NCERT Solutions Class 9 Mathematics Ganita Manjari Chapter 07 The Mathematics of Maybe: Introduction to Probability

Question 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.
Answer:
(i) Certain (Probability = 1) - The days of the week always follow a set order. Sunday is consistently followed by Monday, so this outcome will happen without fail, giving it a probability of 1.

(ii) Impossible (Probability = 0) - Mumbai experiences a tropical climate and faces the monsoon season in July with warm and moist conditions. Snow cannot fall there, making this outcome impossible with a probability of 0.

(iii) Impossible (Probability = 0) - Under typical circumstances, an elephant cannot enter a school classroom. This outcome practically cannot occur, so its probability is 0 or extremely near to 0.

(iv) More likely (Probability is close to 1, but not exactly 1) - Schools bring together many friends. The chances of greeting at least one friend are very strong, making this outcome "more likely." It is not "certain" because rare circumstances (such as being absent or the school being shut) could prevent it, though normally it will take place.
In simple words: Some things always happen (certain), some never happen (impossible), and some might or might not happen (likely or unlikely). We give them numbers from 0 to 1 to show how sure we are.

Exam Tip: When ranking events, be clear about whether the event is controlled by nature, personal choice, or time - and use real reasons from the world around you, not just opinions.

Question 2. 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, 8 green, 7 yellow, 5 blue. (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.
Answer:
(i) In the sample, there are 8 green sweets and 30 sweets total. So the probability is 8 divided by 30, which simplifies to 4/15 or about 0.267.

(ii) In the sample, 7 out of 30 sweets are yellow. Using this ratio on the full bag: (7/30) multiplied by 600 equals 7 times 20, which gives 140. So we would expect roughly 140 yellow sweets in the entire bag.
In simple words: In the sample, some sweets are green and some are yellow. We find what fraction they are, then use that same fraction to guess how many would be that colour in the whole bag.

Exam Tip: Always simplify fractions to lowest terms and show your multiplication step by step - examiners check both process and final answer.

Question 3. 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.
Answer:
(i) Of the 40 students surveyed, 11 prefer the Arts Club. Dividing 11 by 40 gives a probability of 0.275.

(ii) In the sample, 9 out of 40 students favour the Sports Club. Applying this fraction to the full school population: (9/40) multiplied by 800 gives 9 times 20, which equals 180. We estimate that about 180 students across the school would prefer the Sports Club.
In simple words: We ask some students what they like. Then we use those answers to guess what all students in the school would choose.

Exam Tip: When estimating for a larger population, always multiply the sample probability by the population size - this is called statistical projection.

Question 4. 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?
Answer:
(i) In this experiment, heads appeared 11 times.

(ii) Tails appeared 9 times.

(iii) The experimental probability is the number of heads divided by the total tosses: 11 divided by 20 equals 0.55.

(iv) This next toss is a fresh, independent event. For a fair coin, the theoretical probability of getting tails on any single toss is always 1/2, regardless of earlier results.
In simple words: When you toss a coin many times, you count how often heads shows up. Each new toss is fresh - what happened before does not change the next toss.

Exam Tip: Remember that experimental probability (based on what you actually observed) can differ from theoretical probability (what should happen in theory) - especially with small numbers of trials.

Question 5. 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. Assign probabilities to the outcomes by using experimental probability.
Answer: This is a hands-on activity. From one experiment, the results were: bottom landed 35 times, top landed 15 times, side landed 50 times. Using these observations:

P(bottom) = 35 divided by 100 = 35/100

P(top) = 15 divided by 100 = 15/100

P(side) = 50 divided by 100 = 50/100
In simple words: Drop a cup 100 times and count how it lands. Each landing position has a probability based on how many times it actually happened.

Exam Tip: Conduct the experiment carefully and record each result accurately - experimental probability depends entirely on good data collection.

Question 6. What is the probability of getting an even number when rolling a fair 6-sided die?
Answer: The set of all possible results is {1, 2, 3, 4, 5, 6}. The even numbers are {2, 4, 6}, giving us 3 favourable outcomes. Therefore, the probability of rolling an even number is 3 divided by 6, which equals 1/2.
In simple words: Half of the numbers on a die are even (2, 4, and 6), so the chance of rolling an even number is half or 0.5.

