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

2 Mark Questions

Question 1. A fruit seller has 100 fruits: 40 apples, 30 bananas, 20 oranges and 10 mangoes. If one fruit is chosen at random, is the probability of selecting a banana or an orange greater than that of selecting an apple?
Answer:
P(banana or orange) = \( \frac{30 + 20}{100} = \frac{50}{100} = \frac{1}{2} \)
P(apple) = \( \frac{40}{100} = \frac{2}{5} \)
Since \( \frac{50}{100} > \frac{40}{100} \), yes, banana or orange is more likely than apple.
In simple words: Bananas and oranges together are 50 fruits, more than the 40 apples.

Exam Tip: When all probabilities have the same denominator, just compare the numerators.

Question 2. A cinema hall has 300 seats: 120 balcony, 100 middle and 80 front. If one seat is chosen at random, is the probability of selecting a balcony seat greater than \( \frac{1}{3} \)?
Answer:
P(balcony) = \( \frac{120}{300} = \frac{2}{5} = 0.4 \)
\( \frac{1}{3} \approx 0.333 \)
Since 0.4 > 0.333, yes, the probability of a balcony seat is greater than \( \frac{1}{3} \).
In simple words: Balcony seats are 120 of 300, which is 40%, more than one-third.

Exam Tip: Convert fractions to decimals, or to a common denominator, to compare them easily.

Question 3. During a shopping festival, 80 coupons were distributed: 30 for clothes, 25 for electronics, 15 for groceries and 10 for toys. If one coupon is picked at random:
(i) Identify which category has the lowest probability of being selected, after computing the probability of each category.
(ii) Is the probability of selecting a clothes coupon greater than \( \frac{1}{3} \)? Justify.
Answer:
P(clothes) = \( \frac{30}{80} = \frac{3}{8} \); P(electronics) = \( \frac{25}{80} = \frac{5}{16} \); P(groceries) = \( \frac{15}{80} = \frac{3}{16} \); P(toys) = \( \frac{10}{80} = \frac{1}{8} \)
(i) Toys have the lowest probability, \( \frac{1}{8} \).
(ii) \( \frac{3}{8} = 0.375 \) and \( \frac{1}{3} \approx 0.333 \). Yes, 0.375 > 0.333.
In simple words: Toy coupons are the fewest, and clothes coupons make up more than a third of all coupons.

Exam Tip: Show every probability in (i) even though only the lowest is asked; the question requires it.

Question 4. Compare the probability of getting a prime number and the probability of getting a composite number when a die is thrown.
Answer:
Primes: 2, 3, 5, so P(prime) = \( \frac{3}{6} = \frac{1}{2} \)
Composites: 4, 6, so P(composite) = \( \frac{2}{6} = \frac{1}{3} \)
\( \frac{1}{2} > \frac{1}{3} \), so getting a prime is more likely than getting a composite. (1 is neither prime nor composite.)
In simple words: Three faces are prime but only two are composite.

Exam Tip: Remember that 1 is neither prime nor composite, so the two probabilities do not add up to 1.

Question 5. In a test, 40 students appeared: 10 got full marks, 15 above average, 10 below average and 5 failed. Is the probability of selecting a student who did not fail greater than 0.85?
Answer:
Students who did not fail = 40 − 5 = 35
P(not failed) = \( \frac{35}{40} = 0.875 \)
Since 0.875 > 0.85, yes, the probability is greater than 0.85.
In simple words: 35 out of 40 students passed, which is 87.5%, above 85%.

Exam Tip: The complement (1 − P(failed) = 1 − 0.125) gives the same answer quickly.

3 Mark Questions

Question 6. A cricket team played 12 matches: won 7, lost 4 and drew 1.
(i) Find the probability of selecting a match they won.
(ii) Compare the probability of losing with drawing.
(iii) State whether the probability of winning is greater than \( \frac{1}{2} \).
Answer:
(i) P(won) = \( \frac{7}{12} \)
(ii) P(lost) = \( \frac{4}{12} = \frac{1}{3} \) and P(draw) = \( \frac{1}{12} \). Losing is 4 times as likely as drawing.
(iii) \( \frac{7}{12} \approx 0.583 > 0.5 \). Yes, it is greater than \( \frac{1}{2} \).
In simple words: The team won 7 of 12 matches, which is more than half, and lost far more often than it drew.

Exam Tip: Check that the probabilities add up to 1: 7 + 4 + 1 = 12 ✓.

