CBSE Class 10 · Maths
CBSE Class 10 Maths Probability Competency Based Questions
Help your child excel in CBSE Class 10 Maths Probability with these expert-verified competency based questions, fully aligned with the latest CBSE board exam pattern. Each question — from MCQs to case studies and assertion-reason types — comes with step-by-step solutions to strengthen your child’s conceptual understanding and exam readiness.
CBSE Class 10 Maths Probability — Questions with Solutions
Which of the following numbers cannot be the probability of an event?
Solution
(c) $\frac{1}{0.5} = \frac{10}{5} = 2$ cannot be the value of probability, because the probability of any event must be greater than or equal to $0$ and less than or equal to $1$.
A dice is thrown once. Find the probability of getting a number less than $7$.
Solution
(b) Possible outcomes $= \{1, 2, 3, 4, 5, 6\}$. Total outcomes $= 6$. All numbers on a die are less than $7$, so favourable outcomes $= 6$. Therefore, $P = \frac{6}{6} = 1$.
For an event $E$, $P(E) + P(\overline{E}) = x$, then the value of $x^3 – 3$ is:
Solution
(a) Given: $P(E) + P(\overline{E}) = x$. By the complementary rule, $x = 1$. Now, $x^3 – 3 = (1)^3 – 3 = 1 – 3 = -2$.
The probability of getting a bad egg in a lot of $400$ is $0.035$. The number of bad eggs in the lot is:
Solution
(b) Total eggs $= 400$, $P(\text{bad egg}) = 0.035$. Number of bad eggs $= 0.035 \times 400 = 14$.
When a die is thrown, the probability of getting an even number less than $4$ is:
Solution
(d) Possible outcomes $= \{1, 2, 3, 4, 5, 6\}$. Even numbers less than $4 = \{2\}$. Favourable outcomes $= 1$. Therefore, $P = \frac{1}{6}$.
A die is thrown once. The probability of getting an odd prime number is:
Solution
(c) Possible outcomes $= \{1, 2, 3, 4, 5, 6\}$. Odd prime numbers $= \{3, 5\}$. Favourable outcomes $= 2$. Therefore, $P = \frac{2}{6} = \frac{1}{3}$.
A die is rolled once. The probability that a composite number comes up is:
Solution
(c) Possible outcomes $= \{1, 2, 3, 4, 5, 6\}$. Composite numbers $= \{4, 6\}$. Favourable outcomes $= 2$. Therefore, $P = \frac{2}{6} = \frac{1}{3}$.
The probability of getting an even number or a multiple of $3$ if an unbiased die is thrown, is:
Solution
(c) Total outcomes $= 6$. Even numbers $= \{2, 4, 6\}$, Multiples of $3 = \{3, 6\}$. Union $= \{2, 3, 4, 6\}$. Favourable outcomes $= 4$. Therefore, $P = \frac{4}{6} = \frac{2}{3}$.
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?
Solution
Total balls $= 5 + 8 + 7 = 20$. Favourable outcomes $=$ yellow balls $= 7$. Therefore, $P(\text{neither blue nor pink}) = \frac{7}{20}$.
A card is drawn at random from a well-shuffled deck of $52$ cards. The probability of getting a red card is:
Solution
(d) Total possible outcomes $= 52$. Red cards in a deck $= 26$. Therefore, $P = \frac{26}{52} = \frac{1}{2}$.
A card is drawn at random from a well-shuffled deck of $52$ playing cards. The probability of getting an ace of spade is:
Solution
Total possible outcomes $= 52$. There is only one ace of spade in the deck. Therefore, $P = \frac{1}{52}$.
A card is drawn at random from a well-shuffled deck of $52$ playing cards. The probability that it is a red king is:
Solution
(c) Total possible outcomes $= 52$. There are $26$ red cards, of which $2$ are red kings. Therefore, $P = \frac{2}{52} = \frac{1}{26}$.
A card is drawn at random from a well shuffled deck of $52$ playing cards. The probability of getting a face card is:
Solution
(a) Number of face cards in a deck $= 12$ (Jack, Queen, King in all 4 suits). Total outcomes $= 52$. Therefore, $P = \frac{12}{52} = \frac{3}{13}$.
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:
Solution
(b) $P(\text{ace}) = \frac{4}{52} = \frac{1}{13}$. Therefore, $P(\text{not ace}) = 1 – \frac{1}{13} = \frac{12}{13}$.
A card is drawn from a well shuffled deck of cards. What is the probability that the card drawn is neither a king nor a queen?
Solution
(a) Total cards $= 52$. Kings $= 4$, Queens $= 4$. Favourable outcomes $= 52 – 4 – 4 = 44$. Therefore, $P = \frac{44}{52} = \frac{11}{13}$.
$2$ cards of hearts and $4$ cards of spades are missing from a pack of $52$ cards. What is the probability of getting a black card from the remaining pack?
Solution
Total remaining cards $= 52 – 2 – 4 = 46$. Black cards originally $= 26$ (13 clubs + 13 spades). After removing $4$ spades, black cards $= 26 – 4 = 22$. Therefore, $P(\text{black card}) = \frac{22}{46} = \frac{11}{23}$.
Two fair coins are tossed. What is the probability of getting at most one head?
Solution
(a) Possible outcomes $= \{(HH), (HT), (TH), (TT)\}$. Total $= 4$. Favourable outcomes (at most one head) $= \{(HT), (TH), (TT)\} = 3$. Therefore, $P = \frac{3}{4}$.
Two coins are tossed together. The probability of getting at most two heads is:
Solution
(a) Possible outcomes $= \{(HH), (HT), (TH), (TT)\}$. All outcomes have at most $2$ heads. Favourable outcomes $= 4$. Therefore, $P = \frac{4}{4} = 1$.
