CBSE Class 10 Maths HOTs Probability Set 03

More Practice Questions

Question. The probability of throwing a number greater than 2 with a fair die is
(a) \( \frac{2}{3} \)
(b) \( \frac{5}{6} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{2}{5} \)
Answer: (a) \( \frac{2}{3} \)

Question. The set of all possible outcomes of a random experiment is called
(a) sample space
(b) elementary events
(c) complementary events
(d) favourable events
Answer: (a) sample space

Question. A card is drawn from a pack of cards numbered 1 to 52. The probability that the number on the card is a perfect square is
(a) \( \frac{1}{13} \)
(b) \( \frac{3}{13} \)
(c) \( \frac{5}{13} \)
(d) \( \frac{7}{52} \)
Answer: (d) \( \frac{7}{52} \)

Question. Two unbiased dice are thrown. The probability that the total score is more than 5 is
(a) \( \frac{1}{18} \)
(b) \( \frac{5}{18} \)
(c) \( \frac{7}{18} \)
(d) \( \frac{13}{18} \)
Answer: (d) \( \frac{13}{18} \)

Question. The probability that a leap year should have at least 52 Tuesday is
(a) \( \frac{2}{7} \)
(b) \( \frac{3}{7} \)
(c) 1
(d) \( \frac{5}{7} \)
Answer: (c) 1

Question. For an event E, \( P(E) + P(\overline{E}) = x \), then the value of \( x^3 - 3 \) is
(a) –2
(b) 2
(c) 1
(d) –1
Answer: (a) –2

Question. Which of the following cannot be the probability of an event?
(a) 0.01
(b) 3%
(c) \( \frac{16}{17} \)
(d) \( \frac{17}{16} \)
Answer: (d) \( \frac{17}{16} \)

Question. A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears prime-number less than 23 is
(a) \( \frac{7}{45} \)
(b) \( \frac{23}{45} \)
(c) \( \frac{4}{45} \)
(d) \( \frac{11}{45} \)
Answer: (c) \( \frac{4}{45} \)

Question. How many outcomes are possible when three dice are thrown together?
(a) 18
(b) 36
(c) 54
(d) 216
Answer: (d) 216

Question. One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. The probability that the card is red and a king is
(a) \( \frac{1}{13} \)
(b) \( \frac{1}{26} \)
(c) \( \frac{4}{13} \)
(d) \( \frac{5}{26} \)
Answer: (b) \( \frac{1}{26} \)

Question. From a well-shuffled pack of 52 cards one card is drawn at random. The probability that it is a King or a Queen is
(a) \( \frac{1}{13} \)
(b) \( \frac{3}{26} \)
(c) \( \frac{2}{13} \)
(d) \( \frac{3}{13} \)
Answer: (c) \( \frac{2}{13} \)

Question. A card is drawn at random from a well shuffled pack of 52 playing cards. The probability of getting neither a red card nor a queen is
(a) \( \frac{1}{13} \)
(b) \( \frac{3}{13} \)
(c) \( \frac{5}{13} \)
(d) \( \frac{6}{13} \)
Answer: (d) \( \frac{6}{13} \)

Question. A letter is chosen at random from the English alphabets. The probability that the letter chosen succeeds V is
(a) \( \frac{2}{13} \)
(b) \( \frac{5}{26} \)
(c) \( \frac{1}{26} \)
(d) \( \frac{1}{2} \)
Answer: (a) \( \frac{2}{13} \)

Question. A number x is chosen at random from the numbers – 2, – 1, 0, 1, 2. What is the probability that \( x^2 < 2 \)?
(a) \( \frac{1}{5} \)
(b) \( \frac{2}{5} \)
(c) \( \frac{3}{5} \)
(d) \( \frac{4}{5} \)
Answer: (c) \( \frac{3}{5} \)

Question. A number is selected at random from the numbers 2, 4, 4, 6, 6, 6, 8, 8, 8, 8. The probability that the selected number is their average is
(a) \( \frac{1}{10} \)
(b) \( \frac{3}{10} \)
(c) \( \frac{7}{10} \)
(d) \( \frac{9}{10} \)
Answer: (b) \( \frac{3}{10} \)

Question. A letter is chosen at random from the English alphabet. Probability that it is a letter of the word “SIMULTANEOUSLY” is
(a) \( \frac{14}{26} \)
(b) \( \frac{11}{26} \)
(c) \( \frac{10}{26} \)
(d) \( \frac{15}{26} \)
Answer: (b) \( \frac{11}{26} \)

