Probability — Class 10 Maths PYQs

121 previous-year board questions (2023–2025) with marking-scheme solutions, grouped by topic and marks.

Try each question first, then press (or tap Show Solution) to reveal the answer. Press again for the next question.

Probability of events

1 Mark Questions
11 Mark · July 2023 · Standardopen ↗
In a family of two children, the probability of having at least one girl is :
  • (a)$\frac{1}{2}$
  • (b)$\frac{2}{5}$
  • (c)$\frac{3}{4}$
  • (d)$\frac{1}{4}$
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(c) $\frac{3}{4}$
21 Mark · July 2023 · Standardopen ↗
In a family of two children, the probability of having at least one girl is:
  • (a)$\frac{1}{2}$
  • (b)$\frac{3}{4}$
  • (c)$\frac{2}{5}$
  • (d)$\frac{1}{4}$
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(c) $\frac{3}{4}$
31 Mark · July 2023 · Standardopen ↗
If a letter of English alphabet is chosen at random, then the probability of this letter to be a consonant is:
  • (a)$\frac{5}{26}$
  • (b)$\frac{10}{13}$
  • (c)$\frac{21}{26}$
  • (d)$\frac{11}{13}$
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(b) $\frac{21}{26}$
41 Mark · July 2023 · Standardopen ↗
The probability that a leap year selected at random will contain $53$ Sundays and $53$ Mondays, is :
  • (a)$\frac{1}{7}$
  • (b)$\frac{2}{7}$
  • (c)$\frac{3}{7}$
  • (d)$\frac{4}{7}$
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Ans. (a) $\frac{1}{7}$
51 Mark · March 2023 · Standardopen ↗
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)$\frac{3}{4}$
  • (b)$\frac{1}{3}$
  • (c)1
  • (d)$\frac{1}{4}$
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(A) 3/4
61 Mark · March 2023 · Standardopen ↗
In a survey, it is found that every fifth person has a vehicle. The probability of a person NOT having a vehicle is,
  • (a)$\frac{1}{5}$
  • (b)$5\%$
  • (c)$\frac{4}{5}$
  • (d)$95\%$
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(C) $\frac{4}{5}$
71 Mark · March 2023 · Standardopen ↗
In a lottery, there are $5$ prizes and $20$ blanks. The probability of getting a prize is:
  • (a)$\frac{1}{4}$
  • (b)$\frac{1}{20}$
  • (c)$\frac{1}{25}$
  • (d)$\frac{1}{5}$
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(d) $\frac{1}{5}$
81 Mark · March 2023 · Standardopen ↗
Assertion (A) : The probability that a leap year has $53$ Sundays is $\frac{2}{7}$.
Reason (R) : The probability that a non-leap year has $53$ Sundays is $\frac{5}{7}$.
  • (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.
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Sol. (c) Assertion (A) is true but Reason (R) is false
91 Mark · March 2023 · Standardopen ↗
A bag contains 5 pink, 8 blue and 7 yellow balls. One ball is drawn at random from the bag. What is the probability of getting neither a blue nor a pink ball?
  • (a)$\frac{1}{4}$
  • (b)$\frac{7}{20}$
  • (c)$\frac{2}{5}$
  • (d)$\frac{13}{20}$
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(C) $\frac{7}{20}$
101 Mark · July 2024 · Standardopen ↗
The probability for a leap year (selected at random) to have $52$ Mondays and $53$ Sundays is :
  • (a)$\frac{1}{366}$
  • (b)$\frac{1}{52}$
  • (c)$\frac{2}{7}$
  • (d)$\frac{1}{7}$
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Sol. (D) $\frac{1}{7}$
111 Mark · July 2024 · Standardopen ↗
The probability that in a family of three children, there will be at least two boys, is :
  • (a)$\frac{1}{8}$
  • (b)$\frac{7}{8}$
  • (c)$\frac{4}{8}$
  • (d)$\frac{6}{8}$
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Sol. (C) $\frac{4}{8}$
121 Mark · March 2024 · Standardopen ↗
A bag contains $3$ red balls, $5$ white balls and $7$ black balls. The probability that a ball drawn from the bag at random will be neither red nor black is :
  • (a)$\frac{1}{3}$
  • (b)$\frac{1}{5}$
  • (c)$\frac{7}{15}$
  • (d)$\frac{8}{15}$
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Sol.
(A) $\frac{1}{3}$
131 Mark · March 2024 · Standardopen ↗
From the letters of the word "MOBILE", a letter is selected at random. The probability that the selected letter is a vowel, is :
  • (a)$\frac{3}{7}$
  • (b)$\frac{1}{6}$
  • (c)$\frac{1}{2}$
  • (d)$\frac{1}{3}$
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(C) $\frac{1}{2}$
141 Mark · March 2024 · Standardopen ↗
Two friends were born in the year $2000$. The probability that they have the same birthday is :
  • (a)$\frac{1}{365}$
  • (b)$\frac{364}{365}$
  • (c)$\frac{1}{366}$
  • (d)$\frac{365}{366}$
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(C) $\frac{1}{366}$
151 Mark · March 2025 · Standardopen ↗
In a cricket match, a batsman hits the boundary $7$ times out of the $42$ balls he plays. The probability of his not hitting a boundary is :
  • (a)$\frac{1}{7}$
  • (b)$\frac{2}{7}$
  • (c)$\frac{5}{6}$
  • (d)$\frac{1}{6}$
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(C) $\frac{5}{6}$
161 Mark · March 2025 · Standardopen ↗
If in a lottery, there are $10$ prizes and $30$ blanks, then the probability of winning a prize is :
  • (a)$\frac{1}{4}$
  • (b)$\frac{3}{4}$
  • (c)$\frac{1}{3}$
  • (d)$\frac{2}{3}$
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Sol. (A)$\frac{1}{4}$
171 Mark · March 2025 · Standardopen ↗
Letters A to F are mentioned on six faces of a die such that each face has a different letter. Two such die are thrown simultaneously. The probability that vowels turn up on both the dice is :
  • (a)$\frac{1}{4}$
  • (b)$\frac{1}{3}$
  • (c)$\frac{1}{9}$
  • (d)$\frac{1}{36}$
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(c) $\frac{1}{9}$