Cards marked with numbers $2, 4, 6, 8, 10, \ldots, 50$ are placed in a bag and mixed thoroughly. One card is then drawn. What is the probability that the card is marked with a prime number?
Solution
(a) Total cards $= 25$ (even numbers from $2$ to $50$). Among these, the only prime number is $2$. Favourable outcomes $= 1$. Therefore, $P = \frac{1}{25}$.
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?
Solution
(c) Total outcomes $= 100$. Perfect cubes from $1$ to $100 = \{1, 8, 27, 64\}$. Favourable outcomes $= 4$. Therefore, $P = \frac{4}{100} = \frac{1}{25}$.
In the month of May, the weather forecast department gives the prediction of weather for the month of June. The given table shows the probabilities of forecast of different days:
Based on the above information, solve the following questions:
Q1. The number of sunny days in June is:
(a) $5$ (b) $10$ (c) $15$ (d) $20$
Q2. If the number of cloudy days in June is $5$, then $x =$
(a) $\frac{1}{4}$ (b) $\frac{1}{6}$ (c) $\frac{1}{8}$ (d) $\frac{1}{10}$
Q3. The probability that the day is not rainy, is:
(a) $\frac{13}{15}$ (b) $\frac{11}{15}$ (c) $\frac{1}{15}$ (d) None of these
Q4. If the sum of $x$ and $y$ is $\frac{3}{10}$, then the number of rainy days in June is:
(a) $1$ (b) $2$ (c) $3$ (d) $4$
Q5. Find the number of partially cloudy days:
(a) $2$ (b) $4$ (c) $6$ (d) $8$
Based on the above information, solve the following questions:
Q1. The number of sunny days in June is:
(a) $5$ (b) $10$ (c) $15$ (d) $20$
Q2. If the number of cloudy days in June is $5$, then $x =$
(a) $\frac{1}{4}$ (b) $\frac{1}{6}$ (c) $\frac{1}{8}$ (d) $\frac{1}{10}$
Q3. The probability that the day is not rainy, is:
(a) $\frac{13}{15}$ (b) $\frac{11}{15}$ (c) $\frac{1}{15}$ (d) None of these
Q4. If the sum of $x$ and $y$ is $\frac{3}{10}$, then the number of rainy days in June is:
(a) $1$ (b) $2$ (c) $3$ (d) $4$
Q5. Find the number of partially cloudy days:
(a) $2$ (b) $4$ (c) $6$ (d) $8$
Answer
Q1. (c): Total days in June $= 30$. Sunny days $= P(\text{sunny}) \times 30 = \frac{1}{2} \times 30 = 15$.
Q2. (b): Cloudy days $= 5$. Therefore, $x = \frac{5}{30} = \frac{1}{6}$.
Q3. (a): $P(\text{not rainy}) = \frac{1}{2} + \frac{1}{6} + \frac{1}{5} = \frac{15 + 5 + 6}{30} = \frac{26}{30} = \frac{13}{15}$.
Q4. (d): $x + y = \frac{3}{10}$, and $x = \frac{1}{6}$. So $y = \frac{3}{10} – \frac{1}{6} = \frac{9-5}{30} = \frac{4}{30} = \frac{2}{15}$. Rainy days $= \frac{2}{15} \times 30 = 4$.
Q5. (c): Partially cloudy days $= \frac{1}{5} \times 30 = 6$.
Q2. (b): Cloudy days $= 5$. Therefore, $x = \frac{5}{30} = \frac{1}{6}$.
Q3. (a): $P(\text{not rainy}) = \frac{1}{2} + \frac{1}{6} + \frac{1}{5} = \frac{15 + 5 + 6}{30} = \frac{26}{30} = \frac{13}{15}$.
Q4. (d): $x + y = \frac{3}{10}$, and $x = \frac{1}{6}$. So $y = \frac{3}{10} – \frac{1}{6} = \frac{9-5}{30} = \frac{4}{30} = \frac{2}{15}$. Rainy days $= \frac{2}{15} \times 30 = 4$.
Q5. (c): Partially cloudy days $= \frac{1}{5} \times 30 = 6$.
Two friends Soniya and Deepak have some savings in their piggy bank. They decided to count the total coins they both had. After counting they find they have fifty ₹1 coins, forty eight ₹2 coins, thirty six ₹5 coins, twenty eight ₹10 coins and eight ₹20 coins. Now, they said to Annu, their another friend, to choose a coin randomly. Find the probability that the coin chosen is:
Q1. ₹5 coin: (a) $\frac{17}{55}$ (b) $\frac{36}{85}$ (c) $\frac{18}{85}$ (d) $\frac{1}{5}$
Q2. ₹20 coin: (a) $\frac{13}{85}$ (b) $\frac{4}{85}$ (c) $\frac{3}{85}$ (d) $\frac{4}{15}$
Q3. Not a ₹10 coin: (a) $\frac{15}{31}$ (b) $\frac{36}{85}$ (c) $\frac{1}{5}$ (d) $\frac{71}{85}$
Q4. Of denomination of at least ₹10: (a) $\frac{18}{85}$ (b) $\frac{36}{85}$ (c) $\frac{1}{17}$ (d) $\frac{16}{85}$
Q5. Of denomination of at most ₹5: (a) $\frac{67}{85}$ (b) $\frac{36}{85}$ (c) $\frac{4}{85}$ (d) $\frac{18}{85}$
Q1. ₹5 coin: (a) $\frac{17}{55}$ (b) $\frac{36}{85}$ (c) $\frac{18}{85}$ (d) $\frac{1}{5}$
Q2. ₹20 coin: (a) $\frac{13}{85}$ (b) $\frac{4}{85}$ (c) $\frac{3}{85}$ (d) $\frac{4}{15}$
Q3. Not a ₹10 coin: (a) $\frac{15}{31}$ (b) $\frac{36}{85}$ (c) $\frac{1}{5}$ (d) $\frac{71}{85}$
Q4. Of denomination of at least ₹10: (a) $\frac{18}{85}$ (b) $\frac{36}{85}$ (c) $\frac{1}{17}$ (d) $\frac{16}{85}$
Q5. Of denomination of at most ₹5: (a) $\frac{67}{85}$ (b) $\frac{36}{85}$ (c) $\frac{4}{85}$ (d) $\frac{18}{85}$
Answer
Total coins $= 50 + 48 + 36 + 28 + 8 = 170$
Q1. (c): ₹5 coins $= 36$. $P = \frac{36}{170} = \frac{18}{85}$.
