Probability — Class 10 Maths PYQs

Previous-year board questions (2023–2026) 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.

Year:Marks:Level:

Probability of events

1 Mark Questions
11 Mark · 🔁 March 2023 & March 2026 · Standardopen ↗
Assertion (A) : The probability that a leap year has $53$ Mondays is $\frac{2}{7}$.
Reason (R) : The probability that a non-leap year has $53$ Mondays is $\frac{5}{7}$.
  • (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
(c) Assertion (A) is true, but Reason (R) is false.
21 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)$\frac{2}{5}$
  • (c)$\frac{7}{20}$
  • (d)$\frac{13}{20}$
Show SolutionHide Solution
(C) $\frac{7}{20}$
31 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)$\frac{8}{15}$
Show SolutionHide Solution
(A) $\frac{1}{3}$
41 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)$\frac{1}{7}$
Show SolutionHide Solution
(C) 0.7
51 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)$\frac{8}{9}$
Show SolutionHide Solution
(C) $\frac{2}{3}$
61 Mark · March 2026 · Basicopen ↗
Assertion (A): The events "getting $2$" and "not getting $2$" in a single throw of an unbiased die are not equally likely events.
Reason (R): The probability of getting $2$ in a single throw of an unbiased die is $\frac{1}{6}$.
(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.
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(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
71 Mark · March 2026 · Basicopen ↗
Assertion (A): The probability that the date of birth of a man is in the month of June is $\frac{1}{12}$
Reason (R) : There are $12$ months in a year.
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(D) Assertion (A) is false, but Reason (R) is true.
2 Marks Questions
82 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.
Show SolutionHide Solution
Total No of Balls=9
(i) P(drawn ball is red) = $\frac{4}{9}$
(ii) P(drawn ball is yellow) = $\frac{2}{9}$
92 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.
Show SolutionHide Solution
Number of Shirts without major defects = $52$
$P(\text{ Anmol will accept the shirt}) = \frac{52}{60} \text{ or } \frac{13}{15}$
102 Marks · March 2026 · Standardopen ↗
A bag contains $25$ balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is $\frac{3}{5}$, then find the number of yellow balls.
Show SolutionHide Solution
P(getting a yellow ball) = $1$ – P(getting a green ball) (1 Mark)
$\frac{\text{Number of yellow balls}}{25} = 1 - \frac{3}{5} = \frac{2}{5}$ (1 Mark)
Number of yellow balls = $25 \times \frac{2}{5} = 10$
112 Marks · March 2026 · Basicopen ↗
A box consists of 60 wall clocks, out of which 40 are good, 15 have minor
defects and the remaining are broken. A trader will reject the box, if the
clock taken out from the box is broken. The trader randomly takes out one
clock from the box. What is the probability that :
(i) the box will be rejected ?
(ii) the clock taken out of the box has minor defect ?
Show SolutionHide Solution
(i) $P(box \text{ will be rejected}) = \frac{5}{60} \text{ or } \frac{1}{12}$ (1 Mark)