181 Mark · March 2025 · Standardopen ↗
A bag contains red balls and black balls in the ratio $3:7$. A ball is drawn at random. The probability that ball so drawn is black in colour, is
  • (a)$\frac{3}{7}$
  • (b)0.3
  • (c)0.7
  • (d)$\frac{1}{7}$
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(C) 0.7
191 Mark · March 2025 · Standardopen ↗
A piggy bank contains ₹ 1 coins and ₹ 2 coins in the ratio 9 : 11 respectively. The piggy bank is accidently dropped and a coin pops out of it. The probability that it is a ₹ 2 coin is
  • (a)$\frac{9}{11}$
  • (b)0.45
  • (c)0.55
  • (d)$\frac{1}{11}$
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(C) 0.55
201 Mark · March 2025 · Standardopen ↗
A bag contains red coloured, blue coloured and green coloured balls in the ratio $2 : 3 : 4$. A ball is drawn at random from the given bag. The probability that the ball so drawn being not of blue colour is
  • (a)$\frac{1}{9}$
  • (b)$\frac{1}{3}$
  • (c)$\frac{2}{3}$
  • (d)$\frac{8}{9}$
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(C) $\frac{2}{3}$
2 Marks Questions
212 Marks · March 2023 · Standardopen ↗
A bag contains 4 red, 3 blue and 2 yellow balls. One ball is drawn at random from the bag. Find the probability that drawn ball is
(i) red (ii) yellow.
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Total No of Balls=9
(i) P(drawn ball is red) = $\frac{4}{9}$
(ii) P(drawn ball is yellow) = $\frac{2}{9}$
222 Marks · March 2024 · Standardopen ↗
A carton consists of $60$ shirts of which $48$ are good, $8$ have major defects and $4$ have minor defects. Nigam, a trader, will accept the shirts which are good but Anmol, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. Find the probability that it is acceptable to Anmol.
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Number of Shirts without major defects = $52$
$P(\text{ Anmol will accept the shirt}) = \frac{52}{60} \text{ or } \frac{13}{15}$
232 Marks · March 2025 · Standardopen ↗
Two friends Anil and Ashraf were born in the December month in the year $2010$. Find the probability that :
(i) they share same date of birth.
(ii) they have different dates of birth.
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Number of days in December $2010 = 31$
(i) P (same date of birth) = $\frac{1}{31}$
(ii) P (different dates of birth) = $\frac{30}{31}$
3 Marks Questions
243 Marks · March 2025 · Standardopen ↗
If $65\%$ of the population has black eyes, $25\%$ have brown eyes and the remaining have blue eyes, what is the probability that a person selected at random has :
(a) blue eyes?
(b) brown or black eyes?
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Population having blue eyes = $$\begin{aligned}& (100 - 65 - 25)\% = 10\% \\ & (a) \text{ P(blue eyes)} = 10\% \text{ or } \frac{1}{10} \\ & (b) \text{ P(brown or black eyes)} = 90\% \text{ or } \frac{9}{10}\end{aligned}$$
4 Marks Questions
254 Marks · March 2023 · Standardopen ↗
Case Study - 3
Computer-based learning (CBL) refers to any teaching methodology that makes use of computers for information transmission. At an elementary school level, computer applications can be used to display multimedia lesson plans. A survey was done on $1000$ elementary and secondary schools of Assam and they were classified by the number of computers they had.
Number of Computers
Number of Schools
1-10
250
11-20
200
21-50
290
51-100
180
101 and more
80
One school is chosen at random. Then :
(i) Find the probability that the school chosen at random has more than $100$ computers.
(ii) (a) Find the probability that the school chosen at random has $50$ or fewer computers.
OR
(ii) (b) Find the probability that the school chosen at random has no more than $20$ computers.
(iii) Find the probability that the school chosen at random has $10$ or less than $10$ computers.
figure for this question
Show SolutionHide Solution
(i) P (more than $100$ computers) = $\frac{80}{1000}$ or $0.08$
(ii)(a) $50$ or fewer computers = $250 + 200 + 290 = 740$
Required probability = $\frac{740}{1000}$ or $0.74$
OR
(ii)(b) No more than $20$ computers = $250 + 200 = 450$
Required probability = $\frac{450}{1000}$ or $0.45$
(iii) P ($10$ or less than $10$ computer) = $\frac{250}{1000}$ or $0.25$
264 Marks · March 2024 · Standardopen ↗
In a survey on holidays, $120$ people were asked to state which type of transport they used on their last holiday. The following pie chart shows the results of the survey.
Observe the pie chart and answer the following questions :
(i) If one person is selected at random, find the probability that he/she travelled by bus or ship.
(ii) Which is most favourite mode of transport and how many people used it?
(iii) (a) A person is selected at random. If the probability that he did not use train is $4/5$, find the number of people who used train.
OR
(iii) (b) The probability that randomly selected person used aeroplane is $7/60$. Find the revenue collected by air company at the rate of ₹5,000 per person.
figure for this question
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(i) $P$ (travelling by bus or ship) $= \frac{36+33}{360} = \frac{69}{360}$ or $\frac{23}{120}$
(ii) Car
Number of people who used car $= \frac{177}{360} \times 120 = 59$
(iii) (a) $P$ (person used train)$= 1 - \frac{4}{5} = \frac{1}{5}$
$\therefore$ Number of people who used train $= \frac{1}{5} \times 120 = 24$
OR
(iii) (b) Number of people who used aeroplane $= \frac{7}{60} \times 120 = 14$
$\therefore$ Revenue generated$= 14 \times 5000 = \text{Rs}70,000$