Q2. (b): ₹20 coins $= 8$. $P = \frac{8}{170} = \frac{4}{85}$.
Q3. (d): $P(\text{₹10 coin}) = \frac{28}{170}$. $P(\text{not ₹10}) = 1 – \frac{28}{170} = \frac{142}{170} = \frac{71}{85}$.
Q4. (a): Coins of ₹10 and ₹20 $= 28 + 8 = 36$. $P = \frac{36}{170} = \frac{18}{85}$.
Q5. (a): Coins of ₹1, ₹2, ₹5 $= 50 + 48 + 36 = 134$. $P = \frac{134}{170} = \frac{67}{85}$.
Q1. (c): ₹5 coins $= 36$. $P = \frac{36}{170} = \frac{18}{85}$.
Q2. (b): ₹20 coins $= 8$. $P = \frac{8}{170} = \frac{4}{85}$.
Q3. (d): $P(\text{₹10 coin}) = \frac{28}{170}$. $P(\text{not ₹10}) = 1 – \frac{28}{170} = \frac{142}{170} = \frac{71}{85}$.
Q4. (a): Coins of ₹10 and ₹20 $= 28 + 8 = 36$. $P = \frac{36}{170} = \frac{18}{85}$.
Q5. (a): Coins of ₹1, ₹2, ₹5 $= 50 + 48 + 36 = 134$. $P = \frac{134}{170} = \frac{67}{85}$.
Blood group describes the type of blood a person has. It is a classification of blood based on the presence or absence of inherited antigenic substances on the surface of red blood cells. Blood types predict whether a serious reaction will occur in a blood transfusion.
In a sample of $50$ people, $21$ had type O blood, $22$ had type A, $5$ had type B, and the rest had type AB blood group.
Based on the given information, solve the following questions:
Q1. What is the probability that a person chosen at random had type O blood?
Q2. What is the probability that a person chosen at random had type AB blood group?
Q3. What is the probability that a person chosen at random had neither type A nor type B blood group?
Or
What is the probability that the person chosen at random had either type A, type B, or type O blood group?
In a sample of $50$ people, $21$ had type O blood, $22$ had type A, $5$ had type B, and the rest had type AB blood group.
Based on the given information, solve the following questions:
Q1. What is the probability that a person chosen at random had type O blood?
Q2. What is the probability that a person chosen at random had type AB blood group?
Q3. What is the probability that a person chosen at random had neither type A nor type B blood group?
Or
What is the probability that the person chosen at random had either type A, type B, or type O blood group?
Answer
Total people $= 50$, Type O $= 21$, Type A $= 22$, Type B $= 5$.
Number with type AB $= 50 – (21 + 22 + 5) = 2$.
Q1. Total outcomes $= 50$. Favourable outcomes $= 21$.
$$P(\text{type O}) = \frac{21}{50}$$
Q2. Favourable outcomes $= 2$.
$$P(\text{type AB}) = \frac{2}{50} = \frac{1}{25}$$
Q3. Favourable outcomes (neither A nor B) $= 50 – (22 + 5) = 23$.
$$P(\text{neither A nor B}) = \frac{23}{50}$$
Or: Favourable outcomes (A or B or O) $= 22 + 5 + 21 = 48$.
$$P(\text{A or B or O}) = \frac{48}{50} = \frac{24}{25}$$
Number with type AB $= 50 – (21 + 22 + 5) = 2$.
Q1. Total outcomes $= 50$. Favourable outcomes $= 21$.
$$P(\text{type O}) = \frac{21}{50}$$
Q2. Favourable outcomes $= 2$.
$$P(\text{type AB}) = \frac{2}{50} = \frac{1}{25}$$
Q3. Favourable outcomes (neither A nor B) $= 50 – (22 + 5) = 23$.
$$P(\text{neither A nor B}) = \frac{23}{50}$$
Or: Favourable outcomes (A or B or O) $= 22 + 5 + 21 = 48$.
$$P(\text{A or B or O}) = \frac{48}{50} = \frac{24}{25}$$
Sunny goes to a store to purchase juice cartons for his shop. The store has $80$ cartons of litchi juice, $90$ cartons of pineapple juice, $38$ cartons of mango juice and $42$ cartons of banana juice. If Sunny chooses a carton at random:
Q1. Find the probability that the selected carton is of pineapple juice.
Q2. What is the probability that the selected carton is of banana juice?
Q3. Sunny buys $4$ cartons of pineapple juice, $3$ cartons of litchi juice and $3$ cartons of banana juice. A customer picks a tetrapack of juice at random. Find the probability that the customer picks a banana juice, if each carton has $10$ tetrapacks.
Or
If the storekeeper bought $14$ more cartons of pineapple juice, find the probability of selecting a tetrapack of pineapple juice from the store.
Q1. Find the probability that the selected carton is of pineapple juice.
Q2. What is the probability that the selected carton is of banana juice?
