Previous Year Questions: Probability

# Class 10 Maths Chapter 14 Previous Year Questions - Probability

## 2023

Q1: In a group of 20 people. 5 can't swim. If one person is selected at random, then the probability that he/she can swim, is
(a) 3/4
(b) 1/3
(c) 1
(d) 1/4        [2023, 1 Mark]
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

Q2: 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
(a) p + q = 1
(b) p = 1, q = l
(c) p = q - 1
(d) p + p + 1 = 0        [2023, 1 Mark]
Ans:
(a)
Probability of happening of an event + Probability of non - happening of an event
∴ p +q = 1

Q3: 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?
(a) 40
(b) 240
(c) 480
(d) 750        [2023, 1 Mark]
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

Q4: Two dice are thrown together. The probability of getting the difference of numbers on their upper faces equals to 3 is
(a) 1/9
(b) 2/9
(c) 1/6
(d) 1/12        [2023, 1 Mark]
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

Q5: 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
(a) 1/13
(b) 9/13
(c) 4/13
(d) 12/13       [2023, 1 Mark]
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

Q6: 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:
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.      [2023, 1 Mark]
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.

Q7: 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
(a) 18
(b) 15
(c) 10
(d) 20     [2023, 1 Mark]
Ans:
(b)
Probability of drawing a green ball = 3 x Probability of drawing a red ball

∴ n = 15

Q8: 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, 2 Marks]
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

Q9: If a fair coin is tossed twice, find the probability of getting 'almost one head'.   [2023, 2 Marks]
Ans:
Let A be the event of gettirig 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

## 2022

Q1: Two dice are rolled simultaneously. What is the probability that 6 will tome up at least once?
(a) 1/6
(b) 7/36
(c) 11/36
(d) 13/36   [2022, 1 Mark]
Ans:
(c)
When two dice are rolled simultaneously then sample space n(S} = 36
Number of event in which 6 will come up at least once is;
E = {(1 ,6 ), (6,1), (2,6). (6,2) , ( 3, 6 ),(6 ,3), (4, 6), (4 ,4) , (5,6), (6,5), (6,6)}

∴ n(E) = 12
So, required probability P(E) = n(E) / n(S) = 11/36

Q2: Two coins are tossed simultaneously. What is the probability of getting
(ii) At most one tail?
(iii) A head and a tail?   [2022, 3 Marks]
Ans:
Total number of possible outcomes = 4
(i) Let E be the event for at least one head. Then E = {(IH, T), (T, H], {H, H)} = 3, n(S) = 4.
∴ P(E) = 3/4
(ii) Let E be the event for at most one tail. Then E=  {(IH, T), (T, H], {H, H)} = 3, n(S) = 4.
∴ P(E) = 3/4
(iii) Let E be the event for a head and a tail Then ;
E = {(H,T),(T,H)}
n(E)= 2, n(S) = 4
∴ P(E) = 2/4 = 1/2

## 2021

Q1: The probability of getting two heads when two fair coins are tossed together, is
(a) 1/3
(b) 1/4
(c) 1/2
(d) 1    [2021, 1 Mark]
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

Q2: In a single throw of a die. the probability of getting a composite number is
(a) 1/3
(b) 1/2
(c) 2/3
(d) 5/6    [2021, 1 Mark]
Ans:
(a)
Sample space = (12.3.4. 5.6)
∴ Number of total outcomes = 6
Favourable outcomes = (4, 6)
∴ Number of favourable outcomes = 2
∴ Required probability = 2/6 = 1/3

Q3: The probability that a non-leap year has 53 Wednesdays, is
(a) 1/7
(b) 2/7
(c) 5/7
(d) 6/7    [2021, 1 Mark]
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

Q4: From the letters of the word 'MANGO", a letter is selected at random. The probability that the letter is a vowel, is
(a) 1/5
(b) 3/5
(c) 2/5
(d) 4/5    [2021, 1 Mark]
Ans:
(c)
Total number ot 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

Q5: Two fair coins are tossed. What is the probability of getting at the most one bead?
(a) 3/4
(b) 1/4
(c) 1/2
(d) 3/8    [2021, 1 Mark]
Ans:
(a)
Possible outcomes are
HH, HT, TH,  TT
Favourable outcomes (at the most one head} are
HT, TH, TT
So, probability of getting at the most one head = 3/4

Q6: A Setter of English alphabets is chosen at random. What is the probability that it is a letter of the word MATHEMATICS?
(a) 4/13
(b) 9/26
(c) 5/13
(d) 11/26    [2021, 1 Mark]
Ans:
(a)
Total number of possibfeoutcomes = 26
Favourable outcomes are M, A, T, H, E, I, C, S
Therefore, number of favourable outcomes = 8
∴ Required probability =8/26 = 4/13

Q7: A card is drawn from a well shuffled deck of cards. What is the probability that the card drawn is neither a king nor a queen?
(a) 11/13
(b) 12/13
(c) 11/26
(d) 11/52    [2021, 1 Mark]
Ans:
(a)
Total number of cards =52
Total number of kings and queens in the deck of cards = 4 + 4 = 8
∴ Probability that the card drawn is neither a king nor a queen
=