Exam Tip: Always list all possible outcomes and all favourable outcomes clearly - this prevents counting errors.

Question 7. 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?
Answer:
(i) From the actual rolling, you got a 3 three times out of 12 rolls. The experimental probability is 3/12, which simplifies to 1/4.

(ii) Using mathematical reasoning, each face of a fair die has an equal chance. The theoretical probability of rolling a 3 is 1/6.

(iii) These two probabilities differ because experimental results depend on chance variations in a limited number of trials. With only 12 rolls, randomness can cause bigger swings from the theoretical value. A perfectly fair die assumes all outcomes are equally likely. However, as you increase the number of trials, your observed results move closer to theory: At 60 rolls, the experimental probability will be nearer to 1/6 but may still show some variation. At 600 rolls, it will be very near to 1/6. At 6000 rolls, it will be almost exactly 1/6. This pattern is the Law of Large Numbers - the more trials you run, the more your experimental results approach the theoretical value.
In simple words: Sometimes you get lucky or unlucky in a small number of tries. But if you keep trying many more times, your results start to match what the math says should happen.

Exam Tip: Understand the Law of Large Numbers thoroughly - examiners often ask why experimental and theoretical probabilities differ and how sample size affects the gap.

Question 8. When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?
Answer: When you roll a 6-sided die, the possible outcomes are 1, 2, 3, 4, 5, 6. This set, called the sample space S, is written as {1, 2, 3, 4, 5, 6}. The total count of outcomes is 6, written as n(S) = 6.
In simple words: A die has 6 faces, so there are 6 possible numbers it can show. That is the size of the sample space.

Exam Tip: Always write the sample space as a set and count its elements carefully - this is the foundation for all probability calculations.

Question 9. For the following experiments write down the sample space S. (i) Rolling a die and tossing a coin together.
Answer: The die gives outcomes 1, 2, 3, 4, 5, 6. The coin gives outcomes H (Heads) or T (Tails). Each final outcome pairs one die result with one coin result: S = {(1,H), (1,T), (2,H), (2,T), (3,H), (3,T), (4,H), (4,T), (5,H), (5,T), (6,H), (6,T)}. The total number of outcomes is n(S) = 12.
In simple words: When you do two things together, multiply their outcomes. A die has 6 results and a coin has 2, so together they have 6 times 2 equals 12 possible pairs.

Exam Tip: For compound experiments, list outcomes as ordered pairs and use the multiplication rule to find total outcomes without listing them all.

Question 10. For the following experiments write down the sample space S. (ii) Choosing a random integer between -5 and +5.
Answer: The integers that fall between -5 and +5, including the endpoints, are: -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5. Written as a set: S = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}. This sample space has n(S) = 11 elements.
In simple words: Count all the whole numbers from -5 to +5, including both ends. That gives you 11 numbers in total.

Exam Tip: When the question says "between" two values, always check whether the endpoints are included - the phrasing matters.

Question 11. For the following experiments write down the sample space S. (iii) A box containing 5 green and 7 red balls. One ball is drawn at random.
Answer:
Case 1: If balls are identical (only colour matters). There are two types: Green (G) and Red (R). The sample space is S = {Green ball, Red ball} or S = {G, R}. This gives n(S) = 2.

Case 2: If balls are not identical (each ball is distinct). If every ball is individually distinguishable, then S would contain 12 elements: G₁, G₂, G₃, G₄, G₅, R₁, R₂, R₃, R₄, R₅, R₆, R₇. However, at this level, the colour-based sample space (Case 1) is the standard approach used in probability calculations.
In simple words: If we only care about colour, a sample space has two outcomes - green or red. If each ball is different (like having serial numbers), there would be 12 different outcomes.

Exam Tip: Read the question carefully to determine whether objects are treated as identical (by type/colour only) or as distinct (individually labelled).

Question 12. 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.'
Answer:
(i) Each person picks one snack and one drink. The sample space lists all pairings: S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}. The total is n(S) = 6.

(ii) The event "Selecting Samosa as a snack" includes all pairs that have Samosa, regardless of the drink: E = {(Samosa, Chai), (Samosa, Lassi)}. This event has 2 outcomes. The probability of selecting Samosa is P(Samosa) = 2/6 = 1/3.
In simple words: You can match 3 snacks with 2 drinks, giving 6 combinations. If you want Samosa, it pairs with 2 drinks, so the chance is 2 out of 6.

