1 MarkQ7. The sum of the probabilities of all the elementary events of an experiment is:
(a) 1
(b) 0
(c) 2
(d) 0.5
Solution:
The sum of probabilities of all individual elementary outcomes of a random experiment is always 1. Answer: (a) 1
1 MarkQ8. If $P(A)$ denotes the probability of an event $A$, then:
(a) $P(A) + P(\text{not } A) = 0$
(b) $P(A) - P(\text{not } A) = 1$
(c) $P(A) + P(\text{not } A) = 1$
(d) $P(\text{not } A) = P(A) - 1$
Solution:
An event and its complement cover all possibilities, so their probabilities sum up to 1 ($P(A) + P(\text{not } A) = 1$). Answer: (c) $P(A) + P(\text{not } A) = 1$
1 MarkQ9. If a letter is chosen at random from the English alphabet, the probability that the letter is a vowel is:
(a) $\frac{5}{26}$
(b) $\frac{21}{26}$
(c) $\frac{1}{26}$
(d) $\frac{5}{27}$
Solution:
Total letters = 26. Vowels (A, E, I, O, U) = 5. Probability = $\frac{5}{26}$. Answer: (a) $\frac{5}{26}$
1 MarkQ10. A number is chosen at random from numbers 1 to 50. The probability that the number is a prime number is:
(a) $\frac{3}{10}$
(b) $\frac{7}{25}$
(c) $\frac{3}{25}$
(d) $\frac{13}{50}$
Solution:
Prime numbers between 1 and 50 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 (Total 15 primes).
Probability = $\frac{15}{50} = \frac{3}{10}$. Answer: (a) $\frac{3}{10}$
1 MarkQ11. A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random from the box, the probability that it is not white is:
(a) $\frac{2}{9}$
(b) $\frac{4}{9}$
(c) $\frac{7}{9}$
(d) $\frac{5}{9}$
Solution:
Total marbles = $3 + 2 + 4 = 9$. White marbles = 2. Marbles that are not white = $9 - 2 = 7$.
Probability = $\frac{7}{9}$. Answer: (c) $\frac{7}{9}$
1 MarkQ12. Which of the following is an example of a random experiment?
(a) Tossing a fair coin
(b) Boiling water at $100^\circ\text{C}$
(c) Sun rising from the east
(d) Adding 2 and 2
Solution:
Tossing a coin has multiple unpredictable individual outcomes, making it a random experiment, whereas options (b), (c), and (d) are deterministic. Answer: (a) Tossing a fair coin
1 MarkQ13. If $P(A) = \frac{2}{3}$, then $P(\text{not } A)$ is:
1 MarkQ14. The probability of getting a number between 1 and 100 which is divisible by 1 is:
(a) 0
(b) 0.5
(c) 1
(d) $\frac{1}{100}$
Solution:
Every integer is divisible by 1, making this a certain/sure event with a probability of 1. Answer: (c) 1
1 MarkQ15. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, the number of blue balls in the bag is:
(a) 10
(b) 5
(c) 20
(d) 15
Solution:
Let blue balls be $x$. Total balls = $5 + x$.
$P(\text{Red}) = \frac{5}{5+x}$, $P(\text{Blue}) = \frac{x}{5+x}$.
Given: $\frac{x}{5+x} = 2 \times \frac{5}{5+x} \implies x = 10$. Answer: (a) 10
1 MarkQ16. Two players, Sangeeta and Rashmi, play a tennis match. The probability of Sangeeta winning the match is $0.62$. What is the probability that Rashmi wins the match?
1 MarkQ18. In the previous question, if the bulb drawn is defective and is not replaced, what is the probability that the next bulb drawn is not defective?
(a) $\frac{15}{19}$
(b) $\frac{16}{19}$
(c) $\frac{4}{19}$
(d) $\frac{3}{19}$
Solution:
Remaining total bulbs = $20 - 1 = 19$. Remaining non-defective bulbs = $16 - 0 = 16$ (since 1 defective was removed).
Probability = $\frac{16}{19}$. Answer: (b) $\frac{16}{19}$
1 MarkQ19. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears a two-digit number is:
(a) $\frac{9}{10}$
(b) $\frac{1}{10}$
(c) $\frac{81}{90}$
(d) $\frac{89}{90}$
Solution:
Total outcomes = 90. One-digit numbers (1 to 9) are 9. Two-digit numbers = $90 - 9 = 81$.
