CBSE Class 9 Maths Ganita Manjari Part 1 Ch 07 The Mathematics of Maybe Introduction to Probability Long Answer and Case Study Questions with Answers

5 Mark Questions

Question 1. At an airport, 200 passengers are checked: 90 carry handbags, 70 carry backpacks and 40 carry suitcases. If one passenger is chosen at random, find the probability that the passenger is:
(i) carrying a handbag
(ii) carrying a backpack
(iii) carrying a suitcase
(iv) not carrying a suitcase
(v) carrying either a handbag or a backpack
Answer:
(i) \( \frac{90}{200} = \frac{9}{20} \)
(ii) \( \frac{70}{200} = \frac{7}{20} \)
(iii) \( \frac{40}{200} = \frac{1}{5} \)
(iv) \( 1 - \frac{1}{5} = \frac{4}{5} \)
(v) \( \frac{90 + 70}{200} = \frac{160}{200} = \frac{4}{5} \)
Note: (iv) and (v) are the same event here, because every passenger carries exactly one type of bag.
In simple words: Divide each group by 200; 'not suitcase' and 'handbag or backpack' mean the same passengers here.

Exam Tip: Check: \( \frac{9}{20} + \frac{7}{20} + \frac{4}{20} = 1 \).

Question 2. A restaurant served 150 meals in a day: 60 vegetarian, 50 non-vegetarian and 40 vegan. If one meal order is chosen at random, find the probability that it is:
(i) vegetarian
(ii) non-vegetarian
(iii) vegan
(iv) not vegan
(v) either vegetarian or vegan
Answer:
(i) \( \frac{60}{150} = \frac{2}{5} \)
(ii) \( \frac{50}{150} = \frac{1}{3} \)
(iii) \( \frac{40}{150} = \frac{4}{15} \)
(iv) \( 1 - \frac{4}{15} = \frac{11}{15} \)
(v) \( \frac{60 + 40}{150} = \frac{100}{150} = \frac{2}{3} \)
In simple words: Divide each count by 150 and simplify; 'not vegan' is 1 minus the vegan probability.

Exam Tip: Check: \( \frac{60 + 50 + 40}{150} = 1 \).

Question 3. A bus company issues 100 tickets: 25 window, 35 aisle and 40 middle seats. If one ticket is chosen at random, find the probability that it is:
(i) a window seat (ii) an aisle seat (iii) a middle seat (iv) not a window seat (v) either an aisle or a middle seat
Answer:
(i) \( \frac{25}{100} = \frac{1}{4} \)
(ii) \( \frac{35}{100} = \frac{7}{20} \)
(iii) \( \frac{40}{100} = \frac{2}{5} \)
(iv) \( 1 - \frac{1}{4} = \frac{3}{4} \)
(v) \( \frac{35 + 40}{100} = \frac{75}{100} = \frac{3}{4} \) (the same as (iv))
In simple words: Each seat type's count over 100 is its probability; aisle or middle is the same as not window.

Exam Tip: Always simplify fractions fully, for example \( \frac{35}{100} = \frac{7}{20} \).

Question 4. In a shopping mall, 250 customers bought items: 100 bought clothes, 80 bought electronics, 50 bought groceries and 20 bought books. If one customer is chosen at random, find the probability that the customer bought:
(i) clothes (ii) electronics (iii) groceries (iv) not books (v) either clothes or electronics
Answer:
(i) \( \frac{100}{250} = \frac{2}{5} \)
(ii) \( \frac{80}{250} = \frac{8}{25} \)
(iii) \( \frac{50}{250} = \frac{1}{5} \)
(iv) \( \frac{250 - 20}{250} = \frac{230}{250} = \frac{23}{25} \)
(v) \( \frac{100 + 80}{250} = \frac{180}{250} = \frac{18}{25} \)
In simple words: Divide each group by 250 and reduce to lowest terms.

Exam Tip: \( \frac{100}{250} \) simplifies to \( \frac{2}{5} \), not \( \frac{2}{25} \); divide both parts by 50.

Case Study Based Questions

Question 5. A school is organising a picnic for 60 students. Out of them, 25 chose sandwiches, 20 chose fruits and 15 chose snacks.
(i) Find the probability of the most chosen food item.
(ii) State whether the probability of choosing fruits is less than \( \frac{1}{3} \).
(iii) If one student is chosen at random, find the probability that the student chose either sandwiches or snacks.
Answer:
(i) Sandwiches are the most chosen: P = \( \frac{25}{60} = \frac{5}{12} \)
(ii) P(fruits) = \( \frac{20}{60} = \frac{1}{3} \) exactly. So it is not less than \( \frac{1}{3} \); it is equal to it.
(iii) P(sandwiches or snacks) = \( \frac{25 + 15}{60} = \frac{40}{60} = \frac{2}{3} \)
In simple words: Sandwiches lead with 5/12, fruits are exactly one-third, and two-thirds chose sandwiches or snacks.

Exam Tip: Read 'less than' carefully: an equal value does not count as less.

Question 6. A parking lot has 80 vehicles: 30 cars, 25 bikes, 15 scooters and 10 vans.
(i) Which vehicle type has the lowest probability of being selected?
(ii) Is the probability of selecting a car greater than \( \frac{1}{3} \)?
(iii) If one vehicle is chosen at random, find the probability that it is either a bike or a scooter.
Answer:
(i) Vans: P = \( \frac{10}{80} = \frac{1}{8} \), the lowest.
(ii) P(car) = \( \frac{30}{80} = \frac{3}{8} = 0.375 > 0.333 \). Yes.
(iii) P(bike or scooter) = \( \frac{25 + 15}{80} = \frac{40}{80} = \frac{1}{2} \)
In simple words: Vans are the fewest, cars are more than a third, and bikes plus scooters make up half.

Exam Tip: For 'either … or' with separate groups, add the counts before dividing.

Question 7. An online store sold 100 items in a day: 40 clothes, 30 electronics, 20 books and 10 toys.
(i) Which category has a probability of exactly \( \frac{1}{5} \) of being selected?
(ii) Is the probability of selecting electronics equal to the probability of selecting books and toys together?
(iii) If one item is chosen at random, find the probability that it is not clothes.
Answer:
(i) \( \frac{1}{5} = \frac{20}{100} \), so the category is books.
(ii) P(electronics) = \( \frac{30}{100} \) and P(books or toys) = \( \frac{20 + 10}{100} = \frac{30}{100} \). Yes, they are equal (\( \frac{3}{10} \)).
(iii) P(not clothes) = \( 1 - \frac{40}{100} = \frac{60}{100} = \frac{3}{5} \)
In simple words: Books are one-fifth of sales, electronics equal books plus toys, and three-fifths of items are not clothes.

Exam Tip: Turn the given probability into 'out of 100' to match the counts quickly.