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) $\frac{1}{5}$
71 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.
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(c) Assertion (A) is true, but Reason (R) is false.
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 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 :
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 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 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
Assertion (A): From a bag containing 5 red balls, 2 white balls and 3 green balls, the probability of drawing a non-white ball is $\frac{4}{5}$ Reason (R) : For any event E, $P(E) + P (not E) = 1$
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(a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of the Assertion (A).
Assertion (A): The probability of a certain event E is 1. Reason (R): The sum of probabilities of all elementary events of an experiment is 1.
(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.
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(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). Note: “The probability of a certain event E is 1” because certain event consists of all the elementary events and “The sum of probabilities of all elementary events of an experiment is 1".
From a bag containing yellow, red and green balls, the probability of drawing a yellow or red ball is equal to that of drawing a green ball. The probability of drawing a green ball from the bag is :
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).
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) = $\frac{2}{9}$
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 = $52$ $P(\text{ Anmol will accept the shirt}) = \frac{52}{60} \text{ or } \frac{13}{15}$
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) = $\frac{30}{31}$
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.
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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 = $25 \times \frac{2}{5} = 10$
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 ?
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(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)
Each letter of the word 'MATHEMATICS' is written on separate slips and slips are put in a box and shuffled. A slip is selected at random from the box. Find the probability that the letter written on the slip is : (i) a vowel, (ii) either M or T.
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(i) $P$ (getting a vowel) $= \frac{4}{11}$ (1 Mark) (ii) $P$ (getting either M or T) $= \frac{4}{11}$ (1 Mark)
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.
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(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)
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 = $$\begin{aligned}& (100 - 65 - 25)\% = 10\% \\ & (a) \text{ P(blue eyes)} = 10\% \text{ or } \frac{1}{10} \\ & (b) \text{ P(brown or black eyes)} = 90\% \text{ or } \frac{9}{10}\end{aligned}$$
A lot consists of $200$ pens of which $180$ are good and the rest are defective. A customer will buy a pen if it is not defective. The shopkeeper draws a pen at random and gives it to the customer. What is the probability that the customer will not buy it ? Another lot of $100$ pens containing $80$ good pens is mixed with the previous lot of $200$ pens. The shopkeeper draws one pen at random from the entire lot and gives it to the customer. What is the probability that the customer will buy the pen ?
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$P(\text{customer will not buy the pen}) = \frac{20}{200} = \frac{1}{10}$ After mixing the two lots Total pens $= 200 + 100 = 300$ Number of good pens $= 180 + 80 = 260$ $P(\text{customer will buy the pen}) = \frac{260}{300}$ or $\frac{13}{15}$
A lot consists of 200 pens of which 180 are good and the rest are defective. A customer will buy a pen if it is not defective. The shopkeeper draws a pen at random and gives it to the customer. What is the probability that the customer will not buy it ? Another lot of 100 pens containing 80 good pens is mixed with the previous lot of 200 pens. The shopkeeper now draws one pen at random from the entire lot and gives it to the customer. What is the probability that the customer will buy the pen ?
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$P(\text{customer will not buy the pen}) = \frac{20}{200} = \frac{1}{10}$ After mixing the two lots Total pens $= 200 + 100 = 300$ Number of good pens $= 180 + 80 = 260$ $P(\text{customer will buy the pen}) = \frac{260}{300}$ or $\frac{13}{15}$
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
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(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]
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.
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(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 $0.25$
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.
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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
411 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 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 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
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 ?
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 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
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:
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 :
One ticket is drawn at random from a bag containing $50$ tickets numbered $1$ to $50$. The probability that the drawn ticket has a number which is a multiple of $7$ is:
Probability of getting an irrational number at random from the numbers $\sqrt{3}$, $\sqrt{4}$, $3\sqrt{9}$, $3\sqrt{8}$, $\sqrt{5}$, $0$, $4^{\frac{3}{2}}$ is :
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 $\frac{15}{16}$
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 $1$
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.
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(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)
A piggy bank contains fifty ₹1 coins, hundred ₹2 coins and one hundred and fifty ₹5 coins. If it is equally likely any one of the coins will fall out when the bank is turned upside down, find the probability that ₹2 coin has not fallen out, when the bank is turned upside down.
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No. of Rs $ 2$ coins $= 100$ Total no. of coins $= 50 + 100 + 150 = 300$ $\therefore P(\text{not getting a Rs 2 coin}) = \frac{200}{300} = \frac{2}{3}$
Find the probability that a number selected at random from the numbers $30, 31, 32, 33, \text{......}, 60$ is (i) a prime number (ii) a multiple of $6$.