Probability = $\frac{81}{90} = \frac{9}{10}$. Answer: (a) $\frac{9}{10}$
1 MarkQ20. A box contains 90 discs numbered 1 to 90. The probability of getting a square number is:
(a) $\frac{1}{10}$
(b) $\frac{1}{9}$
(c) $\frac{9}{90}$
(d) $\frac{1}{5}$
Solution:
Square numbers between 1 and 90 are: 1, 4, 9, 16, 25, 36, 49, 64, 81 (Total 9 squares).
Probability = $\frac{9}{90} = \frac{1}{10}$. Answer: (a) $\frac{1}{10}$
1 MarkQ21. A box contains 90 discs numbered 1 to 90. The probability of getting a number divisible by 5 is:
(a) $\frac{1}{5}$
(b) $\frac{1}{6}$
(c) $\frac{4}{9}$
(d) $\frac{1}{9}$
Solution:
Numbers divisible by 5 up to 90: $5, 10, 15, \dots, 90$ (Total 18 numbers).
Probability = $\frac{18}{90} = \frac{1}{5}$. Answer: (a) $\frac{1}{5}$
1 MarkQ22. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. The probability that it will point at an odd number is:
1 MarkQ23. In a class, there are 15 boys and 10 girls. One student is selected at random. The probability that the selected student is a girl is:
(a) $\frac{3}{5}$
(b) $\frac{2}{5}$
(c) $\frac{1}{2}$
(d) $\frac{2}{3}$
Solution:
Total students = $15 + 10 = 25$. Number of girls = 10.
Probability = $\frac{10}{25} = \frac{2}{5}$. Answer: (b) $\frac{2}{5}$
1 MarkQ24. A piggy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, the probability that the coin will be a 50p coin is:
(a) $\frac{5}{9}$
(b) $\frac{10}{19}$
(c) $\frac{5}{18}$
(d) $\frac{1}{2}$
Solution:
Total coins = $100 + 50 + 20 + 10 = 180$.
Number of 50p coins = 100.
Probability = $\frac{100}{180} = \frac{5}{9}$. Answer: (a) $\frac{5}{9}$
1 MarkQ25. In the piggy bank mentioned above, the probability that the coin will not be a ₹5 coin is:
(a) $\frac{1}{18}$
(b) $\frac{17}{18}$
(c) $\frac{9}{10}$
(d) $\frac{1}{10}$
Solution:
Total coins = 180. Number of ₹5 coins = 10. Probability of a ₹5 coin = $\frac{10}{180} = \frac{1}{18}$.
Probability of not a ₹5 coin = $1 - \frac{1}{18} = \frac{17}{18}$. Answer: (b) $\frac{17}{18}$
1 MarkQ26. When a coin is tossed once, the probability of getting a head is:
(a) 0
(b) 1
(c) $\frac{1}{2}$
(d) $\frac{1}{4}$
Solution:
Total outcomes = 2 (Head, Tail). Favorable outcome for Head = 1.
Probability = $\frac{1}{2}$. Answer: (c) $\frac{1}{2}$
1 MarkQ27. When two coins are tossed simultaneously, the total number of possible outcomes is:
(a) 2
(b) 6
(c) 4
(d) 8
Solution:
The possible outcomes are {HH, HT, TH, TT}, which totals 4 outcomes. Answer: (c) 4
1 MarkQ28. When two coins are tossed simultaneously, the probability of getting at least one head is:
(a) $\frac{1}{4}$
(b) $\frac{1}{2}$
(c) $\frac{3}{4}$
(d) 1
Solution:
Outcomes with at least one head are {HH, HT, TH} (3 favorable out of 4 total outcomes).
Probability = $\frac{3}{4}$. Answer: (c) $\frac{3}{4}$
1 MarkQ29. When two coins are tossed simultaneously, the probability of getting both heads is:
(a) $\frac{1}{4}$
(b) $\frac{1}{2}$
(c) $\frac{3}{4}$
(d) 0
Solution:
Favorable outcome for both heads is {HH} (1 out of 4).
Probability = $\frac{1}{4}$. Answer: (a) $\frac{1}{4}$
1 MarkQ30. When two coins are tossed simultaneously, the probability of getting no head is:
(a) $\frac{1}{4}$
(b) $\frac{1}{2}$
(c) $\frac{3}{4}$
(d) 0
Solution:
No head means getting both tails {TT} (1 out of 4).