Question 7. A die is rolled once.
(i) Find the probability of getting a prime number.
(ii) Compare the probability of getting a composite number with that of a prime number.
(iii) State whether the probability of getting an odd number is equal to the probability of getting an even number.
Answer:
(i) Primes 2, 3, 5: P(prime) = \( \frac{3}{6} = \frac{1}{2} \)
(ii) Composites 4, 6: P(composite) = \( \frac{2}{6} = \frac{1}{3} \). P(composite) < P(prime).
(iii) Odd: 1, 3, 5 and even: 2, 4, 6, so both are \( \frac{3}{6} = \frac{1}{2} \). Yes, they are equal.
In simple words: Half the faces are prime, a third are composite, and odd and even faces are equally split.

Exam Tip: List the outcomes for each event; it avoids counting mistakes.

Question 8. A factory produces 500 toys: 450 good and 50 defective.
(i) Find the probability of selecting a defective toy.
(ii) Compare the probability of selecting a good toy with a defective toy.
(iii) State whether the probability of selecting a good toy is more than \( \frac{4}{5} \).
Answer:
(i) P(defective) = \( \frac{50}{500} = \frac{1}{10} \)
(ii) P(good) = \( \frac{450}{500} = \frac{9}{10} \), which is 9 times P(defective).
(iii) \( \frac{9}{10} = 0.9 > 0.8 = \frac{4}{5} \). Yes, it is more than \( \frac{4}{5} \).
In simple words: 9 out of every 10 toys are good, so a good toy is very likely.

Exam Tip: P(good) + P(defective) must equal 1; use it to check.

Question 9. In a class of 40 students, 18 study Mathematics, 12 study Science and 10 study both (the 18 and 12 include these 10).
(i) Find the probability of selecting a student who studies only Mathematics.
(ii) Compare the probability of studying Science with studying both subjects.
(iii) Justify: the probability of studying Maths is greater than \( \frac{1}{3} \).
Answer:
(i) Only Maths = 18 − 10 = 8. P = \( \frac{8}{40} = \frac{1}{5} \)
(ii) P(Science) = \( \frac{12}{40} = \frac{3}{10} \) and P(both) = \( \frac{10}{40} = \frac{1}{4} \). P(Science) > P(both), since the Science group includes the 'both' group.
(iii) P(Maths) = \( \frac{18}{40} = 0.45 > 0.333 \). So it is greater than \( \frac{1}{3} \).
In simple words: Subtract the 'both' students to get 'only Maths', then compare the fractions as decimals.

Exam Tip: A Venn diagram with 8 (only Maths), 10 (both), 2 (only Science) and 20 (neither) makes this clear.

Question 10. In a school election, 150 votes were cast: 90 for Candidate A, 50 for Candidate B and 10 NOTA.
(i) Find the probability of selecting a vote for a candidate (valid vote).
(ii) Compare the probability of a vote for A with a vote for B.
(iii) Explain why the chance of picking a NOTA vote at random is smaller than one-twelfth.
Answer:
(i) P(valid) = \( \frac{90 + 50}{150} = \frac{140}{150} = \frac{14}{15} \)
(ii) P(A) = \( \frac{90}{150} = \frac{3}{5} \) and P(B) = \( \frac{50}{150} = \frac{1}{3} \). P(A) > P(B).
(iii) P(NOTA) = \( \frac{10}{150} = \frac{1}{15} \). Since 15 > 12, \( \frac{1}{15} < \frac{1}{12} \).
In simple words: Almost all votes went to a candidate, A got more than B, and NOTA was only 1 in 15.

Exam Tip: With the same numerator, the fraction with the larger denominator is smaller.

Question 11. A hospital has 120 patients: 50 children, 40 adults and 30 senior citizens. If one patient is chosen at random:
(i) Identify which group has the highest probability of being selected.
(ii) Compare the probability of selecting an adult with that of selecting a senior citizen.
(iii) Find the probability that the patient chosen is not a child.
Answer:
(i) P(child) = \( \frac{50}{120} = \frac{5}{12} \), the largest. Children have the highest probability.
(ii) P(adult) = \( \frac{40}{120} = \frac{1}{3} \) and P(senior) = \( \frac{30}{120} = \frac{1}{4} \). P(adult) > P(senior).
(iii) P(not a child) = \( 1 - \frac{5}{12} = \frac{7}{12} \)
In simple words: Children are the biggest group, adults outnumber seniors, and 7 out of 12 patients are not children.

Exam Tip: 'Not' questions are quickest using 1 − P(event).