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Previous Year Questions 2024 | |
Previous Year Questions 2023 | |
Previous Year Questions 2022 | |
Previous Year Questions 2021 | |
Previous Year Questions 2020 | |
Previous Year Questions 2019 |
Q1: A bag contains 3 red balls, 5 white balls and 7 balck balls. The probability that a ball drawn from the bag at random will be neither red nor black is: (2024)
(a) 1/3
(b) 1/5
(c) 7/15
(d) 8/15
Ans: (a)
No. of red balls = 3
No. of white balls = 5
No. of black balls = 7
Total balls = 15
Probability that ball drawn is neither red nor black = 5/15 = 1/3
Q2: The probablity of getting a bad egg in a lot of 400 eggs is 0.045. The number of good eggs in the lot is: (2024)
(a) 18
(b) 180
(c) 382
(d) 220
Ans: (c)
Probability of getting bad in the lot = 0.045
Let the no. of bad eggs = x
∴ Probability of bag eggs
⇒ 0.045 = 400
⇒ 0.045 = x/400
⇒ x = 400 × 0.045
⇒ x = 18
No. of bad eggs = 18
No. of good eggs = 400 – 18
= 382
Q3: Two dice are thrown together. The probablity that they show different numbers is: (2024)
(a) 1/6
(b) 5/6
(c) 1/3
(d) 2/3
Ans: (b)
Total outcomes, when two dice are thrown
= {(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)}
Favourable outcomes= {(1, 2)(1, 3)...(2, 1), (2, 3)... (3, 1)(3, 2)(3, 4)...(4, 1)...
(6, 5)}
No. of favourable outcomes = 30
Total outcomes = 36
So, P(E) = 30/36
= 5/6
Q4: Assertion (A): In a cricket match, a batsman hits a boundary 9 times out of 45 balls he plays. The probalitiliy that in a given ball, he does not hit the boundary is 4/5.
Reason (R): P(E) + P(not E) = 1. (2024)
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Ans: (b)
Total balls (outcomes) = 45
No. of times boundaries hit = 9
(E = hitting the boundry) = 9/45
= 1/5
∴ P(E= not hitting the boundary)
Ans: (a)
Total number of people = 20
Number of people who can't swim = 5
Number of people who can swim = 20 - 5 = 15
∴ Required probability = 15/20 = 3/4
Q6: Probability of happening of an event is denoted by p and probability of non-happening of the event is denoted by q. Relation between p and q is (2023)
(a) p + q = 1
(b) p = 1, q = l
(c) p = q - 1
(d) p + p + 1 = 0
Ans: (a)
Probability of happening of an event + Probability of non - happening of an event
∴ p +q = 1
Q7: A girl calculates that the probability of her winning the first prize In a lottery Is 0.08. If 6000 tickets are sold, how many tickets has she bought? (2023)
(a) 40
(b) 240
(c) 480
(d) 750
Ans: (c)
Probability of winning first prize = Ticket bought by girl / Total ticket sold
⇒ 0.08 = Ticket bought by girl / 6000
⇒ Ticket bought by girl = 0.08 x 6000 = 480
Q8: Two dice are thrown together. The probability of getting the difference of numbers on their upper faces equals to 3 is (2023)
(a) 1/9
(b) 2/9
(c) 1/6
(d) 1/12
Ans: (c)
Total number of outcomes = 6 x 6 = 36
Favourable outcomes are {1,4), (2, 5], (3, 6), (4, 1), |5. 2), (6.3) i.e., 6 in number
∴ Required probability = 6/36 = 1/6
Q9: A card is drawn at random from a well-shuffled pack of 52 cards. The probability that the card drawn is not an ace is (2023)
(a) 1/13
(b) 9/13
(c) 4/13
(d) 12/13
Ans: (d)
Total number of cards = 52
Number of ace card =4
∴ Number of non ace card = 52 - 4 = 48
∴ Required probability = 48/52 = 12/13
Q10: DIRECTIONS; In the question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option out of the following: (2023)
Assertion (A): The probability that a leap year has 53 Sundays is 2/7.
Reason (R): The probability that a non-leap year has 53 Sundays is 5/7.
(a) Both Assertion (A) and Reason [R) are true and Reason (R| is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Ans: (c)
The leap year has 366 days, i.e, 52 weeks and 2 days.