Probability = $\frac{1}{4}$. Answer: (a) $\frac{1}{4}$
1 MarkQ31. When three coins are tossed simultaneously, the total number of outcomes is:
1 MarkQ32. When three coins are tossed simultaneously, the probability of getting exactly two heads is:
(a) $\frac{1}{8}$
(b) $\frac{3}{8}$
(c) $\frac{3}{4}$
(d) $\frac{1}{2}$
Solution:
Outcomes with exactly two heads are {HHT, HTH, THH} (3 favorable out of 8).
Probability = $\frac{3}{8}$. Answer: (b) $\frac{3}{8}$
1 MarkQ33. When three coins are tossed simultaneously, the probability of getting all tails is:
(a) $\frac{1}{8}$
(b) $\frac{3}{8}$
(c) $\frac{1}{4}$
(d) $\frac{7}{8}$
Solution:
Favorable outcome for all tails is {TTT} (1 out of 8).
Probability = $\frac{1}{8}$. Answer: (a) $\frac{1}{8}$
1 MarkQ34. When three coins are tossed simultaneously, the probability of getting at least two heads is:
(a) $\frac{1}{2}$
(b) $\frac{3}{8}$
(c) $\frac{1}{4}$
(d) $\frac{5}{8}$
Solution:
Outcomes with at least two heads are {HHT, HTH, THH, HHH} (4 favorable out of 8).
Probability = $\frac{4}{8} = \frac{1}{2}$. Answer: (a) $\frac{1}{2}$
1 MarkQ35. When a die is thrown once, the probability of getting a composite number is:
(a) $\frac{1}{2}$
(b) $\frac{1}{3}$
(c) $\frac{1}{6}$
(d) $\frac{2}{3}$
Solution:
Composite numbers on a die (1 to 6) are 4 and 6 (Total 2 numbers).
Probability = $\frac{2}{6} = \frac{1}{3}$. Answer: (b) $\frac{1}{3}$
1 MarkQ36. When a die is thrown once, the probability of getting a number less than 3 is:
(a) $\frac{1}{3}$
(b) $\frac{1}{2}$
(c) $\frac{2}{3}$
(d) $\frac{1}{6}$
Solution:
Numbers less than 3 are 1 and 2 (Total 2 numbers).
Probability = $\frac{2}{6} = \frac{1}{3}$. Answer: (a) $\frac{1}{3}$
1 MarkQ37. When a die is thrown once, the probability of getting a factor of 6 is:
(a) $\frac{1}{2}$
(b) $\frac{1}{3}$
(c) 1
(d) $\frac{2}{3}$
Solution:
Factors of 6 on a die are 1, 2, 3, 6 (Total 4 numbers).
Probability = $\frac{4}{6} = \frac{2}{3}$. Answer: (d) $\frac{2}{3}$
1 MarkQ38. When two dice are thrown simultaneously, the total number of outcomes is:
1 MarkQ39. When two dice are thrown simultaneously, the probability of getting a sum of 8 is:
(a) $\frac{5}{36}$
(b) $\frac{1}{6}$
(c) $\frac{1}{12}$
(d) $\frac{7}{36}$
Solution:
Pairs with a sum of 8 are: (2,6), (3,5), (4,4), (5,3), (6,2) (Total 5 pairs).
Probability = $\frac{5}{36}$. Answer: (a) $\frac{5}{36}$
1 MarkQ40. When two dice are thrown simultaneously, the probability of getting a sum of 12 is:
(a) $\frac{1}{36}$
(b) $\frac{1}{18}$
(c) $\frac{1}{12}$
(d) 0
Solution:
The only pair with a sum of 12 is (6,6) (1 favorable outcome out of 36).
Probability = $\frac{1}{36}$. Answer: (a) $\frac{1}{36}$
1 MarkQ41. When two dice are thrown simultaneously, the probability of getting the same number on both dice (doublets) is:
(a) $\frac{1}{6}$
(b) $\frac{1}{36}$
(c) $\frac{1}{3}$
(d) $\frac{5}{36}$
Solution:
Doublets are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) (Total 6 pairs).
Probability = $\frac{6}{36} = \frac{1}{6}$. Answer: (a) $\frac{1}{6}$
1 MarkQ42. When two dice are thrown simultaneously, the probability of getting a sum greater than 10 is:
(a) $\frac{1}{12}$
(b) $\frac{1}{9}$
(c) $\frac{1}{18}$
(d) $\frac{5}{36}$
Solution:
Sums greater than 10 are 11 and 12. Pairs for 11: (5,6), (6,5). Pair for 12: (6,6) (Total 3 pairs).