Question. A month is selected at random from a year. The probability that it is March or June is
(a) \( \frac{1}{12} \)
(b) \( \frac{1}{6} \)
(c) \( \frac{3}{4} \)
(d) \( \frac{1}{3} \)
Answer: (b) \( \frac{1}{6} \)

Question. One ticket is selected at random from 100 tickets numbered 00, 01, 02, ..., 99. Suppose x is sum of the digits and y is product of the digits. Then, probability of getting \( x = 8 \) and \( y = 0 \)
(a) \( \frac{2}{17} \)
(b) \( \frac{3}{17} \)
(c) \( \frac{1}{50} \)
(d) \( \frac{1}{25} \)
Answer: (c) \( \frac{1}{50} \)

Question. A school has five houses P, Q, R, S and T. A class has 23 students, 4 from house P, 8 from house Q, 5 from house R, 2 from house S and rest from house T. A single student is selected at random to be the class monitor. The probability that the selected student is not from P, Q and R is
(a) \( \frac{2}{23} \)
(b) \( \frac{6}{23} \)
(c) \( \frac{8}{23} \)
(d) \( \frac{17}{23} \)
Answer: (b) \( \frac{6}{23} \)

Question. Tony and Monu go to an icecream parlour where they find 5 different varieties of icecreams. If they order one icecream each, then the probability that they both order the same variety of icecreams is
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{1}{5} \)
Answer: (d) \( \frac{1}{5} \)

Question. A box contains 5 apples, 6 oranges and ‘k’ bananas. If the probability of selecting an apple from the box is \( \frac{1}{3} \), then the value of k is
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (d) 4

Question. At a party, there is one last pizza slice and two people who want it. To decide who gets the last slice, two fair six-sided dice are rolled. If the largest number in the roll is:
• 1, 3 or 6, Ananya would get the last slice, and
• 2, 4 or 5, Pranit would get it.
(Note: If the number on both the dice is the same, then consider that number as the larger number.)
In a random roll of dice, who has a higher chance of getting the last pizza slice?
(a) Ananya
(b) Pranit
(c) Both have an equal chance
(d) (cannot be answered without knowing the exact numbers in a roll.)
Answer: (b) Pranit

Question. A number was selected at random from 1 to 100 (inclusive of both numbers) and it was found to be a multiple of 10. What is the probability that the selected number is a multiple of 5?
(a) \( \frac{1}{10} \)
(b) \( \frac{1}{5} \)
(c) \( \frac{1}{2} \)
(d) 1
Answer: (d) 1

Question. A coin is tossed two times. Find the probability of getting at least one head.

Answer: Sample space \( S = \{HH, HT, TH, TT\} \). Total outcomes = 4.
Event of getting at least one head \( E = \{HH, HT, TH\} \). Favourable outcomes = 3.
Probability \( P(E) = \frac{3}{4} \)

Question. A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting
(i) a red king (ii) a queen or a jack.

Answer: Total cards = 52
(i) Number of red kings = 2 (King of Hearts and King of Diamonds). Probability = \( \frac{2}{52} = \frac{1}{26} \)
(ii) Number of queens = 4, Number of jacks = 4. Number of queens or jacks = 8. Probability = \( \frac{8}{52} = \frac{2}{13} \)

Question. A card is drawn from a well-shuffled deck of playing cards. Find the probability of drawing
(i) a face card (ii) a red face card.

Answer: Total cards = 52
(i) Number of face cards = 12 (4 Kings, 4 Queens, 4 Jacks). Probability = \( \frac{12}{52} = \frac{3}{13} \)
(ii) Number of red face cards = 6 (2 Kings, 2 Queens, 2 Jacks of red suits). Probability = \( \frac{6}{52} = \frac{3}{26} \)

Question. 1000 tickets of a lottery were sold and there are 5 prizes on these tickets. If John has purchased one lottery ticket, what is the probability of winning a prize?

Answer: Total tickets = 1000. Favourable tickets (with prizes) = 5.
Probability of winning = \( \frac{5}{1000} = \frac{1}{200} \)

Question. The probability of guessing the correct answer to a certain test is \( \frac{p}{12} \). If the probability of not guessing the correct answer to this question is \( \frac{1}{3} \) then find the value of p.

Answer: \( P(\text{correct}) + P(\text{not correct}) = 1 \)
\( \implies \) \( \frac{p}{12} + \frac{1}{3} = 1 \)
\( \implies \) \( \frac{p}{12} = 1 - \frac{1}{3} \)
\( \implies \) \( \frac{p}{12} = \frac{2}{3} \)
\( \implies \) \( p = \frac{2}{3} \times 12 = 8 \)

Question. In a medical centre, 780 randomly selected people were observed to find if there is a relationship between age and the likelihood of getting a heart attack. The following results were observed.