∴ Required probability = 2/7
The non-leap year has 365 days. i.e.. 52 weeks and 1 day.
∴ Required probability = 1/7
Therefore, assertion is true but reason is false.
Q11: A bag contains 5 red balls and pi green bails. If the probability of drawing a green ball is three times that of a red ball, then the value of n is (2023)
(a) 18
(b) 15
(c) 10
(d) 20
Ans: (b)
Probability of drawing a green ball = 3 x Probability of drawing a red ball
∴ n = 15
Q12: A bag contains 4 red, 3 blue and 2 yellow balls. One ball Fs drawn at random from the bag. Find the probability that drawn ball is (i) red (ii) yellow. (2023)
Ans: Number of red balls =4
Number of blue balls = 3
Number of yellow balls = 2
Total number of balls = 4 + 3 + 2= 9
(i) P(drawing a red balI) = 4/9
(ii) P(drawing a yellow ball) = 2/9
Q13: If a fair coin is tossed twice, find the probability of getting 'almost one head'. (2023)
Ans: Let A be the event of getting atmost one head, and 5 be the sample space.
S = (HH, HT, TH, TT) and A = (HT, TH, TT)
⇒ n(S) = 4
Also, n(A) = 3
Required probability = n(A) / n(S)
= 3/4
Ans: (b)
Sample space = {(H,H), (H,T), (T,H), (T,T)}
∴ Number of total outcomes = 4
Favourable outcomes = {(H,H)}
∴ Number of favourable outcomes = 1
∴ Required probability = 1/4
Q15: In a single throw of a die. the probability of getting a composite number is (2022)
(a) 1/3
(b) 1/2
(c) 2/3
(d) 5/6
Ans: (a)
Sample space = (1, 2, 3, 4, 5, 6)
∴ Number of total outcomes = 6
Favourable outcomes = (4, 6)
∴ Number of favourable outcomes = 2
∴ Required probability = 2/6 = 1/3
Q16: The probability that a non-leap year has 53 Wednesdays, is (2022)
(a) 1/7
(b) 2/7
(c) 5/7
(d) 6/7
Ans: (a)
We know that there are 52 complete weeks m 364 days
Since, it is non leap year.
So. there will be 52 Wednesdays and remaining 365th day may be any of the days of week
So, total number of ways = 7
∴ Number of favourable outcomes = 1
∴ Required probability = 1/7
Q17: From the letters of the word "MANGO", a letter is selected at random. The probability that the letter is a vowel, is (2022)
(a) 1/5
(b) 3/5
(c) 2/5
(d) 4/5
Ans: (c)
Total number of letters in the word MANGO are 5.
So, number of total outcomes = 5
Vowels in the word ‘MANGO’ are A, O
So, number of favourable outcomes = 2
∴ Required probability = 2/5
Q18: Case study based question is compulsory. Attempt any 4 sub-parts from question. Each sub-part carries 1 mark. (2021)
During summer break, Harish wanted to play with his friends but it was too hot outside, so he decided to play some indoor game with his friends. He collects 20 identical cards and writes the numbers 1 to 20 on them (one number on one card). He puts them in a box. He and his friends make a bet for the chances of drawing various cards out of the box. Each was given a chance to tell the probability of picking one card out of the box.