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(i) Prime numbers are $31, 37, 41, 43, 47, 53, 59$ $\text{P (a prime number)} = \frac{7}{31}$ (1 Mark) (ii) Multiples of $6$ are $30, 36, 42, 48, 54, 60$ $\text{P (a multiple of 6)} = \frac{6}{31}$ (1 Mark)
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.
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Total outcomes $$\begin{aligned}& = 90 \\ & \text{(i) P (2 digit number less than 40) } = \frac{30}{90} \text{ or } \frac{1}{3} \\ & \text{(ii) P (a number divisible by 5 and greater than 50) } = \frac{8}{90} \text{ or } \frac{4}{45} \\ & \text{(iii) P (a perfect square number) } = \frac{9}{90} \text{ or } \frac{1}{10}\end{aligned}$$
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 $5$?
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(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 $5$) = $\frac{7}{65}$
Assertion (A) : If probability of happening of an event is $0.2p$, $p > 0$, then $p$ can't be more than $5$. Reason (R) : $P(E) = 1 - P(\bar{E})$ for an event $E$. (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.
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(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
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?
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Total number of outcomes $$\begin{aligned}& = 8 \\ & \text{(i) P (at least one head) } = \frac{7}{8} \\ & \text{(ii) P (exactly 2 tails) } = \frac{3}{8} \\ & \text{(iii) P (at most one tail) } = \frac{4}{8} \text{ or } \frac{1}{2}\end{aligned}$$
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.
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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) $P (\text{2 heads and one tail}) = \frac{3}{8}$
Two different coins are tossed simultaneously. What is the probability of getting: (i) at least one head? (ii) at most one tail ? (iii) a head and a tail ?
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Possible Outcomes are $HH, HT, TH, TT$ (i) $P$ (at least one head) $= \frac{3}{4}$ (I) (1 Mark) (ii) $P$ (at most one tail) $= \frac{3}{4}$ (II) (1 Mark) (iii) $P$ (a head and a tail) $= \frac{2}{4}$ or $\frac{1}{2}$ (III) (1 Mark)
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 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 :
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 :
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:
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 :
Meena calculates that the probability of her winning the first prize in a lottery is $0.08$. If total $800$ tickets were sold, the number of tickets bought by her, is
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 $x$.
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.
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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).
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 $= 16$
In a pre-primary class, a teacher put cards numbered $20$ to $59$ in a bowl. A student picked up a card at random and read the number. Find the probability that the number read was (i) a prime number (ii) a perfect square.
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Total number of cards $= 40$ (i) $P(\text{a prime number}) = \frac{9}{40}$ (ii) $P(\text{no. is perfect square}) = \frac{3}{40}$
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 ?
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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 = $12$
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 $m$.
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(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
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:
Slips of letters of the word 'BACKGROUND' are put in a bowl and thoroughly mixed. One slip is picked up at random. Find the probability that picked up slip's letter is (i) a vowel (ii) present in the word 'BALL'.
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(i) $\text{P (a vowel)} = \frac{3}{10}$ (1 Mark) (ii) $\text{P (present in the word 'BALL')} = \frac{2}{10}$ or $\frac{1}{5}$ (1 Mark)
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 = $99$)
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(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 $= 88 \times 5 - 11 \times 5 = \text{Rs}385$
"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?
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(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.) $= \frac{4}{15}$
Raghav has a collection of balls of different colours. He has a total of $35$ balls in his basket out of which seven are black in colour and eight are yellow in colour. Out of remaining balls, some are white and the rest are red. Based on the above, answer the following questions : (a) If the probability of drawing a red ball at random from the basket is three times that of a white ball, then find the number of red balls in the basket. (b) Find the probability of drawing a ball at random from the basket which is either a black or a white ball.
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Total number of balls = $35$ Number of black balls = $7$ Number of yellow balls = $8$ (a) Number of white and red balls = $35 - 15 = 20$ ($\frac{1}{2}$ Mark) Let number of white balls = $x$, then number of red balls = $(20 - x)$ ($\frac{1}{2}$ Mark) Since, P (a red ball) = $3 \times$ P (a white ball), ($\frac{1}{2}$ Mark) $\therefore \frac{20 - x}{35} = 3 \times \frac{x}{35}$ (1 Mark) $\Rightarrow x = 5$ ($\frac{1}{2}$ Mark) Number of white balls = $5$ and number of red balls = $15$ ($\frac{1}{2}$ Mark) (b) P (a black ball) = $\frac{7}{35}$ ($\frac{1}{2}$ Mark) P (a white ball) = $\frac{5}{35}$ ($\frac{1}{2}$ Mark) P (either a black or a white ball) = $\frac{7 + 5}{35} = \frac{12}{35}$ (1 Mark)
Complement of event
1 Mark Questions
1051 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.
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(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
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
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.