Q3. Sunny buys $4$ cartons of pineapple juice, $3$ cartons of litchi juice and $3$ cartons of banana juice. A customer picks a tetrapack of juice at random. Find the probability that the customer picks a banana juice, if each carton has $10$ tetrapacks.
Or
If the storekeeper bought $14$ more cartons of pineapple juice, find the probability of selecting a tetrapack of pineapple juice from the store.
Answer
Total cartons in store $= 80 + 90 + 38 + 42 = 250$
Q1. Pineapple cartons $= 90$.
$$P(\text{pineapple}) = \frac{90}{250} = \frac{9}{25}$$
Q2. Banana cartons $= 42$.
$$P(\text{banana}) = \frac{42}{250} = \frac{21}{125}$$
Q3. Total cartons Sunny bought $= 4 + 3 + 3 = 10$. Total tetrapacks $= 10 \times 10 = 100$. Banana tetrapacks $= 3 \times 10 = 30$.
$$P(\text{banana tetrapack}) = \frac{30}{100} = \frac{3}{10}$$
Or: Cartons remaining with storekeeper $= 250 – 10 = 240$. After buying $14$ more: $240 + 14 = 254$. Pineapple cartons now $= (90 – 4 + 14) = 100$. Total tetrapacks $= 254 \times 10$.
$$P(\text{pineapple tetrapack}) = \frac{100 \times 10}{254 \times 10} = \frac{100}{254} = \frac{50}{127}$$
Q1. Pineapple cartons $= 90$.
$$P(\text{pineapple}) = \frac{90}{250} = \frac{9}{25}$$
Q2. Banana cartons $= 42$.
$$P(\text{banana}) = \frac{42}{250} = \frac{21}{125}$$
Q3. Total cartons Sunny bought $= 4 + 3 + 3 = 10$. Total tetrapacks $= 10 \times 10 = 100$. Banana tetrapacks $= 3 \times 10 = 30$.
$$P(\text{banana tetrapack}) = \frac{30}{100} = \frac{3}{10}$$
Or: Cartons remaining with storekeeper $= 250 – 10 = 240$. After buying $14$ more: $240 + 14 = 254$. Pineapple cartons now $= (90 – 4 + 14) = 100$. Total tetrapacks $= 254 \times 10$.
$$P(\text{pineapple tetrapack}) = \frac{100 \times 10}{254 \times 10} = \frac{100}{254} = \frac{50}{127}$$
Vivek is very fond of collecting balls of different colours. He has a total of $25$ balls in his basket, out of which five balls are red in colour and eight are white. Out of the remaining balls, some are green in colour and the rest are pink.
Q1. If the probability of drawing a pink ball is twice the probability of drawing a green ball, then find the number of pink balls.
Q2. Find the probability of drawing a ball of colour other than green.
Q3. Find the probability of drawing either a green or white ball.
Or
What is the probability that the drawn ball is neither a pink nor a white ball?
Q1. If the probability of drawing a pink ball is twice the probability of drawing a green ball, then find the number of pink balls.
Q2. Find the probability of drawing a ball of colour other than green.
Q3. Find the probability of drawing either a green or white ball.
Or
What is the probability that the drawn ball is neither a pink nor a white ball?
Answer
Q1. Red + White $= 5 + 8 = 13$. Green + Pink $= 25 – 13 = 12$.
Let green balls $= g$, pink balls $= p = 12 – g$.
Given $P(\text{pink}) = 2 \times P(\text{green})$, i.e. $\frac{12-g}{25} = 2 \cdot \frac{g}{25}$.
So $12 – g = 2g \Rightarrow 3g = 12 \Rightarrow g = 4$. Pink balls $= 12 – 4 = \mathbf{8}$.
Q2. Green balls $= 4$. Balls other than green $= 25 – 4 = 21$.
$$P(\text{not green}) = \frac{21}{25}$$
Q3. Green $= 4$, White $= 8$. Green or white $= 12$.
$$P(\text{green or white}) = \frac{12}{25}$$
Or: Red $= 5$, Green $= 4$. Neither pink nor white $= 5 + 4 = 9$.
$$P(\text{neither pink nor white}) = \frac{9}{25}$$
Let green balls $= g$, pink balls $= p = 12 – g$.
Given $P(\text{pink}) = 2 \times P(\text{green})$, i.e. $\frac{12-g}{25} = 2 \cdot \frac{g}{25}$.
So $12 – g = 2g \Rightarrow 3g = 12 \Rightarrow g = 4$. Pink balls $= 12 – 4 = \mathbf{8}$.
Q2. Green balls $= 4$. Balls other than green $= 25 – 4 = 21$.
$$P(\text{not green}) = \frac{21}{25}$$
Q3. Green $= 4$, White $= 8$. Green or white $= 12$.
$$P(\text{green or white}) = \frac{12}{25}$$
Or: Red $= 5$, Green $= 4$. Neither pink nor white $= 5 + 4 = 9$.
$$P(\text{neither pink nor white}) = \frac{9}{25}$$
In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): When two coins are tossed together then the probability of getting no tail is $\dfrac{1}{4}$.
Reason (R): The probability of getting a head (i.e., no tail) in one toss of a coin is $\dfrac{1}{2}$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): When two coins are tossed together then the probability of getting no tail is $\dfrac{1}{4}$.
Reason (R): The probability of getting a head (i.e., no tail) in one toss of a coin is $\dfrac{1}{2}$.
Answer
(b) Assertion (A): Total possible outcomes $= 2 \times 2 = 4$. Favourable outcomes of getting no tail $= 1$ i.e., $(H, H)$. $P = \frac{1}{4}$. So, A is true.
Reason (R): Total possible outcomes $= 2$. Favourable outcomes $= 1$ i.e., $(H)$. $P = \frac{1}{2}$. So, R is true.