Based on the above, answer the following questions:
(i) The probability that the number on the card drawn is an odd prime number, is
(a) 3/5
(b) 2/5
(c) 9/20
(d) 7/20
Ans: (d)
Card numbered from {1,2, 3, ...., 20}
Total number of possible outcomes = 20
Odd prime numbers from 1 to 20 = {3, 5, 7, 11,13.17.19)
Total number of favourable outcomes = 7
Hence, the probability that the number on the card drawn is an odd prime number = 7/20
(ii) The probability that the number on the card drawn is a composite number Is
(a) 11/20
(b) 3/5
(c) 4/5
(d) 1/2 [2021, 1 Mark]
Ans: (d)
Total number of composite numbers between 1 to 20 = [4, 6, 8, 9, 10, 12, 14, 15, 16, 18)
Total number of favourable outcomes = 10
So, the probability that the number on the drawn card is a composite number = 10/20
∴ Required Probability = 1/2
(iii) The probability that the number on the card drawn is a multiple of 3, 6 and 9 Is
(a) 1/20
(b) 1/10
(c) 3/20
(d) 0
Ans: (c)
Multiple of 3 = {3, 6, 9,12,15, 18}
Multiple of 6 = (6, 12, 18)
Multiple of 9 = (9, 18)
Total number of favourable outcomes = 1
Hence the probability that the card is a multiple of 3, 6 and 9 = 1/20
∴ Required Probability = 1/20
(iv) The probability that the number on the card drawn is a multiple of 3 and 7 is
(a) 3/10
(b) 1/10
(c) 0
(d) 2/5 [2021, 1 Mark]
Ans: (c)
Multiple of 3 between 1 to 20 = {3, 6, 9, 12,15, 18}
Multiple of 7 between 1 to 20 = (7,14)
∴ Multiple of 3 and 7 = 0
∴ Total number of favourable outcomes = 0
∴ Required Probability = 0
(v) If all cards having odd numbers written on them are removed from (lie box and then one card is drawn from the remaining cards, the probability of getting a card having a prime number is
(a) 1/20
(b) 1/10
(c) 0
(d) 1/5
Ans: (b)
If all odd number cards are removed then remaining cards which are left = {2,4, 6, 8,10,12,14,16, 18, 20}
Now, prime number cards in remaining cards = 1
So, the probability of getting a prime number from the remaining cards = 1/10
Ans: The probability of an event that is sure to happen is 1.
Q20: If the probability of an event E happening is 0.023, then = ________. (2020)
Ans: Given, P(E) =0.023
= 1- P(E) = 1 - 0.023 = 0.977
Q21: A letter of English alphabet Is chosen at random. What Is the probability that the (boson letter Is a consonant? (2020)
Ans: Total number of English alphabets = 26
Number of consonants = 26 - 5 = 21
∴ Number of favourable outcomes = 21
P (chosen letter is a consonant) = 21/26
Q22: A die is thrown once. What is the probability of getting a number less than 3? (2020)
Ans: Total number of outcomes = 6
Favourable outcomes are {1.2} i.e.. 2 in number
∴ Required probability = 2/6 = 1/3
Q23: If tire probability of winning a game is 0.07, what is the probability of losing it? (2020)
Ans: Given, probability of winning a game is 0.07
∴ Probability of losing it = 1 - 0.07 = 0.93
Q24: A jar contains 18 marbles. Some are red arid others are yellow. If a marble is drawn at random from the jar. the probability that it is red is 2/3. Find the number of yellow marbles in the jar. (2020)
Ans: There a re 18 marbles in the jar.
∴ Number of possible outcomes = 18
Let there are x yellow marbles in the jar.
∴ Number of red marbles = 18 - x
⇒ Number of favourable outcomes = (18 - x)
∴ Probability of drawing a red marble = (18 - x) / 18
Now. according to the question, = (18 - x) / 18 = 2/3
⇒ 3(18 - x ) = 2 x 13
⇒ 54 -3x = 36
⇒ 3x = 18
⇒ x = 6
So, number of ye How marbles in jar = 6
Q25: A die is thrown twice. What is the probability that
(i) 5 will come up at least once, an
(ii) 5 will not come up either time? (2020)
Ans: Since, throwing a die twice or throwing two dice simultaneously are same.
Possible outcomes are:
(i) Let N be the event t hat 5 wiII come up at least once, the n number of favourable outcomes
= 5 + 6
= 11
(ii) Let E be the event that 5 does not come up either time, then number of favourable outcomes
= [36 - (5 + 6)]
= 25
Q26: If a number x is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What is the probability that x2 ≤ 4? (2020)
Ans: Total number of outcomes = {-3, -2, -1,0, 1, 2, 3} i.e. 7.
∴ Number of favourable outcomes = (4, 1, 0, 1, 4) i.e., 5.