Number Based

1 Mark Questions
271 Mark · July 2023 · Standardopen ↗
The probability for a randomly selected number out of $1, 2, 3, 4, \ldots, 25$ to be a prime number is :
  • (a)$\frac{8}{25}$
  • (b)$\frac{10}{25}$
  • (c)$\frac{11}{25}$
  • (d)$\frac{9}{25}$
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Ans. (d) $\frac{9}{25}$
281 Mark · March 2023 · Standardopen ↗
A bag contains $100$ cards numbered $1$ to $100$. A card is drawn at random from the bag. What is the probability that the number on the card is a perfect cube ?
  • (a)$\frac{1}{20}$
  • (b)$\frac{3}{50}$
  • (c)$\frac{1}{25}$
  • (d)$\frac{7}{100}$
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(c) $\frac{1}{25}$
291 Mark · March 2023 · Standardopen ↗
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 a prime number less than $23$ is
  • (a)$\frac{7}{90}$
  • (b)$\frac{1}{9}$
  • (c)$\frac{4}{45}$
  • (d)$\frac{9}{89}$
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(C) $\frac{4}{45}$
301 Mark · March 2023 · Standardopen ↗
Cards bearing numbers $3$ to $20$ are placed in a bag and mixed thoroughly. A card is taken out of the bag at random. What is the probability that the number on the card taken out is an even number ?
  • (a)$\frac{9}{17}$
  • (b)$\frac{1}{2}$
  • (c)$\frac{5}{9}$
  • (d)$\frac{7}{18}$
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(B) $\frac{1}{2}$
311 Mark · March 2024 · Standardopen ↗
From the data $1, 4, 7, 9, 16, 21, 25$, if all the even numbers are removed, then the probability of getting at random a prime number from the remaining is :
  • (a)$\frac{2}{5}$
  • (b)$\frac{1}{5}$
  • (c)$\frac{1}{7}$
  • (d)$\frac{2}{7}$
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Sol. (b) $\frac{1}{5}$
321 Mark · March 2024 · Standardopen ↗
A box contains cards numbered $6$ to $55$. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square, is
  • (a)$\frac{7}{50}$
  • (b)$\frac{1}{10}$
  • (c)$\frac{7}{55}$
  • (d)$\frac{5}{49}$
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(C) $\frac{1}{10}$
331 Mark · March 2024 · Standardopen ↗
A box contains cards numbered 6 to 55. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square, is
  • (a)$\frac{7}{50}$
  • (b)$\frac{7}{55}$
  • (c)$\frac{1}{10}$
  • (d)$\frac{5}{49}$
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(C) $\frac{1}{10}$
341 Mark · March 2024 · Standardopen ↗
If a digit is chosen at random from the digits $1, 2, 3, 4, 5, 6, 7, 8, 9$; then the probability that this digit is an odd prime number is :
  • (a)$\frac{1}{3}$
  • (b)$\frac{2}{3}$
  • (c)$\frac{4}{9}$
  • (d)$\frac{5}{9}$
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(A) $\frac{1}{3}$
351 Mark · March 2024 · Standardopen ↗
If a digit is chosen at random from the digits $1, 2, 3, 4, 5, 6, 7, 8, 9$; then the probability that this digit is an odd prime number is :
  • (a)$\frac{1}{3}$
  • (b)$\frac{4}{9}$
  • (c)$\frac{2}{3}$
  • (d)$\frac{5}{9}$
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(A) $\frac{1}{3}$
361 Mark · March 2024 · Standardopen ↗
One ticket is drawn at random from a bag containing tickets numbered $1$ to $40$. The probability that the selected ticket has a number which is a multiple of $7$ is:
  • (a)$\frac{1}{7}$
  • (b)$\frac{1}{5}$
  • (c)$\frac{1}{8}$
  • (d)$\frac{7}{40}$
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(B) $\frac{1}{8}$
371 Mark · March 2024 · Standardopen ↗
What is the probability that a number selected randomly from the numbers $1, 2, 3, ..., 15$ is a multiple of $4$ ?
  • (a)$\frac{4}{15}$
  • (b)$\frac{3}{15}$
  • (c)$\frac{6}{15}$
  • (d)$\frac{5}{15}$
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(C) $\frac{3}{15}$
381 Mark · July 2025 · Standardopen ↗
Cards numbered $10, 11, 12, \dots, 30$ are kept in a box and shuffled thoroughly. Rohit draws a card at random from the box. The probability that the number on the card is a multiple of $4$ or $5$ is :
  • (a)$\frac{9}{20}$
  • (b)$\frac{9}{21}$
  • (c)$\frac{10}{20}$
  • (d)$\frac{10}{21}$
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(B) $\frac{9}{21}$
391 Mark · July 2025 · Standardopen ↗
Cards numbered 10, 11, 12, ..., 30 are kept in a box and shuffled thoroughly. Rohit draws a card at random from the box. The probability that the number on the card is a multiple of 4 or 5 is :
  • (a)$\frac{9}{20}$
  • (b)$\frac{10}{20}$
  • (c)$\frac{9}{21}$
  • (d)$\frac{10}{21}$
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(B) $\frac{9}{21}$
401 Mark · March 2025 · Standardopen ↗
The probability of drawing an even prime number out of numbers from $1$ to $30$ is:
  • (a)$\frac{1}{30}$
  • (b)$\frac{4}{15}$
  • (c)$\frac{7}{30}$
  • (d)$0$
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(A) $\frac{1}{30}$
411 Mark · March 2025 · Standardopen ↗
The probability of getting a composite number greater than 3 on throwing a die is
  • (a)$\frac{1}{6}$
  • (b)$\frac{1}{3}$
  • (c)$\frac{1}{2}$
  • (d)$\frac{2}{3}$
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(B) $\frac{1}{3}$
421 Mark · March 2025 · Standardopen ↗
The probability that a $2$-digit number less than $20$, selected at random will be a multiple of $2$ and not a multiple of $3$, is
  • (a)$\frac{1}{2}$
  • (b)$\frac{1}{5}$
  • (c)$\frac{3}{10}$
  • (d)$\frac{3}{11}$
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(C) $\frac{3}{10}$
2 Marks Questions
432 Marks · March 2025 · Standardopen ↗
A bag contains cards which are numbered from $5$ to $100$ such that each card bears a different number. A card is drawn at random. Find the probability that number on the card is (i) a perfect square (ii) a $2$-digit number
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Total possible outcomes = $96$
(i) Number of favourable outcomes for perfect square = $8$
$P(\text{perfect square}) = \frac{8}{96}$ or $\frac{1}{12}$
(ii) Number of favourable outcomes for a $2$-digit number = $90$
$P(\text{a } 2\text{-digit number}) = \frac{90}{96}$ or $\frac{15}{16}$
442 Marks · March 2025 · Standardopen ↗
A bag contains balls numbered $2$ to $91$ such that each ball bears a different number. A ball is drawn at random from the bag. Find the probability that (i) it bears a $2$-digit number (ii) it bears a multiple of $1$.
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Total possible outcomes = $90$
(i) Number of favourable outcomes for a $2$-digit number = $82$
$P(2\text{-digit number}) = \frac{82}{90}$ or $\frac{41}{45}$
(ii) Number of favourable outcomes for multiple of $1 = 90$
$P(\text{a number multiple of } 1) = \frac{90}{90}$ or $1$
3 Marks Questions
453 Marks · March 2024 · Standardopen ↗
A box contains $90$ discs which are numbered $1$ to $90$. If one disc is drawn at random from the box, find the probability that it bears a :
(i) $2$-digit number less than $40$.
(ii) number divisible by $5$ and greater than $50$.
(iii) a perfect square number.
Show SolutionHide Solution
Total outcomes $$\begin{aligned}& = 90 \\ & \text{(i) P (2 digit number less than 40) } = \frac{30}{90} \text{ or } \frac{1}{3} \\ & \text{(ii) P (a number divisible by 5 and greater than 50) } = \frac{8}{90} \text{ or } \frac{4}{45} \\ & \text{(iii) P (a perfect square number) } = \frac{9}{90} \text{ or } \frac{1}{10}\end{aligned}$$
4 Marks Questions
464 Marks · March 2025 · Standardopen ↗
Rahul is a lucky charm for his cricket team. He has a jar of cards with numbers from $10$ to $74$. Before each match, he draws a card from the jar. If the card bears an even number, the team wins. If the number is even and divisible by $5$, they win by a big margin. If the number is an odd number less than $30$, they win by a small margin. And if the number is a prime number between $50$ and $74$, they lose.
Answer the following questions if Rahul draws a card today:
(i) What is the probability that Rahul draws a card with an even number?
(ii) What is the probability that Rahul draws a card with an odd number less than $30$?
(iii) (a) What is the probability that Rahul draws a card with a prime number between $50$ and $74$?
OR
(b) What is the probability that Rahul draws a card with an even number divisible by $5$?
figure for this question
Show SolutionHide Solution
(i) Total possible outcomes = $74 - 10 + 1 = 65$
P (even number) = $\frac{33}{65}$
(ii) P (odd number less than $30$) = $\frac{10}{65}$ or $\frac{2}{13}$
(iii) (a) Favourable outcomes are $53, 59, 61, 67, 71, 73$
Number of favourable outcomes = $6$
P (prime number between $50$ and $74$) = $\frac{6}{65}$
OR
(b) Favourable outcomes are $10, 20, 30, 40, 50, 60, 70$
Number of favourble outcomes = $7$
P (even number divisble by $5$) = $\frac{7}{65}$