(ii) $P(clock \text{ has minor defect}) = \frac{15}{60} \text{ or } \frac{1}{4}$ (1 Mark)
122 Marks · March 2026 · Basicopen ↗
A bag contains $10$ red pens, out of which $3$ are defective and $8$ blue pens, out of which $4$ are defective. A pen is taken out from the bag at random. Find the probability that
(i) it is a defective pen,
(ii) it is a non-defective red pen.
Show SolutionHide Solution
(i) $P (\text{getting a defective pen}) = \frac{7}{18}$ (1 Mark)
(ii) $P (\text{getting a non-defective red pen}) = \frac{7}{18}$ (1 Mark)
3 Marks Questions
133 Marks · March 2025 · Basicopen ↗
A box contains $6$ blue, $4$ white and $8$ red marbles. A marble is drawn at random from this box. Find the probability that the marble so drawn is :
(i) white
(ii) white or red
(iii) not red
Show SolutionHide Solution
(i) $P(\text{white marble}) = \frac{4}{18} \text{ or } \frac{2}{9}$ [$1$ mark]
(ii) $P(\text{white or red marble}) = \frac{12}{18} \text{ or } \frac{2}{3}$ [$1$ mark]
(iii) $P(\text{not a red marble}) = \frac{10}{18} \text{ or } \frac{5}{9}$ [$1$ mark]
4 Marks Questions
144 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
Show SolutionHide Solution
(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
151 Mark · 🔁 July 2023 & March 2026 · 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)$\frac{9}{25}$
Show SolutionHide Solution
(d) $\frac{9}{25}$
161 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)$\frac{7}{100}$
Show SolutionHide Solution
(c) $\frac{1}{25}$
171 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) $\frac{4}{45}$
181 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)$\frac{5}{49}$
Show SolutionHide Solution
(C) $\frac{1}{10}$
191 Mark · March 2026 · Basicopen ↗
A card is drawn from a packet of $50$ identical cards numbered from $1$ to $50$. The probability of drawing a number which is a perfect square, is :
  • (a)$\frac{7}{50}$
  • (b)$\frac{8}{50}$
  • (c)$\frac{6}{50}$
  • (d)$\frac{10}{50}$
Show SolutionHide Solution
(A) $\frac{7}{50}$
2 Marks Questions
202 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
Show SolutionHide Solution
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 $\frac{15}{16}$
212 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 $1$.
Show SolutionHide Solution
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 $1$
222 Marks · March 2025 · Basicopen ↗
A box contains 120 discs, which are numbered from 1 to 120. If one disc is drawn at random from the box, find the probability that
(i) it bears a 2-digit number
(ii) the number is a perfect square.
Show SolutionHide Solution
(i) $P(\text{2-digit number}) = \frac{90}{120} \text{ or } \frac{3}{4}$ (1 mark)
(ii) $P(\text{the number is a perfect square}) = \frac{10}{120} \text{ or } \frac{1}{12}$ (1 mark)
3 Marks Questions
233 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 $$\begin{aligned}& = 90 \\ & \text{(i) P (2 digit number less than 40) } = \frac{30}{90} \text{ or } \frac{1}{3} \\ & \text{(ii) P (a number divisible by 5 and greater than 50) } = \frac{8}{90} \text{ or } \frac{4}{45} \\ & \text{(iii) P (a perfect square number) } = \frac{9}{90} \text{ or } \frac{1}{10}\end{aligned}$$