Exam Tip: An event is a subset of the sample space - always identify which outcomes satisfy the event condition before counting.

Question 13. 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?
Answer:
(i) A tree diagram would show: First branch (Basket A) splits into Apple and Orange. From each, second branches (Basket B) split into Banana and Mango, giving four endpoints: (Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango).

(ii) The sample space lists all pairs: S = {(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango)}. There are 4 total outcomes.

(iii) The favourable outcome is (Apple, Banana), which is just 1 pair. With 4 total outcomes, the probability is 1/4.
In simple words: You pick one fruit from each basket. The tree shows all the ways you can do this. Only 1 way gives you an apple and banana together.

Exam Tip: Tree diagrams are visual tools that make listing outcomes easier - always match the branches to the experiment structure.

Question 14. 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?
Answer:
(i) The possible colours are Red (R), Black (B), and Green (G). Since you replace the pen before your friend picks, each person independently chooses from all three colours. The sample space includes all ordered pairs of colours: S = {(R,R), (R,B), (R,G), (B,R), (B,B), (B,G), (G,R), (G,B), (G,G)}. A tree diagram would show a first pick splitting into three branches (R, B, G), and from each, a second pick also splitting into three branches (R, B, G), creating 9 endpoints total.

(ii) The pairs where both picks are the same colour are: (R,R), (B,B), and (G,G). That is 3 favourable outcomes. With 9 total outcomes, the probability is 3/9, which simplifies to 1/3.
In simple words: You pick a pen and put it back. Your friend picks too. Both of you might pick the same colour or different colours. The chance of matching is 1 out of 3.

Exam Tip: With replacement, the second pick is independent of the first - the sample space is always n × n outcomes for two independent picks from n items.

Exercise Set 7.4

Question 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?
Answer:
(i) A tree diagram would show: First branch (Basket A) splits into Apple and Orange. From each, second branches (Basket B) split into Banana and Mango, creating four endpoints.

Possible pairs: (Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango)

(ii) The complete sample space is S = {(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango)}.

(iii) The outcome (Apple, Banana) represents 1 favourable outcome out of 4 total outcomes. The probability is therefore 1/4.
In simple words: You pick one fruit from each basket, making four possible combinations. Only one combination gives you an apple and a banana.

Exam Tip: For compound experiments with replacement, the sample size is the product of individual choices - here 3 × 2 = 6, but listing shows 4 because you have specific fruit counts.

Question 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 you 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 are your friend pick pens of the same colour?
Answer:
(i) The possible colours are Red (R), Black (B), and Green (G). The sample space showing all colour pairs is: S = {(R,R), (R,B), (R,G), (B,R), (B,B), (B,G), (G,R), (G,B), (G,G)}. A tree diagram starts with a first pick splitting into three colour branches, and from each branch, a second pick also splits into three colour branches, producing 9 total endpoints.

(ii) The outcomes where both picks match in colour are (R,R), (B,B), and (G,G) - a total of 3 favourable outcomes. Since there are 9 total outcomes, the probability is 3/9 = 1/3. This means the chance that both you and your friend pick the same colour is 1/3.
In simple words: Each of you picks a pen from three colours. There are 9 ways this can happen overall. Only 3 ways result in you both picking the same colour.

Exam Tip: Since the pen is returned, your second pick is completely independent - the probabilities for the second pick are identical to the first, making this a simple 3 × 3 grid.

End-of-Chapter Exercises

Question 1. Fill in the blanks. (i) The probability of an important 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.
Answer:
(i) The probability of an important 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 1/2 for getting heads.
In simple words: A certain event has probability 1, an impossible event has probability 0, and all sample spaces contain all possible outcomes of an experiment.

Exam Tip: These are key definitions - memorise them and use them consistently across all probability questions.

Question 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).
Answer: The number of students who like football is 15, and the relative frequency is 15/50 = 3/10 = 0.3.
In simple words: Frequency is just the count (15 students). Relative frequency is what fraction or percentage of all students that is (3 out of 10, or 30%).