↳ nature 0 <= p(E) <= 1

1 Mark Questions
471 Mark · March 2023 · Standardopen ↗
Which of the following numbers cannot be the probability of happening of an event?
  • (a)$0$
  • (b)$\frac{7}{0.01}$
  • (c)$0.07$
  • (d)$\frac{0.07}{3}$
Show SolutionHide Solution
(b) $\frac{7}{0.01}$
481 Mark · March 2024 · Standardopen ↗
Which of the following is not probability of an event ?
  • (a)$0.89$
  • (b)$52\%$
  • (c)$\frac{1}{13}\%$
  • (d)$\frac{1}{0.89}$
Show SolutionHide Solution
(d) $\frac{1}{0.89}$

Coin Based

1 Mark Questions
491 Mark · July 2023 · Standardopen ↗
Lali tosses two different coins simultaneously. The probability that she gets at most one head is:
  • (a)$1$
  • (b)$\frac{1}{2}$
  • (c)$\frac{3}{4}$
  • (d)$\frac{1}{7}$
Show SolutionHide Solution
(b) $\frac{3}{4}$
501 Mark · March 2023 · Standardopen ↗
If three coins are tossed simultaneously, what is the probability of getting at most one tail?
  • (a)$\frac{3}{8}$
  • (b)$\frac{4}{8}$
  • (c)$\frac{5}{8}$
  • (d)$\frac{7}{8}$
Show SolutionHide Solution
(b) $\frac{4}{8}$
511 Mark · March 2023 · Standardopen ↗
Two coins are tossed together. The probability of getting at least one tail is:
  • (a)$\frac{1}{4}$
  • (b)$\frac{1}{2}$
  • (c)$\frac{3}{4}$
  • (d)$1$
Show SolutionHide Solution
(c) $\frac{3}{4}$
521 Mark · March 2024 · Standardopen ↗
Two coins are tossed simultaneously. The probability of getting at most one tail is:
  • (a)$\frac{1}{2}$
  • (b)$\frac{3}{4}$
  • (c)$\frac{1}{4}$
  • (d)$1$
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(C) $\frac{3}{4}$
531 Mark · March 2024 · Standardopen ↗
The probability of throwing a number greater than $2$ with a fair die is :
  • (a)$\frac{2}{3}$
  • (b)$\frac{1}{3}$
  • (c)$\frac{1}{2}$
  • (d)$\frac{5}{6}$
Show SolutionHide Solution
(A) $\frac{2}{3}$
541 Mark · March 2025 · Standardopen ↗
Two coins are tossed simultaneously. The probability of getting atleast one head is
  • (a)$\frac{1}{4}$
  • (b)$\frac{1}{2}$
  • (c)$\frac{3}{4}$
  • (d)1
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(C) $\frac{3}{4}$
2 Marks Questions
552 Marks · March 2023 · Standardopen ↗
If a fair coin is tossed twice, find the probability of getting 'atmost one head'.
Show SolutionHide Solution
Sol. Total outcomes are HH, HT, TH, TT
Favourable outcomes are HT, TH, TT
P (at most one head) = $\frac{3}{4}$
562 Marks · July 2025 · Standardopen ↗
Three different coins are tossed together. Find the probability of getting at least two tails.
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Total number of possible outcomes = $8$
P (atleast two tails) = $\frac{4}{8}$ or $\frac{1}{2}$
3 Marks Questions
573 Marks · July 2023 · Standardopen ↗
Three different coins are tossed simultaneously. Find the probability of getting :
(i) At least one head, (ii) At most two heads.
Show SolutionHide Solution
Total possible outcomes = $2^3 = 8$ (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT)
(i) P (at least one head) = $1 - P(\text{no head}) = 1 - P(TTT) = 1 - \frac{1}{8} = \frac{7}{8}$
(ii) P (at most two heads) = $1 - P(\text{three heads}) = 1 - P(HHH) = 1 - \frac{1}{8} = \frac{7}{8}$
583 Marks · March 2024 · Standardopen ↗
Three coins are tossed simultaneously. What is the probability of getting
(i) at least one head?
(ii) exactly two tails ?
(iii) at most one tail?
Show SolutionHide Solution
Total number of outcomes $$\begin{aligned}& = 8 \\ & \text{(i) P (at least one head) } = \frac{7}{8} \\ & \text{(ii) P (exactly 2 tails) } = \frac{3}{8} \\ & \text{(iii) P (at most one tail) } = \frac{4}{8} \text{ or } \frac{1}{2}\end{aligned}$$
593 Marks · March 2024 · Standardopen ↗
Three unbiased coins are tossed simultaneously. Find the probability of getting :
(i) at least one head.
(ii) exactly one tail.
(iii) two heads and one tail.
Show SolutionHide Solution
Total number of possible outcomes = $8$
(i) $P(\text{at least one head}) = \frac{7}{8}$
(ii) $P (\text{exactly one tail}) = \frac{3}{8}$
(iii) $P (\text{2 heads and one tail}) = \frac{3}{8}$
4 Marks Questions
604 Marks · March 2025 · Standardopen ↗
Three unbiased coins are tossed simultaneously. Find the probability of getting :
(a) exactly two tails
(b) at least one head
(c) at most two heads
Show SolutionHide Solution
Sol. Possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
(a) P(exactly two tails) $= \frac{3}{8}$
(b) P(atleast one head) $= \frac{7}{8}$
(c) P(atmost two heads) $= \frac{7}{8}$