↳ nature 0 <= p(E) <= 1

1 Mark Questions
241 Mark · March 2025 · Basicopen ↗
In a random experiment of throwing a die, which of the following is a sure event ?
  • (a)Getting a number between 1 and 6
  • (b)Getting an odd number < 7
  • (c)Getting an even number < 7
  • (d)Getting a natural number < 7
Show SolutionHide Solution
(D) Getting a natural number < 7

Coin Based

1 Mark Questions
251 Mark · 🔁 March 2024 & March 2025 · Standardopen ↗
Two coins are tossed simultaneously. The probability of getting at most one tail is:
  • (a)$\frac{1}{2}$
  • (b)$\frac{1}{4}$
  • (c)$\frac{3}{4}$
  • (d)$1$
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(C) $\frac{3}{4}$
3 Marks Questions
263 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 $$\begin{aligned}& = 8 \\ & \text{(i) P (at least one head) } = \frac{7}{8} \\ & \text{(ii) P (exactly 2 tails) } = \frac{3}{8} \\ & \text{(iii) P (at most one tail) } = \frac{4}{8} \text{ or } \frac{1}{2}\end{aligned}$$

Reverse question of probability

1 Mark Questions
271 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
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(C) 480
281 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) $15$
2 Marks Questions
292 Marks · March 2025 · Basicopen ↗
A bag contains $40$ marbles out of which some are white and others are black. If the probability of drawing a black marble is $\frac{3}{5}$, then find the number of white marbles.
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(a) Let the number of black marbles be $n$.
$P(\text{drawing a black marble}) = \frac{n}{40}$
$\therefore \frac{3}{5} = \frac{n}{40} \Rightarrow n = 24$
Hence, number of white marbles $= 16$
3 Marks Questions
303 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 = $12$
313 Marks · March 2026 · Standardopen ↗
A bag contains $30$ balls out of which '$m$' number of balls are blue in colour.
(i) Find the probability that a ball drawn at random from the bag is not blue.
(ii) If $6$ more blue balls are added in the bag, then the probability of drawing a blue ball will be $\frac{5}{4}$ times the probability of drawing a blue ball in the first case. Find the value of $m$.
Show SolutionHide Solution
(i) $P(\text{ball drawn is not blue}) = \frac{30 - m}{30}$ or $1 - \frac{m}{30}$ (I) (1 Mark)
(ii) Total number of balls now $= 36$
Number of blue balls now $= m + 6$
$P(\text{ball drawn is blue}) = \frac{m+6}{36}$ (II) (1 Mark)
According to question, $\frac{m+6}{36} = \frac{5}{4} \times \frac{m}{30}$ (III) ($\frac{1}{2}$ Mark)
$\Rightarrow m = 12$ (IV) ($\frac{1}{2}$ Mark)

Creating sample space from given situation and find probability

1 Mark Questions
321 Mark · March 2025 · Basicopen ↗
The total number of outcomes in the experiment of simultaneous throw of three dice is :
  • (a)$6$
  • (b)$18$
  • (c)$36$
  • (d)$216$
Show SolutionHide Solution
(D) $216$

Complement of event

1 Mark Questions
331 Mark · 🔁 July 2023 & March 2024 · Standardopen ↗
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 $1$.
  • (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).
341 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): $P(E) + P(\text{not } E) = 1$
(a) Both Assertion (A) and Reason (R) are true, and (R) is the correct explanation of (A).
(b) Both Assertion (A) and Reason (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (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).
351 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)$-5$
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(A) $-3$
361 Mark · 🔁 March 2025 & March 2026 · 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)$1$
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(B) $- 2$
371 Mark · March 2025 · Basicopen ↗
A bag contains 3 red, 4 white and 7 green balls. A ball is drawn at random. The probability that the ball drawn is $not$ of red colour is :
  • (a)$\frac{1}{11}$
  • (b)$\frac{3}{14}$
  • (c)$\frac{11}{14}$
  • (d)$\frac{3}{11}$
Show SolutionHide Solution
(C) $\frac{11}{14}$
381 Mark · March 2026 · Basicopen ↗
A bag contains some red and some white balls. A ball is drawn at random from the bag. If the probability of getting a red ball is $\frac{2}{7}$, then the probability of getting a white ball is
  • (a)$\frac{1}{14}$
  • (b)$\frac{5}{7}$
  • (c)$\frac{1}{7}$
  • (d)$\frac{2}{7}$
Show SolutionHide Solution
(B) $\frac{5}{7}$

Dice Based

1 Mark Questions
391 Mark · March 2026 · Basicopen ↗
Assertion (A): The events "getting $2$" and "not getting $2$" in a single throw of an unbiased die are not equally likely events.
Reason (R): The probability of getting $2$ in a single throw of an unbiased die is $\frac{1}{6}$.
  • (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 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 Assertion (A).
3 Marks Questions
403 Marks · March 2026 · Standardopen ↗
Two dice of different colours are thrown at the same time. Write down all the possible outcomes. What is the probability that :
(i) same number appears on both the dice ?
(ii) different number appears on both the dice ?
Show SolutionHide Solution
Possible outcomes are
$(1,1)(1,2), (1,3), (1,4), (1,5), (1,6)$
$(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)$
$(3,1)(3,2), (3,3), (3,4), (3,5), (3,6)$
$(4,1)(4,2), (4,3), (4,4), (4,5), (4,6)$
$(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)$
$(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)$ (1 Mark)
(i) $P(\text{same number appears on both the dice}) = \frac{6}{36}$ or $\frac{1}{6}$ (1 Mark)
(ii) $P(\text{different numbers appear on both the dice}) = 1 - \frac{1}{6} = \frac{5}{6}$ (1 Mark)