Exam Tip: Relative frequency is always the part divided by the whole - it shows proportion, not just a raw count.

Question 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.
Answer:
(i) Not equally likely. The car starting depends on factors like the car's condition, fuel level, and battery strength. Under normal circumstances, the two outcomes (starts or does not start) are not equally likely.

(ii) Yes, equally likely. Heads and Tails each have probability 1/2. A fair coin treats both outcomes equally.

(iii) Yes, equally likely. Each of the 6 faces has an equal probability of 1/6 appearing.

(iv) Not equally likely. There are more blue marbles than red ones. P(red) = 3/10 and P(blue) = 7/10, so the two colours have different probabilities.
In simple words: Outcomes are equally likely when each has the same chance of happening. A fair coin or die is equally likely. A bag with different numbers of colours is not.

Exam Tip: Equal likelihood requires symmetry - all outcomes must have the same probability, which is common in fair games but rare in real-world situations.

Question 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?
Answer: When two coins are tossed, the sample space is S = {HH, HT, TH, TT}, so n(S) = 4. The event "at least one head" includes the outcomes {HH, HT, TH}, giving n(E) = 3. Therefore, P(at least one head) = 3/4.
In simple words: Out of four possible results when tossing two coins, three of them have at least one head.

Exam Tip: "At least one" means one or more - it includes cases with one head and cases with more than one head, but excludes the case with zero heads.

Question 5. Write the sample space and calculate the probability based on the given information. (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?
Answer: The sample space is S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, so n(S) = 10. The even numbers are {2, 4, 6, 8, 10}, giving n(E) = 5. Therefore, P(even number) = 5/10 = 1/2.
In simple words: Half of the numbers from 1 to 10 are even, so the chance of drawing an even number is 1 out of 2.

Exam Tip: When counting outcomes, list them systematically - this prevents missing elements or double-counting.

Question 6. Write the sample space and calculate the probability based on the given information. (iii) A die is rolled once. What is the probability of getting a number greater than 4?
Answer: The sample space is S = {1, 2, 3, 4, 5, 6}, so n(S) = 6. Numbers greater than 4 are {5, 6}, giving n(E) = 2. Therefore, P(number greater than 4) = 2/6 = 1/3.
In simple words: Only 2 out of 6 numbers on a die are bigger than 4, so the probability is 1 out of 3.

Exam Tip: Pay careful attention to the wording - "greater than 4" means 5 and 6, not including 4 itself.

Question 7. Write the sample space and calculate the probability based on the given information. (iv) A bag contain 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?
Answer: The total number of balls is 3 + 2 + 1 = 6. Balls that are not red are the blue and green ones: 2 + 1 = 3. Therefore, P(not red) = 3/6 = 1/2.
In simple words: Half the balls are red and half are not red, so the chance of not picking red is 1 out of 2.

Exam Tip: "Not red" includes all colours except red - count all non-red items, not just one specific colour.

Question 8. Write the sample space and calculate the probability based on the given information. (v) Three coins are tossed simultaneously. What is the probability of getting exactly two heads?
Answer: When three coins are tossed, the sample space is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, so n(S) = 8. Exactly two heads appear in {HHT, HTH, THH}, giving n(E) = 3. Therefore, P(exactly two heads) = 3/8.
In simple words: Out of eight possible results, only three have exactly two heads and one tail.

Exam Tip: "Exactly two" means precisely that number, not two or more - list all such outcomes carefully.

Question 9. A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?
Answer: The sample space is S = {strawberry, lemon, mint}, so n(S) = 3. Only one outcome is strawberry, so n(E) = 1. Therefore, P(strawberry) = 1/3.
In simple words: There are 3 different candies, and only 1 is strawberry, so the probability is 1 out of 3.

Exam Tip: Simple outcomes where there is one favourable result have probability = 1 divided by total outcomes.

Question 10. 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.
Answer:

The total number of possible combinations is 6 outfits. This comes from 2 shirt choices multiplied by 3 pant choices.
In simple words: With 2 shirts and 3 types of pants, you can make 2 times 3 equals 6 different outfits by pairing each shirt with each pant type.

Exam Tip: Use the multiplication principle for counting - if you have m choices for one category and n for another, total combinations = m × n.