Reverse question of probability

1 Mark Questions
611 Mark · March 2023 · Standardopen ↗
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
Show SolutionHide Solution
(C) 480
621 Mark · March 2023 · Standardopen ↗
A bag contains $5$ red balls and $n$ green balls. 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)$20$
Show SolutionHide Solution
(b) $15$
631 Mark · March 2024 · Standardopen ↗
The probability of getting a chocolate flavoured ice cream at random, in a lot of $600$ ice creams is $0.055$. The number of chocolate flavoured ice creams in the lot is :
  • (a)$33$
  • (b)$11$
  • (c)$55$
  • (d)$44$
Show SolutionHide Solution
(A) $33$
641 Mark · March 2024 · Standardopen ↗
The probability of guessing the correct answer to a certain test question is $\frac{x}{6}$. If the probability of not guessing the correct answer to this question is $\frac{2}{3}$, then the value of $x$ is:
  • (a)$2$
  • (b)$3$
  • (c)$4$
  • (d)$6$
Show SolutionHide Solution
(A) $2$
651 Mark · March 2024 · Standardopen ↗
The probability of getting a bad egg in a lot of $400$ eggs is $0.045$. The number of good eggs in the lot is:
  • (a)$18$
  • (b)$382$
  • (c)$180$
  • (d)$220$
Show SolutionHide Solution
Sol.
(C) $382$
661 Mark · March 2025 · Standardopen ↗
The number of red balls in a bag is $10$ more than the number of black balls. If the probability of drawing a red ball at random from this bag is $\frac{3}{5}$, then the total number of balls in the bag is :
  • (a)$50$
  • (b)$60$
  • (c)$80$
  • (d)$40$
Show SolutionHide Solution
(a) $50$
2 Marks Questions
672 Marks · March 2025 · Standardopen ↗
The probability of guessing the correct answer of a certain test question is $\frac{x}{12}$. If the probability of not guessing the correct answer is $\frac{5}{6}$, then find the value of $x$.
Show SolutionHide Solution
Sol. $\frac{x}{12} + \frac{5}{6} = 1$
$x = 2$
682 Marks · March 2025 · Standardopen ↗
The number of red balls in a bag is three more than the number of black balls. If the probability of drawing a red ball at random from the given bag is $\frac{12}{23}$, find the total number of balls in the given bag.
Show SolutionHide Solution
Let number of black balls = $x$, then number of red balls = $x+3$. $\therefore$ total number of balls = $2x+3$ ($\frac{1}{2}$ mark). ATQ, $\frac{x+3}{2x+3} = \frac{12}{23}$ ($\frac{1}{2}$ mark). $x = 33$ ($\frac{1}{2}$ mark). Total number of balls = 69 ($\frac{1}{2}$ mark).
3 Marks Questions
693 Marks · March 2024 · Standardopen ↗
A jar contains $54$ marbles, each of which is blue, green or white. The probability of selecting a blue marble at random from the jar is $\frac{1}{3}$, and the probability of selecting a green marble at random is $\frac{4}{9}$. How many white marbles does this jar contain ?
Show SolutionHide Solution
Let number of white marbles in the jar = $x$
$\therefore P(\text{white marbles}) = \frac{x}{54}$ (1 Mark)
$P(\text{white}) = 1 - \frac{1}{3} - \frac{4}{9} = \frac{9-3-4}{9} = \frac{2}{9}$ (1 Mark)
$\frac{x}{54} = \frac{2}{9}$
$x = 12$ (1 Mark)
Hence, the number of white marbles = $12$

Creating sample space from given situation and find probability

1 Mark Questions
701 Mark · July 2023 · Standardopen ↗
A number is chosen from the numbers $1, 2, 3$ and denoted as $x$, and a number is chosen from the numbers $1, 4, 9$ and denoted as $y$. Then $P(xy < 9)$ is:
  • (a)$\frac{1}{9}$
  • (b)$\frac{5}{9}$
  • (c)$\frac{3}{9}$
  • (d)$\frac{7}{9}$
Show SolutionHide Solution
(c) $\frac{5}{9}$
4 Marks Questions
714 Marks · March 2023 · Standardopen ↗
A middle school decided to run the following spinner game as a fund-raiser on Christmas Carnival.
Making Purple: Spin each spinner once. Blue and red make purple. So, if one spinner shows Red (R) and another Blue (B), then you 'win'. One such outcome is written as 'RB'.
Based on the above, answer the following questions :
(i) List all possible outcomes of the game.
(ii) Find the probability of 'Making Purple'.
(iii) (a) For each win, a participant gets ₹10, but if he/she loses, he/she has to pay ₹5 to the school. If $99$ participants played, calculate how much fund could the school have collected.
OR
(iii) (b) If the same amount of ₹5 has been decided for winning or losing the game, then how much fund had been collected by school? (Number of participants = $99$)
figure for this question
Show SolutionHide Solution
(i) All possible outcomes: RR, RG, RB, GR, GB, GG, YR, YB, YG
(ii) Number of favourable outcome (RB) $= 1$
$P (\text{Making purple}) = \frac{1}{9}$
(iii)(a) As $P(\text{winning}) = \frac{1}{9}$
Therefore, number of people must win $= \frac{1}{9} \times 99 = 11$
$\therefore$ Game lost by $88$ persons.
Funds collected $= 5 \times 88 - 10 \times 11 = \text{Rs}330$
OR
(iii)(b) Number of participants $= 99$
$P(\text{winning the game}) = \frac{1}{9}$
Number of persons won $= 11$
Number of persons lost $= 88$
Funds collected $= 88 \times 5 - 11 \times 5 = \text{Rs}385$
724 Marks · March 2023 · Standardopen ↗
"Eight Ball" is a game played on a pool table with 15 balls numbered 1 to 15 and a "cue ball" that is solid and white. Of the 15 numbered balls, eight are solid (non-white) coloured and numbered 1 to 8 and seven are striped balls numbered 9 to 15.
The 15 numbered pool balls (no cue ball) are placed in a large bowl and mixed, then one ball is drawn out at random.
Based on the above information, answer the following questions :
(i) What is the probability that the drawn ball bears number 8 ?
(ii) What is the probability that the drawn ball bears an even number?
OR
What is the probability that the drawn ball bears a number, which is a multiple of 3?
(iii) What is the probability that the drawn ball is a solid coloured and bears an even number?
figure for this question
Show SolutionHide Solution
(i)P (drawing ball bearing number 8) $= \frac{1}{15}$
(ii)Even numbers = 2, 4, 6, 8, 10, 12, 14
No. of favourable outcomes = 7
P (even number ball) $= \frac{7}{15}$
OR
(ii)Multiples of 3 are 3, 6, 9, 12, 15
No. of favourable outcomes = 5
$\therefore P(\text{multiple of } 3) = \frac{5}{15} = \frac{1}{3}$
(iii) Solid colour and even number 2, 4, 6, 8
P(solid colour and bear an even no.) $= \frac{4}{15}$