Playing Cards

1 Mark Questions
411 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
(d) $\frac{12}{13}$
421 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)$\frac{3}{13}$
  • (c)$\frac{4}{13}$
  • (d)$\frac{1}{13}$
Show SolutionHide Solution
(B) $\frac{3}{13}$
431 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) $\frac{1}{10}$
441 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)$\frac{23}{52}$
Show SolutionHide Solution
(B) $\frac{23}{26}$
451 Mark · March 2026 · Standardopen ↗
A card is drawn at random from a well shuffled deck of 52 playing cards. The probability that it is either a ten or a king is
  • (a)$\frac{1}{26}$
  • (b)$\frac{2}{13}$
  • (c)$\frac{1}{13}$
  • (d)$\frac{8}{26}$
Show SolutionHide Solution
(B) $\frac{2}{13}$
461 Mark · March 2025 · Basicopen ↗
A card is drawn at random from a well shuffled deck of $52$ playing cards. The probability that drawn card shows number '9' is :
  • (a)$\frac{1}{26}$
  • (b)$\frac{4}{13}$
  • (c)$\frac{1}{52}$
  • (d)$\frac{1}{13}$
Show SolutionHide Solution
(d) $\frac{1}{13}$
471 Mark · March 2025 · Basicopen ↗
A black card is lost from a deck of $52$ playing cards. Rest of the cards are shuffled and one card is drawn at random from the available cards. The probability that drawn card is 'king of hearts', is
  • (a)$\frac{1}{52}$
  • (b)$\frac{1}{4}$
  • (c)$\frac{1}{51}$
  • (d)$\frac{1}{26}$
Show SolutionHide Solution
(C) $\frac{1}{51}$
481 Mark · March 2026 · Basicopen ↗
A card is drawn from a well shuffled deck of $52$ cards. The probability that it is not a diamond card is :
  • (a)$\frac{1}{4}$
  • (b)$0$
  • (c)$\frac{1}{2}$
  • (d)$\frac{3}{4}$
Show SolutionHide Solution
(D) $\frac{3}{4}$
4 Marks Questions
494 Marks · March 2026 · Standardopen ↗
A group of friends wanted to play cards with two identical packs together. While shuffling the cards, three cards are dropped. Rest of the cards are shuffled and one card is drawn at random. Assuming that the dropped cards were a queen of hearts, a ten of spades and an ace of clubs, answer the following questions:
(i) Find the probability that the drawn card is a face card.
(ii) Find the probability that the drawn card is either a king or a queen.
(iii) (a) Do you think that the probability of getting a queen was higher if none of the cards were dropped? Justify your answer.
OR
(iii) (b) Find the probability that the drawn card is a jack. Compare it with the probability when none of the cards were dropped. In which case is the probability of getting a jack higher?
figure for this question
Show SolutionHide Solution
Total number of cards $= 2 \times 52 - 3 = 101$
(i) P (a face card) $= \frac{23}{101}$ (1 Mark)
(ii) P (either a king or a queen) $= \frac{15}{101}$ (1 Mark)
(iii) (a)Yes (1/2 Mark)
P (a queen when no cards were dropped) $= \frac{8}{104}$ (1/2 Mark)
P (a queen when cards were dropped) $= \frac{7}{101}$ (1/2 Mark)
$\therefore \frac{8}{104} > \frac{7}{101}$ as $808 > 728$ (1/2 Mark)
So probability of getting a queen was higher if none of the cards were dropped.
OR
(iii) (b) P (a jack when cards were dropped) $= \frac{8}{101}$ (1/2 Mark)
P (a jack when no cards were dropped) $= \frac{8}{104}$ (1/2 Mark)
Since $\frac{8}{101} > \frac{8}{104}$ as $101 < 104$ (1 Mark)
Therefore probability of getting a jack is higher when $3$ cards were dropped.