Both A and R are true but Reason (R) is not the correct explanation of Assertion (A). (R explains one coin; A is about two coins.)
Reason (R): Total possible outcomes $= 2$. Favourable outcomes $= 1$ i.e., $(H)$. $P = \frac{1}{2}$. So, R is true.
Both A and R are true but Reason (R) is not the correct explanation of Assertion (A). (R explains one coin; A is about two coins.)
In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): The probability that a leap year has $53$ Sundays is $\dfrac{2}{7}$.
Reason (R): The probability that a non-leap year has $53$ Sundays is $\dfrac{5}{7}$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): The probability that a leap year has $53$ Sundays is $\dfrac{2}{7}$.
Reason (R): The probability that a non-leap year has $53$ Sundays is $\dfrac{5}{7}$.
Answer
(c) Assertion (A): A leap year has $366$ days $= 52$ complete weeks $+ 2$ extra days. These $2$ extra days can be any two consecutive days of the week, so there are $7$ possibilities and $2$ of them include Sunday. $P = \frac{2}{7}$. So, A is true.
Reason (R): A non-leap year has $365$ days $= 52$ weeks $+ 1$ extra day. Only $1$ out of $7$ days can be Sunday, so $P = \frac{1}{7}$, not $\frac{5}{7}$. So, R is false.
Hence, A is true but R is false.
Reason (R): A non-leap year has $365$ days $= 52$ weeks $+ 1$ extra day. Only $1$ out of $7$ days can be Sunday, so $P = \frac{1}{7}$, not $\frac{5}{7}$. So, R is false.
Hence, A is true but R is false.
In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): Two players Sania and Deepika play a tennis match. If the probability of Sania winning the match is $0.68$, then the probability of Deepika winning the match is $0.32$.
Reason (R): The sum of the probability of two complementary events is $1$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): Two players Sania and Deepika play a tennis match. If the probability of Sania winning the match is $0.68$, then the probability of Deepika winning the match is $0.32$.
Reason (R): The sum of the probability of two complementary events is $1$.
Answer
(a) Assertion (A): Let $E$ = event of Sania winning. $P(E) = 0.68$. Since Sania and Deepika’s winning are complementary events, $P(E’) = 1 – 0.68 = 0.32$. So, A is true.
Reason (R): By the complement rule, $P(E) + P(E’) = 1$. So, R is true.
Both A and R are true and Reason (R) is the correct explanation of Assertion (A).
Reason (R): By the complement rule, $P(E) + P(E’) = 1$. So, R is true.
Both A and R are true and Reason (R) is the correct explanation of Assertion (A).
In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): Cards numbered $5$ to $102$ are placed in a box. If a card is selected at random from the box, then the probability that the card selected has a number which is a perfect square is $\dfrac{4}{49}$.
Reason (R): Probability of an event $E$ is a number $P(E)$ such that $0 \leq P(E) \leq 1$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): Cards numbered $5$ to $102$ are placed in a box. If a card is selected at random from the box, then the probability that the card selected has a number which is a perfect square is $\dfrac{4}{49}$.
Reason (R): Probability of an event $E$ is a number $P(E)$ such that $0 \leq P(E) \leq 1$.
Answer
(a) Assertion (A): Total cards $= 102 – 5 + 1 = 98$. Perfect squares from $5$ to $102 = \{9, 16, 25, 36, 49, 64, 81, 100\}$ i.e., $8$. $P(E) = \frac{8}{98} = \frac{4}{49}$. So, A is true.
Reason (R): The probability of any event always lies between $0$ and $1$ inclusive. So, R is true.
Both A and R are true and R is the correct explanation of A.
Reason (R): The probability of any event always lies between $0$ and $1$ inclusive. So, R is true.
Both A and R are true and R is the correct explanation of A.
In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): The probability of getting a prime number, when a die is thrown once, is $\dfrac{2}{3}$.
Reason (R): On the faces of a die, prime numbers are $2, 3, 5$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): The probability of getting a prime number, when a die is thrown once, is $\dfrac{2}{3}$.
Reason (R): On the faces of a die, prime numbers are $2, 3, 5$.
Answer
(d) Assertion (A): Total outcomes $= 6$. Prime numbers on a die $= \{2, 3, 5\}$, i.e., $3$ outcomes. $P = \frac{3}{6} = \frac{1}{2}$, not $\frac{2}{3}$. So, A is false.
Reason (R): Prime numbers on a standard die are indeed $2, 3, 5$. So, R is true.
Hence, A is false but R is true.
Reason (R): Prime numbers on a standard die are indeed $2, 3, 5$. So, R is true.
Hence, A is false but R is true.