Complement of event

1 Mark Questions
731 Mark · July 2023 · Standardopen ↗
Questions number $19$ and $20$ are Assertion and Reason based questions carrying $1$ mark each. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.
Assertion (A): Two players, Sania and Ashnam play a tennis match. The probability of Sania winning the match is $0.79$ and that of Ashnam winning the match is $0.21$.
Reason (R): The sum of probabilities of two complementary events is $1$.
Show SolutionHide Solution
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
741 Mark · March 2023 · Standardopen ↗
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)$p =q-1$
  • (c)$p = 1,q = 1$
  • (d)$p+q+1= 0$
Show SolutionHide Solution
(A) $p + q = 1$
751 Mark · March 2023 · Standardopen ↗
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)$p = 1,q = 1$
  • (c)$p = q -1$
  • (d)$p + q + 1 = 0$
Show SolutionHide Solution
(A) $p + q = 1$
761 Mark · March 2024 · Standardopen ↗
If the probability of a player winning a game is $0.79$, then the probability of his losing the same game is :
  • (a)$1.79$
  • (b)$0.31$
  • (c)$0.21\%$
  • (d)$0.21$
Show SolutionHide Solution
Sol. (d) $0.21$
771 Mark · March 2024 · Standardopen ↗
Assertion (A): In a cricket match, a batsman hits a boundary $9$ times out of $45$ balls he plays. The probability that in a given ball, he does not hit the boundary is $\frac{4}{5}$.
Reason (R): $P(E) + P(\text{not } E) = 1$
Show SolutionHide Solution
(A) Both Assertion (A) and Reason(R) are true and Reason (R) is the correct explanation of the Assertion (A).
781 Mark · March 2024 · Standardopen ↗
Directions : Questions number $19$ and $20$ are Assertion and Reason based questions carrying $1$ mark each. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below:
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
Assertion (A): In a cricket match, a batsman hits a boundary $9$ times out of $45$ balls he plays. The probability that in a given ball, he does not hit the boundary is $\frac{4}{5}$.
Reason (R): $P(E) + P(\text{not } E) = 1$
Show SolutionHide Solution
(A) Both Assertion (A) and Reason(R) are true and Reason (R) is the correct explanation of the Assertion (A).
791 Mark · March 2024 · Standardopen ↗
For an event $E$, if $P(E) + P(\overline{E}) = q$, then the value of $q^2-4$ is:
  • (a)$-3$
  • (b)$3$
  • (c)$5$
  • (d)$-5$
Show SolutionHide Solution
(A) $-3$
801 Mark · March 2024 · Standardopen ↗
For an event E, if $P(E) + P(\bar{E}) = q$, then the value of $q^2-4$ is :
  • (a)$-3$
  • (b)$3$
  • (c)$5$
  • (d)$-5$
Show SolutionHide Solution
(A) $-3$
811 Mark · March 2025 · Standardopen ↗
Assertion (A): The probability of selecting a number at random from the numbers $1$ to $20$ is $1$. Reason (R): For any event $E$, if $P(E) = 1$, then $E$ is called a sure event.
  • (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • (b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • (c)Assertion (A) is true, but Reason (R) is false.
  • (d)Assertion (A) is false, but Reason (R) is true.
Show SolutionHide Solution
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
821 Mark · March 2025 · Standardopen ↗
Questions number $19$ and $20$ are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
Assertion (A): The probability of selecting a number at random from the numbers $1$ to $20$ is $1$.
Reason (R): For any event E, if P(E) = 1, then E is called a sure event.
Show SolutionHide Solution
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
831 Mark · March 2025 · Standardopen ↗
If for any event E, $P(E) + P(\bar{E}) = q$, then the value of $q^2-3$ is:
  • (a)$0$
  • (b)$- 2$
  • (c)$2$
  • (d)$1$
Show SolutionHide Solution
(B) $- 2$
841 Mark · March 2025 · Standardopen ↗
In an experiment of throwing a die, Assertion (A): Event $E_1$: getting a number less than 3 and Event $E_2$: getting a number greater than 3 are complementary events. Reason (R): If two events $E$ and $F$ are complementary events, then $P(E) + P(F) = 1$.
  • (a)Both Assertion (A) and Reason (R) are true and Reason (R) is 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.
Show SolutionHide Solution
(D) Assertion (A) is false, but Reason (R) is true.