∴ Required Probability = 5/7
Q27: Cards numbered 7 to 40 were put In a box. Poonam selects a card at random. What is the probability that Poonam selects a card which is a multiple of 7? (2019)
Ans: Cards are numbered from 7 to 40. i.e. {7,8,9, ......, 40}
So, total number of outcomes = 34
Multiple of 7 lies between 7 to 40 are {7, 14, 21, 28, 35}
∴ Total number of favourable outcomes= 5
∴ Required probability = 5/34
Q28: A card is drawn at random from a pack of 52 playing cards. Find the probability of drawing a card which is neither a spade nor a king. (2019)
Ans: Total number of cards = 52
Total number of spade cards = 13
Total number of king cards = 4
Total number of spade cards and king cards = 13 + 4 - 1 = 16
[One card is subtracted as it is already included as a king of spade]
∴ Probability of drawing a spade or king card = 16/52
So, probability of drawing a card which is neither a spade nor a king = 1- 16/52
= 9/13
Q29: A pair of dice is thrown once. Find the probability of getting
(i) even number on each dice
(ii) a total of 9. (2019)
Ans: If a pair of dice is thrown once, then possible outcomes are:
∴ Number of possible outcomes are 36.
(i) Total possible outcomes of getting even number on each die
= {( 2, 2), ( 2, 4 ), ( 2, 6 ), ( 4, 4 ), [4, 6), (6, 6). (6, 2), (6, 4 ), (4, 2)}
Number of favourable outcomes = 9
∴ Required probability of getting an even number on each die = 9/36 = 1/4
(ii) Total possible outcomes of getting a total of 9
= {(3, 6), (4, 5), ( 5, 4), (6, 3)} which are 4 in number.
∴ Probability of getting a total of 9 = 9/36 = 1/4
Q30: A bag contains some balls of which x are white, 2x are black and 3xare red. A ball is selected at random. What is the probability that it is (2019)
(i) not red
(ii) white?
Ans: We have, total number of balls = x + 2x + 3x = 6x
Total number of outcomes =6x
(i) Number of favourable outcomes = 3x
∴ Probability of getting red ball = 3x /6x = 1/2
Now, probability of not getting red ball = 1-1/2 = 1/2
∴ Requited probability = 1/2
(ii) Total nu m be r of favourable outcomes = x
∴ Probability of getting white ball = x / 6x
∴ Required probability = 1/6
Q31: A die is thrown once. Find the probability of getting a number which
(i) is a prime number
(ii) lies between 2 and 6. (2019)
Ans: Total possible outcomes are f 1, 2, 3, 4, 5, i.e., 6 in number.
(i) Favourable outcomes are {2, 3, 5} i.e.. 3 in number.
∴ P (getting a prime number) = 3/6 = 1/2
(ii) Favourable outcomes are {3, 4, 5} i.e., 3 in number.
∴ P(getting a number lying between 2 and 6) = 3/6 = 1/2
Q32: A game consists of tossing a coin 3 times and noting the outcome each time. If getting the same result in all the tosses is a success, find the probability of losing the game. (2019)
Ans: When a coin is tossed 3 times, then total possible outcomes are
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
∴ Total number of possible outcomes = 8
Possible outcomes to lose the game are {HHT, HTH, THH, HTT,THT, TTH}
∴ Number of favourable outcomes = 6
∴ Required Probability = 6/8 = 3/4
Q33: Cards marked with numbers 5 to 50 (one number on one card) are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card taken out is
(i) a prime number less than 10,
(ii) a number which is a perfect square. (2019)
Ans: Total number of cards = 50 - 5 + 1 = 46
∴ Total number of possible outcomes = 46
(i) Prime numbers less than 10 are 5, 7.
So, number of favourable outcomes = 2
∴ P(getting a prime number less than 10) = 2/46 = 1/23
(ii) Per feet squares from 5 to 50 are 9, 16, 25, 36, 49 i.e., 5 in number.
∴ P (getting a number which is a perfect square) =5/46
Q34: A child has a die whose 6 faces show the letters given below:
The die is thrown once: What is the probability of getting (i) A (ii) B? (2019)
Ans: Total number of faces in a die = 6
(i) Number of favourable outcomes = 3
∴ P(getting A) = 3/6 = 1/2
(ii) Number of favourable outcomes = 2
∴ P (getting B) = 2/6 = 1/3
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1. What are some key concepts of probability that Year 8 students should understand? |
2. How can I prepare for a probability exam in Year 8? |
3. What types of questions are commonly found in Year 8 probability exams? |
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