Q8: Two fair dice are rolled simultaneously. The probability that 5 will come up at least once is
(a) 5/36
(b) 11/36
(c) 12/36
(d) 23/36    [2021, 1 Mark]
Ans:
(b)
Total number of possible outcomes = 6 x 6 = 36
Now, outcomes when 5 will come up at least once are ! (1, 5), (2, 5), (3, 5), (4,5), (5, 5), (6, 5). (5, 1), (5, 2),(5,3],(5,4)and(5,6)
∴ Probability that 5 will come op at least once = 11/36

## 2020

Q1: The probability of an event that is sure to happen, is  _______.     [2020, 1 Mark]
Ans:
The probability of an event that is sure to happen is 1.

Q2: If the probability of an event E happening is 0.023, then  = ________.     [2020, 1 Mark]
Ans:
Given, P(E) =0.023
= 1- P(E) = 1 - 0.023 = 0.977

Q3: 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, 2 Marks]
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 favourabie 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

Q4: 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, 2 Marks]
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

Q5: If a number n is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What is the probability that x2 ≤4?   [2020, 2 Marks]
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

Q6: Read the following passage and answer the questions given at the end:
Diwali Fair : A game in a booth at a Diwali Fair involves using a spinner first. Them if the spinner stops on an even number, the player is allowed to pick a marble from a bag. The spinner and the marbles in the bag are represented in figure.
Prizes are given, when a black marble is picked. Shweta prays the game once.

(i) What is the probability that she will be allowed to pick a marble from the bag?
(ii) Suppose she is allowed to pick a marble from the bag. what is the probability of getting a prize, when it is given that the bag contains 20 balls out of which 6 are black?   [2020, 4/5/6 Marks]

Ans: (i) Numbers on the spinner are 1, 4, 10, 8, 6, 2 i.e., 6 in number.
∴ n(S) = 6
Even numbers on the spinner are 4 , 10, 8, 6, 2 i.e., 5 in number
∴ n(E) = 5
So, the probability that Shweta will be allowed to pick a marble = 5/6
(ii) Probability of getting a prize =  Number of black balIs / Total-number of balls = 6/20 = 3/10

Q7: Find the probability of getting a doublet in a throw o f a pair of dice.     [2020, 1 Mark]
Ans:
Total number of possible outcomes = 36.
For getting a doublet possible outcomes are (1,1), (2,2), (3,3), (4,4), (5, 5),(6,6).
∴ Probability of getting a doublet = 6/36 = 1/6

Q8: Find the probability of getting a black queen when a card is drawn at random from a well-shuffled pack of 52 cards.     [2020, 1 Mark]
Ans:
Total number of possible outcomes = 52
Favourable number of outcomes = 2
∴ Probability of getting a black queen = 2/52 =1/26

## 2019

Q1: 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, 2 Marks]
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

Q2: 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, 2 Marks]
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

Q3: A pair of dice is thrown once. Find the probability of getting
(i) even number on each dice
(ii) a total of 9.    [2019, 2 Marks]
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

Q4: 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, 2 Marks]
(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

Q5: 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, 2 Marks]
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

Q6: 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, 2 Marks]
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
Q7: 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, 2 Marks]
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

Q8: 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, 2 Marks]
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

The document Class 10 Maths Chapter 14 Previous Year Questions - Probability is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

## Mathematics (Maths) Class 10

118 videos|463 docs|105 tests

## FAQs on Class 10 Maths Chapter 14 Previous Year Questions - Probability

 1. What is the concept of probability in Class 10 Mathematics?
Ans. In Class 10 Mathematics, probability refers to the measure of the likelihood that an event will occur. It is represented as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. This concept helps us analyze and predict the chances of various outcomes in different situations.
 2. How is probability calculated in Class 10 Mathematics?
Ans. Probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. For example, if we want to find the probability of getting a head in a fair coin toss, the favorable outcome is 1 (getting a head) and the total possible outcomes are 2 (head or tail). So, the probability would be 1/2 or 0.5.
 3. What are the different types of events in probability?
Ans. In probability, events are classified into three categories: certain events, impossible events, and uncertain events. Certain events have a probability of 1 and are guaranteed to occur, while impossible events have a probability of 0 and cannot occur. Uncertain events have probabilities between 0 and 1, indicating their likelihood of occurrence.
 4. How can we represent probability on a scale in Class 10 Mathematics?
Ans. In Class 10 Mathematics, probability can be represented on a scale using fractions, decimals, or percentages. For example, a probability of 1/4 can be represented as 0.25 or 25%. This scale helps us compare and understand the likelihood of different events.
 5. What are the applications of probability in real life?
Ans. Probability has various applications in real life, such as weather forecasting, insurance, gambling, sports analysis, and medical research. It helps us make informed decisions, assess risks, and predict outcomes based on available data. By understanding probability, we can analyze uncertain situations and estimate the chances of different events occurring.

## Mathematics (Maths) Class 10

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