Two friends Richa and Sohan have some savings in their piggy bank. They decided to count the total coins they both had. After counting they find that they have fifty ₹1 coins, forty eight ₹2 coins, thirty six ₹5 coins, twenty eight ₹10 coins and eight ₹20 coins. Now, they said to Nisha, their another friend, to choose a coin randomly. Find the probability that the coin chosen is:
(i) ₹5 coin: (a) $\frac{17}{55}$ (b) $\frac{36}{85}$ (c) $\frac{18}{85}$ (d) $\frac{1}{15}$
(ii) ₹20 coin: (a) $\frac{13}{85}$ (b) $\frac{4}{85}$ (c) $\frac{3}{85}$ (d) $\frac{4}{15}$
(iii) Not a ₹10 coin: (a) $\frac{15}{31}$ (b) $\frac{36}{85}$ (c) $\frac{1}{5}$ (d) $\frac{71}{85}$
(iv) Of denomination of at least ₹10: (a) $\frac{18}{85}$ (b) $\frac{36}{85}$ (c) $\frac{1}{17}$ (d) $\frac{16}{85}$
(v) Of denomination of at most ₹5: (a) $\frac{67}{85}$ (b) $\frac{36}{85}$ (c) $\frac{4}{85}$ (d) $\frac{18}{85}$
(i) ₹5 coin: (a) $\frac{17}{55}$ (b) $\frac{36}{85}$ (c) $\frac{18}{85}$ (d) $\frac{1}{15}$
(ii) ₹20 coin: (a) $\frac{13}{85}$ (b) $\frac{4}{85}$ (c) $\frac{3}{85}$ (d) $\frac{4}{15}$
(iii) Not a ₹10 coin: (a) $\frac{15}{31}$ (b) $\frac{36}{85}$ (c) $\frac{1}{5}$ (d) $\frac{71}{85}$
(iv) Of denomination of at least ₹10: (a) $\frac{18}{85}$ (b) $\frac{36}{85}$ (c) $\frac{1}{17}$ (d) $\frac{16}{85}$
(v) Of denomination of at most ₹5: (a) $\frac{67}{85}$ (b) $\frac{36}{85}$ (c) $\frac{4}{85}$ (d) $\frac{18}{85}$
Answer
Total coins $= 50 + 48 + 36 + 28 + 8 = 170$
(i) (c): ₹5 coins $= 36$. $P = \frac{36}{170} = \frac{18}{85}$.
(ii) (b): ₹20 coins $= 8$. $P = \frac{8}{170} = \frac{4}{85}$.
(iii) (d): $P(\text{₹10 coin}) = \frac{28}{170}$. $P(\text{not ₹10}) = 1 – \frac{28}{170} = \frac{142}{170} = \frac{71}{85}$.
(iv) (a): ₹10 and ₹20 coins $= 28 + 8 = 36$. $P = \frac{36}{170} = \frac{18}{85}$.
(v) (a): ₹1, ₹2, ₹5 coins $= 50 + 48 + 36 = 134$. $P = \frac{134}{170} = \frac{67}{85}$.
(i) (c): ₹5 coins $= 36$. $P = \frac{36}{170} = \frac{18}{85}$.
(ii) (b): ₹20 coins $= 8$. $P = \frac{8}{170} = \frac{4}{85}$.
(iii) (d): $P(\text{₹10 coin}) = \frac{28}{170}$. $P(\text{not ₹10}) = 1 – \frac{28}{170} = \frac{142}{170} = \frac{71}{85}$.
(iv) (a): ₹10 and ₹20 coins $= 28 + 8 = 36$. $P = \frac{36}{170} = \frac{18}{85}$.
(v) (a): ₹1, ₹2, ₹5 coins $= 50 + 48 + 36 = 134$. $P = \frac{134}{170} = \frac{67}{85}$.
In a play zone, Nishtha is playing a claw crane game which consists of $58$ teddy bears, $42$ pokemons, $36$ tigers and $64$ monkeys. Nishtha picks a puppet at random. Now, find the probability of getting:
(i) A tiger: (a) $\frac{3}{50}$ (b) $\frac{9}{50}$ (c) $\frac{1}{25}$ (d) $\frac{27}{50}$
(ii) A monkey: (a) $\frac{8}{25}$ (b) $\frac{4}{25}$ (c) $\frac{16}{25}$ (d) $\frac{1}{5}$
(iii) A teddy bear: (a) $\frac{41}{50}$ (b) $\frac{29}{50}$ (c) $\frac{29}{100}$ (d) $\frac{41}{100}$
(iv) Not a monkey: (a) $\frac{1}{25}$ (b) $\frac{8}{25}$ (c) $\frac{13}{25}$ (d) $\frac{17}{25}$
(v) Not a pokemon: (a) $\frac{27}{100}$ (b) $\frac{43}{100}$ (c) $\frac{61}{100}$ (d) $\frac{79}{100}$
(i) A tiger: (a) $\frac{3}{50}$ (b) $\frac{9}{50}$ (c) $\frac{1}{25}$ (d) $\frac{27}{50}$
(ii) A monkey: (a) $\frac{8}{25}$ (b) $\frac{4}{25}$ (c) $\frac{16}{25}$ (d) $\frac{1}{5}$
(iii) A teddy bear: (a) $\frac{41}{50}$ (b) $\frac{29}{50}$ (c) $\frac{29}{100}$ (d) $\frac{41}{100}$
(iv) Not a monkey: (a) $\frac{1}{25}$ (b) $\frac{8}{25}$ (c) $\frac{13}{25}$ (d) $\frac{17}{25}$
(v) Not a pokemon: (a) $\frac{27}{100}$ (b) $\frac{43}{100}$ (c) $\frac{61}{100}$ (d) $\frac{79}{100}$
Answer
Total puppets $= 58 + 42 + 36 + 64 = 200$
(i) (b): $P(\text{tiger}) = \frac{36}{200} = \frac{9}{50}$.
(ii) (a): $P(\text{monkey}) = \frac{64}{200} = \frac{8}{25}$.
(iii) (c): $P(\text{teddy bear}) = \frac{58}{200} = \frac{29}{100}$.
(iv) (d): $P(\text{not monkey}) = 1 – \frac{8}{25} = \frac{17}{25}$.
(v) (d): $P(\text{pokemon}) = \frac{42}{200} = \frac{21}{100}$. $P(\text{not pokemon}) = 1 – \frac{21}{100} = \frac{79}{100}$.
(i) (b): $P(\text{tiger}) = \frac{36}{200} = \frac{9}{50}$.
(ii) (a): $P(\text{monkey}) = \frac{64}{200} = \frac{8}{25}$.
(iii) (c): $P(\text{teddy bear}) = \frac{58}{200} = \frac{29}{100}$.