Question 11. 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.
Answer:
(i) Out of 1000 tyres, 20 lasted less than 4000 km. The probability is P(Less than 4000 km) = 20/1000 = 1/50.

(ii) Tyres lasting between 4001 and 14000 km total 210 + 325 = 535 cases. The probability is P(Between 4000 and 14000 km) = 535/1000.

(iii) Out of 1000 tyres, 445 lasted more than 14000 km. The probability is P(More than 14000 km) = 445/1000.
In simple words: Probability is the number of tyres in each distance group divided by the total of 1000 tyres.

Exam Tip: For grouped data, identify the ranges carefully - "between 4000 and 14000" requires adding the two middle intervals.

Question 12. The letter 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?
Answer:
(i) The word PEACE has 5 letters total: P, E, A, C, E. The letters P, E, and C appear: P once, E twice, C once - totalling 4 cards. The probability is P(P, E or C) = 4/5.

(ii) The letter E appears twice. Cards that are not E total 5 - 2 = 3 (which are P, A, C). The probability is P(not E) = 3/5.
In simple words: Count how many letters match what you want, then divide by the total number of letters.

Exam Tip: When letters repeat, count all instances of each letter - here E appears twice, so there are really 5 letter cards, not 4 unique letters.

Question 13. 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?
Answer:
(i) Only one outcome is 8. With 8 total equally likely outcomes, P(8) = 1/8.

(ii) The odd numbers are {1, 3, 5, 7}, giving 4 favourable outcomes. P(odd) = 4/8 = 1/2.

(iii) Numbers greater than 2 are {3, 4, 5, 6, 7, 8}, giving 6 outcomes. P(number greater than 2) = 6/8 = 3/4.

(iv) All numbers from 1 to 8 are less than 9, so all 8 outcomes are favourable. P(number less than 9) = 8/8 = 1.

(v) Multiples of 3 in the range are {3, 6}, giving 2 outcomes. P(multiple of 3) = 2/8 = 1/4.
In simple words: Count the outcomes that match each condition, then divide by 8 to find the probability.

Exam Tip: A probability of 1 means the event is certain - always verify that all outcomes in the sample space satisfy the condition.

Question 14. A basket contains 4 red ball 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?
Answer:
The basket starts with 4 red and 5 blue balls, totalling 9. Since one ball is removed and not returned, the second draw is affected (without replacement). Possible outcomes are {(R,R), (R,B), (B,R), (B,B)}.

(i) To find P(Red then Blue): First, draw a red ball with probability 4/9. After removing it, 8 balls remain: 3 red and 5 blue. The probability of then drawing blue is 5/8. Multiplying these: P(R,B) = (4/9) × (5/8) = 20/72 = 5/18. So the probability of drawing red first and blue second is 5/18.

(ii) To find P(Blue then Blue): First, draw a blue ball with probability 5/9. After removing it, 8 balls remain: 4 red and 4 blue. The probability of drawing another blue ball is 4/8. Multiplying: P(B,B) = (5/9) × (4/8) = 20/72 = 5/18. So the probability of drawing two blue balls is also 5/18.
In simple words: Without putting the first ball back, the number of balls changes for the second draw. You multiply the probabilities of each step to get the final answer.

Exam Tip: For without-replacement problems, always recalculate the total and remaining items after each draw - the denominators change.

Question 15. 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.
Answer:
Event with probability 0 (Impossible): Getting a sum of 1. When rolling two dice, the minimum possible sum is 1 + 1 = 2, making a sum of 1 impossible. So P(sum = 1) = 0.

Outcome with probability 1 (Certain): Getting a sum that is less than 13. The maximum possible sum with two 6-sided dice is 6 + 6 = 12. Since 12 is always less than 13, this outcome is certain. So P(sum < 13) = 1.
In simple words: Impossible events have probability 0 because they can never happen. Certain events have probability 1 because they always happen.

Exam Tip: Understand the extremes: 0 means "impossible," 1 means "certain" - these anchor the entire probability scale.

Question 16. 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?
Answer: When two dice are rolled, each die shows 1 to 6. The sample space contains all ordered pairs: S = {(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)}. The total is n(S) = 36.

Prime numbers greater than 5, up to the maximum sum of 12, are 7 and 11.