Probability = $\frac{3}{36} = \frac{1}{12}$. Answer: (a) $\frac{1}{12}$
1 MarkQ43. When two dice are thrown simultaneously, the probability of getting a product of numbers equal to 12 is:
(a) $\frac{1}{9}$
(b) $\frac{1}{12}$
(c) $\frac{1}{6}$
(d) $\frac{5}{36}$
Solution:
Pairs with a product of 12 are: (2,6), (3,4), (4,3), (6,2) (Total 4 pairs).
Probability = $\frac{4}{36} = \frac{1}{9}$. Answer: (a) $\frac{1}{9}$
1 MarkQ44. When two dice are thrown simultaneously, the probability that the sum is odd is:
(a) $\frac{1}{2}$
(b) $\frac{1}{3}$
(c) $\frac{1}{4}$
(d) $\frac{3}{4}$
Solution:
Out of 36 total outcomes for two dice, exactly half (18 outcomes) yield an odd sum and half yield an even sum.
Probability = $\frac{18}{36} = \frac{1}{2}$. Answer: (a) $\frac{1}{2}$
1 MarkQ45. When two dice are thrown simultaneously, the probability that 5 will not come up on either of them is:
(a) $\frac{25}{36}$
(b) $\frac{11}{36}$
(c) $\frac{1}{36}$
(d) $\frac{5}{36}$
Solution:
Number of outcomes where 5 does not appear on either die = $5 \times 5 = 25$.
Probability = $\frac{25}{36}$. Answer: (a) $\frac{25}{36}$
1 MarkQ46. When two dice are thrown simultaneously, the probability that 5 will come up at least once is:
(a) $\frac{11}{36}$
(b) $\frac{25}{36}$
(c) $\frac{1}{6}$
(d) $\frac{5}{36}$
Solution:
Complementary to Question 45: $1 - P(\text{5 does not come up}) = 1 - \frac{25}{36} = \frac{11}{36}$. Answer: (a) $\frac{11}{36}$
1 MarkQ47. If a single die is rolled twice, the probability of getting a sum of 7 is:
(a) $\frac{1}{6}$
(b) $\frac{1}{12}$
(c) $\frac{5}{36}$
(d) $\frac{7}{36}$
Solution:
Rolling a single die twice is equivalent to throwing two dice simultaneously. Pairs with a sum of 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) (Total 6 pairs).
Probability = $\frac{6}{36} = \frac{1}{6}$. Answer: (a) $\frac{1}{6}$
1 MarkQ48. The probability of getting 53 Sundays in a non-leap year is:
(a) $\frac{1}{7}$
(b) $\frac{2}{7}$
(c) 0
(d) $\frac{53}{365}$
Solution:
A non-leap year has 365 days = 52 weeks and 1 extra day. This 1 extra day can be any of the 7 days of the week. The probability that it is a Sunday is $\frac{1}{7}$. Answer: (a) $\frac{1}{7}$
1 MarkQ49. The probability of getting 53 Mondays in a leap year is:
(a) $\frac{1}{7}$
(b) $\frac{2}{7}$
(c) 0
(d) $\frac{53}{366}$
Solution:
A leap year has 366 days = 52 weeks and 2 extra days. The possible combinations for these 2 consecutive days are 7 pairs: (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun). Out of these, Monday appears in 2 pairs: (Sun, Mon) and (Mon, Tue).
Probability = $\frac{2}{7}$. Answer: (b) $\frac{2}{7}$
1 MarkQ50. If a coin is tossed 1000 times and head appears 455 times, then the experimental probability of getting a tail is:
(a) $0.455$
(b) $0.545$
(c) $0.5$
(d) $0.445$
Solution:
Number of tails = Total tosses - Number of heads = $1000 - 455 = 545$.
Experimental probability of getting a tail = $\frac{545}{1000} = 0.545$. Answer: (b) $0.545$
Academic Repository Disclaimer & எச்சரிக்கை அறிவிப்பு :
English: These model solutions are structured to global academic standards for self-assessment and practice purposes. Students are advised to verify with official textbooks for board examinations.
தமிழ்: இந்த மாதிரி கணிதத் தீர்வுகள் மாணவர்களின் சுய கற்றல் மற்றும் பயிற்சித் திறனுக்காக உலகளாவிய கல்வித் தரத்தில் உருவாக்கப்பட்டுள்ளன. தேர்வுகளுக்குத் தயாராகும் போது உங்கள் அதிகாரப்பூர்வ பாடப்புத்தகத்துடன் சரிபார்க்கவும்.