(iv) (d): $P(\text{not monkey}) = 1 – \frac{8}{25} = \frac{17}{25}$.
(v) (d): $P(\text{pokemon}) = \frac{42}{200} = \frac{21}{100}$. $P(\text{not pokemon}) = 1 – \frac{21}{100} = \frac{79}{100}$.
Rohit wants to distribute chocolates in his class on his birthday. The chocolates are of three types: Milk chocolate, White chocolate, and Dark chocolate. If the total number of students in the class is $54$ and everyone gets a chocolate, then answer the following questions.
(i) If the probability of distributing milk chocolates is $\frac{1}{3}$, then the number of milk chocolates Rohit has is:
(a) $18$ (b) $20$ (c) $22$ (d) $30$
(ii) If the probability of distributing dark chocolates is $\frac{4}{9}$, then the number of dark chocolates Rohit has is:
(a) $18$ (b) $25$ (c) $24$ (d) $36$
(iii) The probability of distributing white chocolates is:
(a) $\frac{11}{27}$ (b) $\frac{8}{21}$ (c) $\frac{1}{9}$ (d) $\frac{2}{9}$
(iv) The probability of distributing both milk and white chocolates is:
(a) $\frac{3}{17}$ (b) $\frac{5}{9}$ (c) $\frac{1}{3}$ (d) $\frac{1}{27}$
(v) The probability of distributing all the chocolates is:
(a) $0$ (b) $1$ (c) $\frac{1}{2}$ (d) $\frac{3}{4}$
(i) If the probability of distributing milk chocolates is $\frac{1}{3}$, then the number of milk chocolates Rohit has is:
(a) $18$ (b) $20$ (c) $22$ (d) $30$
(ii) If the probability of distributing dark chocolates is $\frac{4}{9}$, then the number of dark chocolates Rohit has is:
(a) $18$ (b) $25$ (c) $24$ (d) $36$
(iii) The probability of distributing white chocolates is:
(a) $\frac{11}{27}$ (b) $\frac{8}{21}$ (c) $\frac{1}{9}$ (d) $\frac{2}{9}$
(iv) The probability of distributing both milk and white chocolates is:
(a) $\frac{3}{17}$ (b) $\frac{5}{9}$ (c) $\frac{1}{3}$ (d) $\frac{1}{27}$
(v) The probability of distributing all the chocolates is:
(a) $0$ (b) $1$ (c) $\frac{1}{2}$ (d) $\frac{3}{4}$
Answer
Since every student gets one chocolate, total chocolates $= 54$.
(i) (a): Let milk chocolates $= x$. $\frac{x}{54} = \frac{1}{3} \Rightarrow x = 18$.
(ii) (c): Let dark chocolates $= y$. $\frac{y}{54} = \frac{4}{9} \Rightarrow y = 24$.
(iii) (d): White chocolates $= 54 – (18 + 24) = 12$. $P = \frac{12}{54} = \frac{2}{9}$.
(iv) (b): Milk + White $= 18 + 12 = 30$. $P = \frac{30}{54} = \frac{5}{9}$.
(v) (b): All $54$ chocolates are distributed. $P = \frac{54}{54} = 1$.
(i) (a): Let milk chocolates $= x$. $\frac{x}{54} = \frac{1}{3} \Rightarrow x = 18$.
(ii) (c): Let dark chocolates $= y$. $\frac{y}{54} = \frac{4}{9} \Rightarrow y = 24$.
(iii) (d): White chocolates $= 54 – (18 + 24) = 12$. $P = \frac{12}{54} = \frac{2}{9}$.
(iv) (b): Milk + White $= 18 + 12 = 30$. $P = \frac{30}{54} = \frac{5}{9}$.
(v) (b): All $54$ chocolates are distributed. $P = \frac{54}{54} = 1$.
In a party, some children decided to play the musical chair game. In the game, the person playing the music has been advised to stop the music at any time in the interval of $3$ mins after he starts the music in each turn. On the basis of the given information, answer the following questions.
(i) What is the probability that the music will stop within the first $30$ secs after starting?
(a) $\frac{1}{6}$ (b) $\frac{1}{5}$ (c) $\frac{1}{4}$ (d) $\frac{1}{3}$
(ii) The probability that the music will stop within $45$ secs after starting is:
(a) $\frac{1}{4}$ (b) $\frac{1}{5}$ (c) $\frac{1}{6}$ (d) $\frac{1}{8}$
(iii) The probability that the music will stop after $2$ mins after starting is:
(a) $\frac{1}{8}$ (b) $\frac{1}{5}$ (c) $\frac{1}{4}$ (d) $\frac{1}{3}$
(iv) The probability that the music will not stop within the first $60$ secs after starting is:
(a) $\frac{1}{3}$ (b) $\frac{2}{3}$ (c) $\frac{4}{5}$ (d) $\frac{8}{9}$
(v) The probability that the music will stop within the first $82$ secs after starting is:
(a) $\frac{11}{30}$ (b) $\frac{41}{90}$ (c) $\frac{31}{35}$ (d) $\frac{41}{93}$
(i) What is the probability that the music will stop within the first $30$ secs after starting?
(a) $\frac{1}{6}$ (b) $\frac{1}{5}$ (c) $\frac{1}{4}$ (d) $\frac{1}{3}$
(ii) The probability that the music will stop within $45$ secs after starting is:
(a) $\frac{1}{4}$ (b) $\frac{1}{5}$ (c) $\frac{1}{6}$ (d) $\frac{1}{8}$
(iii) The probability that the music will stop after $2$ mins after starting is:
(a) $\frac{1}{8}$ (b) $\frac{1}{5}$ (c) $\frac{1}{4}$ (d) $\frac{1}{3}$
(iv) The probability that the music will not stop within the first $60$ secs after starting is:
(a) $\frac{1}{3}$ (b) $\frac{2}{3}$ (c) $\frac{4}{5}$ (d) $\frac{8}{9}$
(v) The probability that the music will stop within the first $82$ secs after starting is:
(a) $\frac{11}{30}$ (b) $\frac{41}{90}$ (c) $\frac{31}{35}$ (d) $\frac{41}{93}$
Answer
Total time $= 3$ min $= 3 \times 60 = 180$ secs.