(i) Based on this table. what is the probability that a randomly chosen person from the same sample is younger than or equal to 55 years and has had a heart attack?
(ii) Looking at the data in the table, Giri says “If a person is randomly chosen, then the probability that the person have had a heart attack is about 12.5%”. Is the statement true or false? Justify your reason.

Answer: (i) Number of persons younger than or equal to 55 with heart attack = 29. Total = 780.
Probability = \( \frac{29}{780} \)
(ii) Total persons with heart attack = 104. Total people = 780.
Probability = \( \frac{104}{780} \approx 0.1333 \) or \( 13.33\% \).
Giri says the probability is about 12.5%. Since 13.33% is significantly different from 12.5% in mathematical terms, the statement is False.

Question. Box A contains 25 slips of which 19 are marked Rs. 1 and other are marked Rs. 5 each. Box B contains 50 slips of which 45 are marked Rs. 1 each and others are marked Rs. 13 each. Slips of both boxes are poured into a third box and reshuffled. A slip is drawn at random. What is the probability that it is marked other than Rs. 1?

Answer: Total slips in Box A = 25 (19 of Rs. 1, 6 of Rs. 5).
Total slips in Box B = 50 (45 of Rs. 1, 5 of Rs. 13).
Total slips in the third box = \( 25 + 50 = 75 \).
Number of slips marked other than Rs. 1 = \( 6 (\text{from A}) + 5 (\text{from B}) = 11 \).
Probability = \( \frac{11}{75} \)

Question. Three different coins are tossed together. Find the probability of getting (i) exactly two heads (ii) at least two heads (iii) at least two tails.

Answer: Sample space \( S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\} \). Total = 8.
(i) Exactly two heads: {HHT, HTH, THH}. Count = 3. Probability = \( \frac{3}{8} \)
(ii) At least two heads: {HHT, HTH, THH, HHH}. Count = 4. Probability = \( \frac{4}{8} = \frac{1}{2} \)
(iii) At least two tails: {HTT, THT, TTH, TTT}. Count = 4. Probability = \( \frac{4}{8} = \frac{1}{2} \)

Question. From a pack of 52 playing cards, Jacks, Queens and Kings of red colours are removed. From the remaining, a card is drawn at random. Find the probability that drawn card is: (i) a black King (ii) a card of red colour (iii) a card of black colour

Answer: Red face cards removed = 2 Jacks + 2 Queens + 2 Kings = 6.
Remaining cards = \( 52 - 6 = 46 \).
(i) Number of black Kings = 2. Probability = \( \frac{2}{46} = \frac{1}{23} \)
(ii) Initial red cards = 26. Red cards removed = 6. Remaining red cards = 20. Probability = \( \frac{20}{46} = \frac{10}{23} \)
(iii) Initial black cards = 26. None removed. Remaining black cards = 26. Probability = \( \frac{26}{46} = \frac{13}{23} \)

Question. There are 100 cards in a bag on which numbers from 1 to 100 are written. A card is taken out from the bag at random. Find the probability that the number on the selected card (i) is divisible by 9 and is a perfect square (ii) is a prime number greater than 80.

Answer: Total cards = 100.
(i) Perfect squares divisible by 9 are 9, 36, 81. Count = 3. Probability = \( \frac{3}{100} \)
(ii) Prime numbers greater than 80: 83, 89, 97. Count = 3. Probability = \( \frac{3}{100} \)

Question. In a single throw of a pair of different dice, what is probability of getting (i) a prime number on each dice? (ii) a total of 9 or 11?

Answer: Total outcomes = 36.
(i) Prime numbers on a die are 2, 3, 5. Pairs are (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). Count = 9. Probability = \( \frac{9}{36} = \frac{1}{4} \)
(ii) Total of 9: (3,6), (4,5), (5,4), (6,3) [4 outcomes]. Total of 11: (5,6), (6,5) [2 outcomes]. Total favourable = 6. Probability = \( \frac{6}{36} = \frac{1}{6} \)

Question. Two different dice are thrown together. Find the probability of:
(i) getting a number greater than 3 on each die.
(ii) getting a total of 6 or 7 of the numbers on two dice.