Outcomes with sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) - that is 6 outcomes.

Outcomes with sum = 11: (5,6), (6,5) - that is 2 outcomes.

Total favourable = 6 + 2 = 8. Therefore, P(sum is a prime number greater than 5) = 8/36 = 2/9.
In simple words: Out of 36 possible outcomes when rolling two dice, 8 of them give a sum that is prime and bigger than 5.

Exam Tip: For dice problems, list all outcomes systematically by row (first die = 1, then 2, then 3, etc.) to avoid missing any.

Question 17. Write the sample space and calculate the probability based on the given information. (ii) A bag contain 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours?
Answer: The bag contains 4 + 3 + 2 = 9 balls. Two balls are drawn without replacement. The first ball can be chosen in 9 ways and the second in 8 ways, giving total outcomes = 9 × 8 = 72.

Favourable outcomes (different colours):

Red and Green: First red then green = 4 × 3 = 12, or first green then red = 3 × 4 = 12. Total = 24.

Red and Blue: First red then blue = 4 × 2 = 8, or first blue then red = 2 × 4 = 8. Total = 16.

Green and Blue: First green then blue = 3 × 2 = 6, or first blue then green = 2 × 3 = 6. Total = 12.

Total favourable outcomes = 24 + 16 + 12 = 52. Therefore, P(different colours) = 52/72 = 13/18.
In simple words: Different colours means the two balls are not the same colour. You count all ways to pick one colour first and another colour second.

Exam Tip: For "different" or "distinct" conditions, count both orders - first-second and second-first - as separate outcomes.

Question 18. Write the sample space and calculate the probability based on the given information. (iii) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?
Answer: When three coins are tossed, the sample space is S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, so n(S) = 8. We need outcomes where the first coin is H AND exactly two heads appear total. The outcomes satisfying this are {HHT, HTH}, giving n(E) = 2. Therefore, the probability is 2/8 = 1/4.
In simple words: The first coin must be heads. Among the remaining two coins, exactly one must be heads (so we have exactly two heads total).

Exam Tip: When a question has multiple conditions (e.g., "first coin shows heads AND exactly two heads total"), filter the sample space step by step.

Question 19. Write the sample space and calculate the probability based on the given information. (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?
Answer: Using the four distinct digits 1, 2, 3, 4 with no repetition, the total number of four-digit arrangements is 4 × 3 × 2 × 1 = 24.

For a number to be even, its last digit must be even. The even digits available are 2 and 4.

Case 1: Last digit = 2. The remaining digits {1, 3, 4} can be arranged in the first three positions in 3 × 2 × 1 = 6 ways.

Case 2: Last digit = 4. The remaining digits {1, 2, 3} can be arranged in the first three positions in 3 × 2 × 1 = 6 ways.

Total even numbers = 6 + 6 = 12. Therefore, P(even number) = 12/24 = 1/2.
In simple words: Half of the digits (2 and 4) are even. If the number ends in an even digit, it is even. That happens in exactly half of all arrangements.

Exam Tip: For permutation problems, identify the key constraint (here, the last digit must be even) and count arrangements satisfying it systematically by cases.

Question 20. Write the sample space and calculate the probability based on the given information. (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 answer correct?
Answer:
For each question: Probability of a correct guess = 1/4, and probability of an incorrect guess = 3/4. There are 3 questions total, and we want exactly 2 correct answers.

Ways to get exactly 2 correct: The student can get the 1st and 2nd correct (and 3rd wrong), or the 1st and 3rd correct (and 2nd wrong), or the 2nd and 3rd correct (and 1st wrong). That is 3 possible patterns: {(C, C, W), (C, W, C), (W, C, C)}.

Probability of each pattern: Each correct answer has probability 1/4, and each wrong answer has probability 3/4. So one pattern = (1/4) × (1/4) × (3/4) = 3/64.

Total probability: Since there are 3 such patterns, P(exactly 2 correct) = 3 × (3/64) = 9/64.
In simple words: Out of 3 questions, exactly 2 must be right and 1 must be wrong. There are 3 ways this can happen (which question is wrong?), and each way has the same probability.

Exam Tip: For "exactly k successes" problems, count all possible positions for the successes, multiply by the probability of one arrangement, then sum.

