i) A driver attempts to start a car. The car starts or does not start
Not equally likely because the chances depend on the condition of the car.

(ii) Tossing a fair coin once
Equally likely, because getting heads or tails has the same probability, $\frac{1}{2}$ .

(iii) Rolling a fair 6-sided die
Equally likely, because each outcome (1, 2, 3, 4, 5, 6) has the same probability, $\frac{1}{6}$ .

(iv) Choosing a marble randomly from a bag that contains 3 red marbles and 7 blue marbles
Not equally likely, because the chance of getting a blue marble $(\frac{7}{10})$ is more than the chance of getting a red marble $(\frac{3}{10})$ .

(v) A baby is born. It is a boy or a girl
Equally likely, because the probability of a boy or a girl is generally equal ( $\frac{1}{2}$ ).

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}
Event: ‘Getting atleast one head’. E = {HH, HT, TH}
Number of favourable outcomes = 3
Total number of outcomes = 4
P(E) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{3}{4}$ .

(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}
Event: ‘Getting an even number’. E = {2, 4, 6, 8, 10}
Number of favourable outcomes = 5
Total number of outcomes = 10
P(E) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{5}{10}$ = $\frac{1}{2}$ .

(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}
Event: ‘Getting a number greater than 4’. E = {5, 6}
Number of favourable outcomes = 2
Total number of outcomes = 6
P(E) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{2}{6}$ = $\frac{1}{3}$ .

(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 = {R, R, R, B, B, G}
Event: ‘Picking a ball that it is not red’. E = {B, B, G}
Number of favourable outcomes = 3
Total number of outcomes = 6
P(E) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{3}{6}$ = $\frac{1}{2}$ .

(v) Three coins are tossed simultaneously. What is the probability of getting exactly two heads?
Solution:
Sample Space: S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Event: ‘Getting exactly two heads’. E = {HHT, HTH, THH}
Number of favourable outcomes = 3
Total number of outcomes = 8
P(E) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{3}{8}$ .

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:
Experiment: Picking a strawberry candy
Sample Space: S = {strawberry, lemon, mint}
Event: ‘Picking a stawberry candy’. E = {strawberry}
Number of favourable outcomes = 1
Total number of outcomes = 3
P(E) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{1}{3}$ .
Hence, the probability of picking a strawberry candy is $\frac{1}{3}$ .

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 combinations of outfits are shown below:

Hence, the total number of possible outfits is 6.

7. A tyre company records distances before replacement in 1000 cases.

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.
(i) Number of cases less than 4000 km = 20
P(less than 4000 km) = $\frac{20}{1000}$ = $\frac{1}{50}$ .

(ii) Number of cases between 4000 km and 14000 km = 210 + 325 = 535.
P(between 4000 and 14000 km) = $\frac{535}{1000}$ = $\frac{107}{200}$ .

(iii) Number of cases more than 14000 km = 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?

(ii) What is the probability that it is not an E?
Solution:
Sample space: S = {P, E, A, C, E}
Total number of cards = 5
(i) Event: ‘Drawing a P, E or C’. E = {P, E, E, C}
Number of favourable outcomes = 4
P(E) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{4}{5}$ .

(ii) Event(F): ‘Drawing a card that is not an E’. F = {P, A, C}
Number of favourable outcomes = 3
P(F) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{3}{5}$ .

9. 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 = 8
(i) Event: ‘Arrow Pointing at 8’. E = {8}
Number of favourable outcomes = 1
P(E) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{1}{8}$ .

(ii) Event: ‘Arrow Pointing at an odd number’. E = {1, 3, 5, 7}
Number of favourable outcomes = 4
P(E) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{4}{8}$ = $\frac{1}{2}$ .

(iii) Event: ‘Arrow Pointing at a number greater than 2’. E = {3, 4, 5, 6, 7, 8}
Number of favourable outcomes = 6
P(E) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{6}{8}$ = $\frac{3}{4}$ .

(iv) Event: ‘Arrow Pointing at a number less than 9’. E = {1, 2, 3, 4, 5, 6, 7, 8}
Number of favourable outcomes = 8
P(E) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{8}{8}$ = 1.

(v) Event: ‘Arrow Pointing at a multiple of 3’. E = {3, 6}
Number of favourable outcomes = 2
P(E) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{2}{8}$ = $\frac{1}{4}$ .