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(A) Both Assertion (A) and Reason(R) are true and Reason (R) is the correct explanation of the Assertion (A).
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.
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(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
1111 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:
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.
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(D) Assertion (A) is false, but Reason (R) is true.
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
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:
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 pair of dice is thrown simultaneously. Let $E$ denote the event that "The sum of numbers obtained on both dice is at least 9." The number of outcomes in favour of event $E$ is :
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.
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(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
A die is thrown twice. What is the probability that (i) difference between two numbers obtained is 3? (ii) sum of the numbers obtained is 8?
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(i) $P(\text{difference between two numbers obtained is 3}) = \frac{6}{36}$ or $\frac{1}{6}$ [$1$ mark] (ii) $P(\text{sum of numbers obtained is 8}) = \frac{5}{36}$ [$1$ mark]
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.
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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 $\frac{2}{9}$
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 ?
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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)
Two dice are thrown at the same time. Determine the probability that the (i) sum of the numbers on the two dice is $5$, and (ii) difference of the numbers on the two dice is $3$.
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Total outcomes $= 36$ (1/2 Mark) (i) Number of outcomes with sum of the numbers on the two dice is $5 = 4$ $ (1,4) (4,1) (2,3) (3,2)$ (1 Mark) P(sum of the numbers on two dice is $5$) $= \frac{4}{36}$ or $\frac{1}{9}$ (1/2 Mark) (ii) Number of outcomes with difference of the numbers on the two dice is $3 = 6$ $ (1,4) (4,1) (5,2) (2,5) (6,3) (3,6)$ (1 Mark) P(difference of the numbers on the two dice is $3$) $= \frac{6}{36}$ or $\frac{1}{6}$ (1/2 Mark)
Two dice are rolled together. Find the probability that : (i) the sum of the numbers obtained is 10. (ii) the product of the numbers obtained is 6.
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(i) P(sum of the numbers obtained is 10) = $\frac{3}{36}$ or $\frac{1}{12}$ (1 1/2 Marks) (ii) P(product of the numbers obtained is 6) = $\frac{4}{36}$ or $\frac{1}{9}$ (1 1/2 Marks)
Two dice are rolled together. Find the probability that at least one of the numbers obtained is a multiple of $3$.
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Total number of outcomes = $36$ (1) Number of favourable outcomes = $20$ (1) P( at least one of the numbers obtained is multiple of $3$) = $\frac{20}{36}$ or $\frac{5}{9}$ (1)
Two dice are rolled together. Find the probability that (i) in the obtained outcomes one number is twice the another. (ii) both the numbers obtained are greater than $4$.
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(i) P ( one number is twice the other) = $\frac{6}{36}$ or $\frac{1}{6}$ (1.5 Mark) (ii) P (both the numbers obtained are greater than $4$) = $\frac{4}{36}$ or $\frac{1}{9}$ (1.5 Mark)
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 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
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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Total number of remaining cards = $51$ P (getting queen of heart) = $\frac{1}{51}$
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 $\frac{12}{13}$
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 $\frac{1}{2}$
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) = $\frac{19}{49}$
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 $\frac{1}{26}$
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 $\frac{9}{10}$
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 $\frac{1}{2}$
All the red face cards are removed from a pack of $52$ playing cards. The remaining cards are well shuffled and then a card is drawn at random. Find the probability of getting a : (i) red card (ii) a king or queen
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Total number of cards = $52 - 6 = 46$ (i) P (a red card) = $\frac{20}{46}$ or $\frac{10}{23}$ (1 Mark) (ii) P (a king or queen) = $\frac{4}{46}$ or $\frac{2}{23}$ (1 Mark)
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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After removing face cards of spades, total number of cards $$\begin{aligned}& = 52 - 3 = 49 \\ & (a) P(\text{a face card}) = \frac{9}{49} \\ & (b) P(\text{an ace or a jack}) = \frac{7}{49} \text{ or } \frac{1}{7}\end{aligned}$$
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?
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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.
A group of friends wanted to play cards with two identical packs together. nWhile shuffling the cards, three cards are dropped. Rest of the cards are nshuffled and one card is drawn at random. Assuming that the dropped ncards were a queen of hearts, a ten of spades and an ace of clubs, answer nthe following questions : n(i) Find the probability that the drawn card is a face card. n(ii) Find the probability that the drawn card is either a king or a nqueen. n(iii) (a) Do you think that the probability of getting a queen was nhigher if none of the cards were dropped? Justify your nanswer. nOR n(iii) (b) Find the probability that the drawn card is a jack. Compare nit with the probability when none of the cards were dropped. nIn which case is the probability of getting a jack higher?
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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 ndropped. 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 $101 < 104$ (1/2 Mark) Therefore probability of getting a jack is higher when $3$ cards were ndropped. (1 Mark)