Dice Based

1 Mark Questions
851 Mark · July 2023 · Standardopen ↗
Assertion (A): A fair die is thrown once. The probability of getting a prime number is $\frac{1}{2}$.
Reason (R): A natural number is a prime number if it has only two factors.
Show SolutionHide Solution
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
861 Mark · July 2023 · Standardopen ↗
In a single throw of two dice, the probability of getting a sum of $10$ is:
  • (a)$\frac{1}{12}$
  • (b)$\frac{1}{36}$
  • (c)(e) $\frac{1}{6}$
  • (d)$\frac{1}{4}$
Show SolutionHide Solution
Sol. (a) $\frac{1}{12}$
871 Mark · March 2023 · Standardopen ↗
In a single throw of two dice, the probability of getting $12$ as a product of two numbers obtained is:
  • (a)$\frac{1}{9}$
  • (b)$\frac{4}{9}$
  • (c)$\frac{2}{9}$
  • (d)$\frac{5}{9}$
Show SolutionHide Solution
(a) $\frac{1}{9}$
881 Mark · March 2023 · Standardopen ↗
Two dice are thrown together. The probability of getting the difference of numbers on their upper faces equals to $3$ is :
  • (a)$\frac{1}{9}$
  • (b)$\frac{2}{9}$
  • (c)$\frac{1}{6}$
  • (d)$\frac{1}{12}$
Show SolutionHide Solution
Sol. (c) $\frac{1}{6}$
891 Mark · March 2023 · Standardopen ↗
Two dice are rolled together. What is the probability of getting a sum greater than $10$?
  • (a)$\frac{1}{9}$
  • (b)$\frac{1}{12}$
  • (c)$\frac{1}{6}$
  • (d)$\frac{5}{18}$
Show SolutionHide Solution
(c) $\frac{1}{12}$
901 Mark · March 2024 · Standardopen ↗
Two dice are thrown at the same time and the product of the numbers appearing on them is noted. The probability that the product of the numbers lies between $8$ and $13$ is:
  • (a)$\frac{7}{36}$
  • (b)$\frac{2}{9}$
  • (c)$\frac{5}{36}$
  • (d)$\frac{1}{4}$
Show SolutionHide Solution
(A) $\frac{7}{36}$
911 Mark · March 2024 · Standardopen ↗
Two dice are rolled together. The probability of getting sum of numbers on the two dice as $2, 3$ or $5$, is :
  • (a)$\frac{7}{36}$
  • (b)$\frac{11}{36}$
  • (c)$\frac{5}{36}$
  • (d)$\frac{4}{9}$
Show SolutionHide Solution
Sol. (a) $\frac{7}{36}$
921 Mark · March 2024 · Standardopen ↗
Two dice are rolled together. The probability of getting the sum of the two numbers to be more than $10$, is
  • (a)$\frac{1}{9}$
  • (b)$\frac{7}{12}$
  • (c)$\frac{1}{6}$
  • (d)$\frac{1}{12}$
Show SolutionHide Solution
(D) $\frac{1}{12}$
931 Mark · March 2024 · Standardopen ↗
Two dice are rolled together. The probability of getting the sum of the two numbers to be more than 10, is
  • (a)$\frac{1}{9}$
  • (b)$\frac{1}{6}$
  • (c)$\frac{7}{12}$
  • (d)$\frac{1}{12}$
Show SolutionHide Solution
(D) $\frac{1}{12}$
941 Mark · March 2024 · Standardopen ↗
Two dice are thrown together. The probability that they show different numbers is :
  • (a)$1/6$
  • (b)$5/6$
  • (c)$1/3$
  • (d)$2/3$
Show SolutionHide Solution
(B) $5/6$
951 Mark · March 2024 · Standardopen ↗
Two dice are tossed simultaneously. The probability of getting odd numbers on both the dice is :
  • (a)$\frac{6}{36}$
  • (b)$\frac{12}{36}$
  • (c)$\frac{3}{36}$
  • (d)$\frac{9}{36}$
Show SolutionHide Solution
(D) $\frac{9}{36}$
961 Mark · March 2024 · Standardopen ↗
The probability of getting a sum of $8$, when two dice are thrown simultaneously, is :
  • (a)$\frac{1}{12}$
  • (b)$\frac{1}{9}$
  • (c)$\frac{1}{6}$
  • (d)$\frac{5}{36}$
Show SolutionHide Solution
(D) $\frac{5}{36}$
971 Mark · July 2025 · Standardopen ↗
Two dice are thrown simultaneously and the product of the numbers appearing on the tops is noted. The probability of the product to be less than $6$ is:
  • (a)$\frac{1}{6}$
  • (b)$\frac{1}{4}$
  • (c)$\frac{5}{18}$
  • (d)$\frac{7}{18}$
Show SolutionHide Solution
(C) $\frac{5}{18}$
981 Mark · July 2025 · Standardopen ↗
Two dice are thrown simultaneously and the product of the numbers appearing on the tops is noted. The probability of the product to be less than 6 is:
  • (a)$\frac{1}{6}$
  • (b)$\frac{5}{18}$
  • (c)$\frac{1}{4}$
  • (d)$\frac{7}{18}$
Show SolutionHide Solution
(C) $\frac{5}{18}$
991 Mark · July 2025 · Standardopen ↗
Two dice are thrown at the same time. The probability that the sum appearing on the tops is $8$, is :
  • (a)$\frac{5}{8}$
  • (b)$\frac{5}{12}$
  • (c)$\frac{5}{36}$
  • (d)$\frac{1}{9}$
Show SolutionHide Solution
(C) $\frac{5}{36}$
1001 Mark · March 2025 · Standardopen ↗
A die is thrown once. The probability of getting a number which is
textbf{not} a factor of $36$, is :
  • (a)$\frac{1}{2}$
  • (b)$\frac{2}{3}$
  • (c)$\frac{1}{6}$
  • (d)$\frac{5}{6}$
Show SolutionHide Solution
Sol. (C) $\frac{1}{6}$
1011 Mark · March 2025 · Standardopen ↗
A pair of dice is thrown. The probability that sum of numbers appearing on top faces is at most $10$ is :
  • (a)$\frac{1}{11}$
  • (b)$\frac{10}{11}$
  • (c)$\frac{5}{6}$
  • (d)$\frac{11}{12}$
Show SolutionHide Solution
(d) $\frac{11}{12}$
1021 Mark · March 2025 · Standardopen ↗
A pair of dice is thrown once. The probability that sum of numbers appearing on top faces is at least $4$ is :
  • (a)$\frac{1}{11}$
  • (b)$\frac{10}{11}$
  • (c)$\frac{5}{6}$
  • (d)$\frac{11}{12}$
Show SolutionHide Solution
(d) $\frac{11}{12}$
3 Marks Questions
1033 Marks · March 2025 · Standardopen ↗
Two dice are thrown at the same time. Determine the probability that the difference of the numbers on the two dice is $2$.
Show SolutionHide Solution
Total outcomes $= 36$. Outcomes with difference $2 = 8$: $(1,3), (3,1), (4,2), (2,4), (5,3), (3,5), (4,6), (6,4)$. $P(\text{difference is } 2) = \frac{8}{36} = \frac{2}{9}$.
1043 Marks · March 2025 · Standardopen ↗
Two dice are rolled together. Find the probability of getting:
(i) a multiple of $2$ on one and a multiple of $3$ on the other die.
(ii) the product of two numbers on the top of the two dice is a perfect square number.
Show SolutionHide Solution
Total outcomes = $36$
(i) $(2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), (6, 6)$
Number of outcomes having multiple of $2$ on one die and a multiple of $3$ on other die = $11$
Hence, $P(E) = \frac{11}{36}$
(ii) $(1, 1), (2, 2), (3, 3), (1, 4), (4, 1), (4, 4), (5, 5), (6, 6)$
Number of outcomes having product of two numbers on the top of the dice is a perfect square number = $8$
$P(E) = \frac{8}{36}$ or $\frac{2}{9}$