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?
Solution:
The basket starts with 9 balls total (4 red, 5 blue). Since the first ball is laid aside (not replaced), the second draw happens from only 8 remaining balls.

Possible outcomes = {(R,R), (R,B), (B,R), (B,B)}
(i) P(drawing a red ball and then a blue ball) = $\frac{4}{9}$ × $\frac{5}{8}$ = $\frac{20}{72}$ = $\frac{5}{18}$ .

(ii) P(drawing two blue balls) = $\frac{5}{9}$ × $\frac{4}{8}$ = $\frac{20}{72}$ = $\frac{5}{18}$ .

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:
When a pair of 6-sided dice is thrown, the possible sums range from 2 to 12.
An event with probability 0 (impossible event):
Getting a sum of 13, since this cannot happen.
An event with probability 1 (certain event):
Getting a sum less than 13, since every possible outcome has a sum from 2 to 12.

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?
(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?
(iii) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?
(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?
(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:
(i) Sample Space: S = {(1, 1), (1, 2), (1, 3) ………. (6, 5), (6, 6)}
Total number of outcome = 36
Prime numbers greater than 5 = 7, 11
Favourable outcome for sum 7 = (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
Favourable outcome for sum 11 = (5, 6), (6, 5)
Total favourable outcomes = 8
P(E) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{8}{36}$ = $\frac{2}{9}$ .

(ii) Sample Space: S = {RR ,RG, RB, GR, GG, GB, BR, BG, BB}
Total balls = 9
P(same color) + P(different colors) = 1
P(different colors) = 1 – P(same color)
P(same color) = $\frac{4}{9}$ × $\frac{3}{8}$ + $\frac{3}{9}$ × $\frac{2}{8}$ + $\frac{2}{9}$ × $\frac{1}{8}$
= $\frac{12}{72}$ + $\frac{6}{72}$ + $\frac{2}{72}$ = $\frac{12 + 6 + 2}{72}$ = $\frac{20}{72}$ = $\frac{5}{18}$
P(different colors) = 1 – $\frac{5}{18}$ = $\frac{18 – 5}{18}$ = $\frac{13}{18}$ .

(iii) Sample Space: S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Event: ‘first coin is head and exactly two heads occur’. E = {{HHT, HTH}
Number of favourable outcomes = 2
Total possible outcomes = 8
P(E) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{2}{8}$ = $\frac{1}{4}$ .

(iv) Sample space: S = All arrangements of {1, 2, 3, 4} = 4! = 24 numbers
For the number to be even, the last digit must be 2 or 4.
If last digit = 2, then remaining digits can be arranged in 3! = 6 ways.
If last digit = 4, then remaining digits can be arranged in 3! = 6 ways.
Total favourable outcomes = 6 + 6 = 12
P(E) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s}$ = $\frac{12}{24}$ = $\frac{1}{2}$ .

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:
(i)

Sample space: S = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}

(ii)

Sample space: S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

(iii) For experiment (i), n(S) = 16
For experiment (ii), n(S) = 12
Hence, the sizes of the sample spaces are 16 and 12, respectively.

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:

Sample Space: 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)}
Hence, the sample space contains 12 outcomes.

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:
Three coins are tossed, and the number of heads is recorded.
The possible numbers of heads = 0, 1, 2, 3.
Hence, the sample space, S = {0, 1, 2, 3}
Therefore, (iv) {0, 1, 2, 3} is the correct sample space.
(i) {1, 2, 3} is not a sample space because 0 heads is possible.
(ii) {0, 1, 2} is not a sample space because 3 heads is possible.
(iii) {0, 1, 2, 3, 4} is not a sample space because 4 heads is not possible when only 3 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:
Area of the rectangle = 3 m × 2 m = 6 m²
Diameter of the circle = 1 m
So, radius of the circle = $\frac{1}{2}$ m
Area of the circle = $π r^{2}$
= $\frac{22}{7}$ × $\frac{1}{2}$ × $\frac{1}{2}$
= $\frac{22}{28}$ = $\frac{11}{14}$ m²
P(dye landing inside the circle) = $\frac{A r e a o f t h e c i r c l e}{A r e a o f t h e r e c t a n g l e}$ = $\frac{11 / 14}{6}$ = $\frac{11}{84}$
Hence, the required probability is $\frac{11}{84}$ .
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