Question 21. 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?
Answer:
(i) With replacement: 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)}. The sample size is n(S) = 16.

(ii) Without replacement: 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)}. All pairs have two different numbers. The sample size is n(S) = 12.

(iii) With replacement: n(S) = 16. Without replacement: n(S) = 12.
In simple words: When you put the ball back, you can pick the same number again (16 pairs). When you don't put it back, all pairs must have two different numbers (12 pairs).

Exam Tip: With replacement: n × n outcomes. Without replacement: n(n-1) outcomes (each first choice has n-1 second choices).

Question 22. 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.
Answer: The coin gives outcomes H or T. The card gives outcomes 1, 2, 3, 4, 5, or 6. The combined sample space pairs each coin result with each card: 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)}. The sample size is n(S) = 12.
In simple words: The coin has 2 outcomes and the card has 6 outcomes. Pairing them together gives 2 times 6 equals 12 total outcomes.

Exam Tip: Use the multiplication principle: for independent experiments, total outcomes = (outcomes in experiment 1) × (outcomes in experiment 2).

Question 23. 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}
Answer:
(i) {1, 2, 3} - Not a valid sample space. This set is missing 0 heads (the outcome TTT, where all three coins show tails).

(ii) {0, 1, 2} - Not a valid sample space. This set misses 3 heads (the outcome HHH, where all three coins show heads).

(iii) {0, 1, 2, 3, 4} - Not a valid sample space. The value 4 cannot occur since only three coins are tossed. The maximum number of heads is 3.

(iv) {0, 1, 2, 3} - Yes, this is the correct sample space. When three coins are tossed, the possible counts of heads are 0 (all tails: TTT), 1 head, 2 heads, or 3 heads (all heads: HHH). This set includes all possible values.
In simple words: A sample space must include every possible outcome and exclude anything impossible. Three coins can give 0, 1, 2, or 3 heads - nothing more, nothing less.

Exam Tip: A valid sample space is both complete (no missing outcomes) and correct (no impossible outcomes) - check both properties before choosing.

Question 24. Suppose you drop a dye at random on the rectangular region shown in figure 7.8. What is the probability that it will land inside the circle with a diameter of 1 m?
Answer: The rectangle has dimensions 3 m by 2 m, so its area is 3 × 2 = 6 m². The circle has a diameter of 1 m, making its radius 0.5 m. The circle's area is π × r² = π × (0.5)² = π × 0.25 = π/4 m². The probability of the dye landing inside the circle is the ratio of areas: P = (π/4) divided by 6 = π/24.
In simple words: The dye is equally likely to land anywhere in the rectangle. The chance of hitting the circle is the area of the circle divided by the area of the rectangle.

Exam Tip: For geometric probability, probability = (area of favourable region) / (area of total region).

Class 9 Maths Chapter 7 Concepts: What is Probability and Why Does It Matter?

Probability is the mathematical study of how likely an event is to happen. Unlike most other math topics that give exact answers, probability deals with uncertainty - it tells you not what will definitely happen, but how confident you can be that something will occur. In daily life, from weather reports to sports predictions to school raffles, we face uncertain situations constantly. Probability gives us a precise, numerical way to discuss them. The chapter distinguishes between subjective probability (formed from personal belief, like "it seems cloudy, so it might rain") and objective probability (based on evidence or math). Chapter 7 aims to help students move away from guesses toward measuring likelihood in a way that is consistent, testable, and mathematical.

Probability is always a number from 0 to 1 (or equivalently from 0% to 100%).

• A random event is one where all possible outcomes are known in advance, but which outcome will actually occur cannot be predicted beforehand.
• Randomness is not chaos - when repeated many times, random events follow patterns, and this is what makes probability valuable.
• The probability scale acts like a number line: the nearer to 1, the more likely the event; the nearer to 0, the less likely.