Playing Cards

1 Mark Questions
1051 Mark · March 2023 · Standardopen ↗
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)$\frac{1}{13}$
  • (b)$\frac{9}{13}$
  • (c)$\frac{4}{13}$
  • (d)$\frac{12}{13}$
Show SolutionHide Solution
Sol. (d) $\frac{12}{13}$
1061 Mark · March 2023 · Standardopen ↗
One card is drawn at random from a well shuffled pack of $52$ playing cards. The probability that the drawn card is a queen, is :
  • (a)$\frac{4}{13}$
  • (b)$\frac{4}{52}$
  • (c)$\frac{2}{13}$
  • (d)$\frac{1}{26}$
Show SolutionHide Solution
(b) $\frac{4}{52}$
1071 Mark · March 2023 · Standardopen ↗
A card is drawn at random from a well shuffled deck of 52 playing cards. The probability of getting a face card is
  • (a)$\frac{1}{4}$
  • (b)$\frac{4}{13}$
  • (c)$\frac{3}{13}$
  • (d)$\frac{1}{13}$
Show SolutionHide Solution
(B) $\frac{3}{13}$
1081 Mark · March 2024 · Standardopen ↗
All queens, jacks and aces are removed from a pack of $52$ playing cards. The remaining cards are well-shuffled and one card is picked up at random from it. The probability of that card to be a king is :
  • (a)$\frac{1}{10}$
  • (b)$\frac{1}{13}$
  • (c)$\frac{3}{10}$
  • (d)$\frac{3}{13}$
Show SolutionHide Solution
(A) $\frac{1}{10}$
1091 Mark · March 2024 · Standardopen ↗
One card is drawn at random from a well shuffled deck of 52 playing cards. The probability that it is a red ace card, is :
  • (a)$\frac{1}{13}$
  • (b)$\frac{1}{26}$
  • (c)$\frac{1}{52}$
  • (d)$\frac{1}{2}$
Show SolutionHide Solution
(b) $\frac{1}{26}$
1101 Mark · March 2025 · Standardopen ↗
A card is selected at random from a deck of $52$ playing cards. The probability of it being a red face card is:
  • (a)$\frac{3}{13}$
  • (b)$\frac{2}{13}$
  • (c)$\frac{1}{2}$
  • (d)$\frac{3}{26}$
Show SolutionHide Solution
(D) $\frac{3}{26}$
1111 Mark · March 2025 · Standardopen ↗
A card is drawn at random from a pack of $52$ cards. What is the probability that the card drawn is a spade or a king?
  • (a)$\frac{1}{13}$
  • (b)$\frac{2}{13}$
  • (c)$\frac{4}{13}$
  • (d)$\frac{9}{13}$
Show SolutionHide Solution
(C) $\frac{4}{13}$
1121 Mark · March 2025 · Standardopen ↗
If all the red face cards are removed from the deck of $52$ playing cards, then the probability of getting a black jack from the remaining cards is :
  • (a)$\frac{2}{46}$
  • (b)$\frac{2}{52}$
  • (c)$\frac{4}{48}$
  • (d)$\frac{2}{23}$
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(A) $\frac{2}{46}$
1131 Mark · March 2025 · Standardopen ↗
A card is drawn at random from a deck of $52$ playing cards. The probability that the drawn card is not a red face card, is
  • (a)$\frac{3}{26}$
  • (b)$\frac{23}{26}$
  • (c)$\frac{7}{52}$
  • (d)$\frac{23}{52}$
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(B) $\frac{23}{26}$
2 Marks Questions
1142 Marks · March 2024 · Standardopen ↗
In a pack of $52$ playing cards one card is lost. From the remaining cards, a card is drawn at random. Find the probability that the drawn card is queen of heart, if the lost card is a black card.
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Sol. Total number of remaining cards = $51$
P (getting queen of heart) = $\frac{1}{51}$
1152 Marks · March 2024 · Standardopen ↗
One card is drawn at random from a well shuffled deck of $52$ cards. Find the probability that the card drawn
(i) is queen of hearts;
(ii) is not a jack.
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Total outcomes = $52$
(i) $P (\text{card is queen of hearts}) = \frac{1}{52}$
(ii) $P (\text{not a jack}) = \frac{48}{52}$ or $\frac{12}{13}$
1162 Marks · March 2024 · Standardopen ↗
This section comprises Very Short Answer (VSA) type questions of $2$ marks each
The king, queen and ace of clubs and diamonds are removed from a deck of $52$ playing cards and the remaining cards are shuffled. A card is randomly drawn from the remaining cards. Find the probability of getting
(i) a card of clubs.
(ii) a red coloured card.
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Total cards left $= 52 - 3 - 3 = 46$
(i) $P$ (card of clubs) $= \frac{10}{46}$ or $\frac{5}{23}$
(ii) $P$ (red coloured card) $= \frac{23}{46}$ or $\frac{1}{2}$
1172 Marks · July 2025 · Standardopen ↗
From a pack of $52$ playing cards, jack, queen and king of diamonds are removed. A card is drawn at random from the remaining cards. Find the probability of getting a face card or a card of spades.
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Number of cards after removing jack, queen and king of diamonds = $52 – 3 = 49$
Favourable outcomes = $19$
P (getting a face card or a card of spades) = $\frac{19}{49}$
1182 Marks · March 2025 · Standardopen ↗
While shuffling a pack of 52 cards, one card was accidentally dropped. Find the probability that the dropped card (i) is not a face card. (ii) is a black king.
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(i) $P(\text{not a face card}) = \frac{40}{52}$ or $\frac{10}{13}$
(ii) $P(\text{black king}) = \frac{2}{52}$ or $\frac{1}{26}$
1192 Marks · March 2025 · Standardopen ↗
All the face cards are removed from the pack of 52 cards and a card is drawn at random from the remaining cards. Find the probability that the card so drawn is
(i) a spade.
(ii) not an ace.
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Remaining cards = 52 - 12 = 40
(i) $P (a \text{ spade}) = \frac{10}{40}$ or $\frac{1}{4}$
(ii) $P (\text{not an ace}) = \frac{36}{40}$ or $\frac{9}{10}$
1202 Marks · March 2025 · Standardopen ↗
From a pack of $52$ cards, all aces and all kings are removed. A card is drawn at random from the remaining cards. Find the probability that the card so drawn is (i) a face card. (ii) a card of red colour.
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Remaining cards = $52 - 8 = 44$
(i) $P(\text{a face card}) = \frac{8}{44}$ or $\frac{2}{11}$
(ii) $P(\text{a card of red colour}) = \frac{22}{44}$ or $\frac{1}{2}$
3 Marks Questions
1213 Marks · March 2025 · Standardopen ↗
All face cards of spades are removed from a pack of $52$ playing cards and the remaining pack is shuffled well. A card is then drawn at random from the remaining pack. Find the probability of getting :
(a) a face card
(b) an ace or a jack
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Sol. After removing face cards of spades, total number of cards $$\begin{aligned}& = 52 - 3 = 49 \\ & (a) P(\text{a face card}) = \frac{9}{49} \\ & (b) P(\text{an ace or a jack}) = \frac{7}{49} \text{ or } \frac{1}{7}\end{aligned}$$