Answer: Total outcomes = 36.
(i) Numbers > 3 are 4, 5, 6. Pairs are (4,4), (4,5), (4,6), (5,4), (5,5), (5,6), (6,4), (6,5), (6,6). Count = 9. Probability = \( \frac{9}{36} = \frac{1}{4} \)
(ii) Total 6: (1,5), (2,4), (3,3), (4,2), (5,1) [5 outcomes]. Total 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) [6 outcomes]. Total favourable = 11. Probability = \( \frac{11}{36} \)

Question. A box consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Ramesh, a shopkeeper will buy only those shirts which are good but ‘Kewal’ another shopkeeper will not buy shirts with major defects. A shirt is taken out of the box at random. What is the probability that
(i) Ramesh will buy the selected shirt?
(ii) ‘Kewal’ will buy the selected shirt?

Answer: Total shirts = 100.
(i) Ramesh buys only good shirts. Favourable = 88. Probability = \( \frac{88}{100} = \frac{22}{25} \)
(ii) Kewal buys everything except major defects. Favourable = \( 88 (\text{good}) + 8 (\text{minor}) = 96 \). Probability = \( \frac{96}{100} = \frac{24}{25} \)

Question. Farah and Sheena are playing a game with number tokens. Each of them has four number tokens, 2, 3, 4 and 5. A token is randomly picked by each of them from their stack simultaneously. If the sum of the numbers picked by each of them is a prime number; Farah wins the game and if it is a composite number; then Sheena wins the game. Find the probability of each of them winning the game and state who has a higher probability of winning the game. Show your work.

Answer: Total pairs from {2, 3, 4, 5} and {2, 3, 4, 5} are \( 4 \times 4 = 16 \).
Sums:
2+2=4 (C), 2+3=5 (P), 2+4=6 (C), 2+5=7 (P)
3+2=5 (P), 3+3=6 (C), 3+4=7 (P), 3+5=8 (C)
4+2=6 (C), 4+3=7 (P), 4+4=8 (C), 4+5=9 (C)
5+2=7 (P), 5+3=8 (C), 5+4=9 (C), 5+5=10 (C)
Prime sums (Farah wins): {5, 7, 5, 7, 7, 7}. Count = 6. P(Farah) = \( \frac{6}{16} = \frac{3}{8} \)
Composite sums (Sheena wins): {4, 6, 6, 8, 6, 8, 9, 8, 9, 10}. Count = 10. P(Sheena) = \( \frac{10}{16} = \frac{5}{8} \)
Sheena has a higher probability of winning.

Question. A dice is thrown twice. Find the probability that
(i) 5 may not come either time.
(ii) same number may not come on the dice thrown two times.

Answer: Total outcomes = 36.
(i) Outcomes where 5 comes at least once: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5). Count = 11.
5 may not come either time: \( 36 - 11 = 25 \). Probability = \( \frac{25}{36} \)
(ii) Outcomes with same number (doublets): (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). Count = 6.
Same number may not come: \( 36 - 6 = 30 \). Probability = \( \frac{30}{36} = \frac{5}{6} \)

Question. A coin is tossed. If it results in a head a coin is tossed, otherwise a die is thrown. Describe the following events:
(i) A = getting atleast one head
(ii) B = getting an even number
(iii) C = getting a tail
(iv) D = getting a tail and an odd number.

Answer: Sample Space \( S = \{HH, HT, T1, T2, T3, T4, T5, T6\} \)
(i) \( A = \{HH, HT\} \)
(ii) \( B = \{T2, T4, T6\} \)
(iii) \( C = \{HT, T1, T2, T3, T4, T5, T6\} \)
(iv) \( D = \{T1, T3, T5\} \)

Question. A number x is selected at random from the numbers 1, 4, 9, 16 and another number y is selected at random from the numbers 1, 2, 3, 4. Find the probability that the value of xy is more than 16.

Answer: Possible products xy:
x=1: 1, 2, 3, 4
x=4: 4, 8, 12, 16
x=9: 9, 18, 27, 36
x=16: 16, 32, 48, 64
Total outcomes = 16. Products > 16: {18, 27, 36, 32, 48, 64}. Count = 6.
Probability = \( \frac{6}{16} = \frac{3}{8} \)

Question. A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is less than 16.

Answer: Total outcomes = 16. Possible products xy:
x=1: 1, 4, 9, 16
x=2: 2, 8, 18, 32
x=3: 3, 12, 27, 48
x=4: 4, 16, 36, 64
Products < 16 are: {1, 4, 9, 2, 8, 3, 12, 4}. Count = 8.
Probability = \( \frac{8}{16} = \frac{1}{2} \)