Class 9 Ganita Manjari Chapter 7 Learning Concepts: Experimental and Theoretical Probability

Chapter 7 introduces two major, complementary approaches to measuring probability. Experimental probability comes from real data - you perform an experiment repeatedly or examine historical records and count how often an event actually happens. Theoretical probability, by contrast, uses logic and mathematics to work out what should happen in an ideal, perfectly fair scenario where all outcomes have equal chances. These are not rivals; they work together. Experimental probability reflects the messy real world with all its quirks, while theoretical probability shows the clean mathematical ideal. Ganita Manjari chapter 7 also brings in a third tool - statistical sampling - where information gathered from a small, representative group predicts likelihood across a much larger population. This is how companies forecast sales, researchers gauge public views, and schools might estimate what all students prefer without surveying everyone.

• The Law of Large Numbers tells us that as you run an experiment more and more times, the experimental probability gets closer and closer to the theoretical value.
• Gambler's Fallacy is the false belief that past random results change future ones - a coin that has landed Heads six times in a row still has exactly a 1/2 chance on the next flip.
• A coin, die, or pick process is fair or unbiased when all outcomes have equal probability and nothing is favoured.
• Statistical estimates become stronger when the sample is both large and truly representative of the whole population.

Class 9 Ganita Manjari Chapter 7 Problem Solving: Sample Spaces and Events

Every probability question starts by defining the sample space clearly - the complete set of all possible outcomes of an experiment, shown as S = { }. The count of elements in this set is the sample size, written n(S). An event is then any specific outcome or group of outcomes you care about - formally, it is a subset of the sample space. Probability of an event is P(E) = n(E) ÷ n(S), where n(E) is the count of outcomes matching the event's rule. The most frequent error students make is an incomplete sample space - for example, writing only {HH, TT} when tossing two coins, missing {HT, TH}. A correct sample space must contain every outcome with no duplicates, and must reflect the level of detail the question asks for.

• The Law of Large Numbers shows that as trial count goes up, experimental probability converges to theoretical probability.
• Gambler's Fallacy is the mistaken idea that past random events influence future ones - each toss is fresh and independent.
• A coin, die, or selection is fair or unbiased when all outcomes are equally likely with no advantage to any outcome.
• Statistical projections are more robust when both the sample size is large and the sample truly mirrors the full population.

Probability: Key Concepts and Exam Strategy

An event E will always exist within the sample space S. To find its probability, apply the formula P(E) = n(E) ÷ n(S).

Before you start any probability calculation, always list out the entire sample space first. This step keeps you from overlooking outcomes and can earn you partial credit on exams even if your final answer is slightly off.

The sample space you write must fit the specific question being asked. For a straightforward yes-or-no question about weather, {Rain, No Rain} is enough. However, if the question asks about how much rain falls, you will need a more detailed set like {No Rain, Drizzle, Light Rain, Heavy Rain}.

Remember that 0 ≤ P(E) ≤ 1 is always true. No probability can ever be a negative number or a number larger than 1.

Tree Diagrams, Exam Strategy and Key Takeaways

When a problem asks you to work through two or more steps one after another, a tree diagram becomes your most useful tool for finding every possible result and making sure you do not leave any out. You start at a single point and draw branches for each outcome of the first step. Then from each of those branches, you draw more branches for the second step - and you keep going this way for each step that follows. Every path that goes from the starting point all the way out to the end of a branch shows one complete outcome in the sample space.

Tree diagrams do much more than just list outcomes. They also help you see probability calculations in a visual way and let you follow your working step by step. This is especially helpful when you are solving multi-step problems that involve picking items with or without putting them back. Chapter 7 teaches tree diagrams by looking at experiments such as flipping a coin two times, choosing from several baskets, and drawing coloured balls one after another - all cases where you need to work through the problem carefully and systematically. Learning to use tree diagrams well in Class 9 sets you up for harder probability work in later classes.

Working Strategy for Probability Problems

• Write out your steps clearly on paper: first state what n(E) equals, then write what n(S) equals, then show P(E) = ?/?. Do not skip straight to your final answer without showing these steps.
• When you draw a tree diagram, make sure every branch has a label. Read through each path from start to finish very carefully so you can build the correct sample space from it.
• A common exam question asks you to explain the difference between experimental probability (what actually happens when you try something) and theoretical probability (what should happen based on maths). Many students should prepare a short written answer for this topic because it comes up often in exams.
• Chapter 7 asks you to truly understand how probability works, not just remember formulas. Students who can give reasons for why probability behaves the way it does almost always do better in exams than students who just memorise the formulas without understanding them.