(i) (a): $P = \frac{30}{180} = \frac{1}{6}$.
(ii) (a): $P = \frac{45}{180} = \frac{1}{4}$.
(iii) (d): $P(\text{within 2 min}) = \frac{120}{180} = \frac{2}{3}$. $P(\text{after 2 min}) = 1 – \frac{2}{3} = \frac{1}{3}$.
(iv) (b): $P(\text{not in first 60 sec}) = 1 – \frac{60}{180} = 1 – \frac{1}{3} = \frac{2}{3}$.
(v) (b): $P = \frac{82}{180} = \frac{41}{90}$.
(i) (a): $P = \frac{30}{180} = \frac{1}{6}$.
(ii) (a): $P = \frac{45}{180} = \frac{1}{4}$.
(iii) (d): $P(\text{within 2 min}) = \frac{120}{180} = \frac{2}{3}$. $P(\text{after 2 min}) = 1 – \frac{2}{3} = \frac{1}{3}$.
(iv) (b): $P(\text{not in first 60 sec}) = 1 – \frac{60}{180} = 1 – \frac{1}{3} = \frac{2}{3}$.
(v) (b): $P = \frac{82}{180} = \frac{41}{90}$.
Three persons toss $3$ coins simultaneously and note the outcomes. Then, they ask a few questions to one another. Help them in finding the answers to the following questions.
(i) The probability of getting at most one tail is:
(a) $0$ (b) $1$ (c) $\frac{1}{2}$ (d) $\frac{1}{4}$
(ii) The probability of getting exactly $1$ head is:
(a) $\frac{1}{2}$ (b) $\frac{1}{4}$ (c) $\frac{1}{8}$ (d) $\frac{3}{8}$
(iii) The probability of getting exactly $3$ tails is:
(a) $0$ (b) $1$ (c) $\frac{1}{4}$ (d) $\frac{1}{8}$
(iv) The probability of getting at most $3$ heads is:
(a) $0$ (b) $1$ (c) $\frac{1}{2}$ (d) $\frac{1}{8}$
(v) The probability of getting at least two heads is:
(a) $0$ (b) $1$ (c) $\frac{1}{2}$ (d) $\frac{1}{4}$
(i) The probability of getting at most one tail is:
(a) $0$ (b) $1$ (c) $\frac{1}{2}$ (d) $\frac{1}{4}$
(ii) The probability of getting exactly $1$ head is:
(a) $\frac{1}{2}$ (b) $\frac{1}{4}$ (c) $\frac{1}{8}$ (d) $\frac{3}{8}$
(iii) The probability of getting exactly $3$ tails is:
(a) $0$ (b) $1$ (c) $\frac{1}{4}$ (d) $\frac{1}{8}$
(iv) The probability of getting at most $3$ heads is:
(a) $0$ (b) $1$ (c) $\frac{1}{2}$ (d) $\frac{1}{8}$
(v) The probability of getting at least two heads is:
(a) $0$ (b) $1$ (c) $\frac{1}{2}$ (d) $\frac{1}{4}$
Answer
Sample space $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$, so $n(S) = 8$.
(i) (c): At most one tail: $\{HHH, HHT, HTH, THH\}$ → $n = 4$. $P = \frac{4}{8} = \frac{1}{2}$.
(ii) (d): Exactly 1 head: $\{HTT, THT, TTH\}$ → $n = 3$. $P = \frac{3}{8}$.
(iii) (d): Exactly 3 tails: $\{TTT\}$ → $n = 1$. $P = \frac{1}{8}$.
(iv) (b): At most 3 heads includes all $8$ outcomes. $P = \frac{8}{8} = 1$.
(v) (c): At least 2 heads: $\{HHT, HTH, THH, HHH\}$ → $n = 4$. $P = \frac{4}{8} = \frac{1}{2}$.
(i) (c): At most one tail: $\{HHH, HHT, HTH, THH\}$ → $n = 4$. $P = \frac{4}{8} = \frac{1}{2}$.
(ii) (d): Exactly 1 head: $\{HTT, THT, TTH\}$ → $n = 3$. $P = \frac{3}{8}$.
(iii) (d): Exactly 3 tails: $\{TTT\}$ → $n = 1$. $P = \frac{1}{8}$.
(iv) (b): At most 3 heads includes all $8$ outcomes. $P = \frac{8}{8} = 1$.
(v) (c): At least 2 heads: $\{HHT, HTH, THH, HHH\}$ → $n = 4$. $P = \frac{4}{8} = \frac{1}{2}$.
In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): Three unbiased coins are tossed together, then the probability of getting exactly $1$ head is $\dfrac{3}{8}$.
Reason (R): Favourable number of outcomes do not lie in the sample space of total number of outcomes.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): Three unbiased coins are tossed together, then the probability of getting exactly $1$ head is $\dfrac{3}{8}$.
Reason (R): Favourable number of outcomes do not lie in the sample space of total number of outcomes.
Answer
(c) Assertion (A): Sample space $= \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$, total $= 8$. Event of exactly $1$ head $= \{HTT, THT, TTH\}$ i.e., $3$. $P = \frac{3}{8}$. So, A is true.
Reason (R): Favourable outcomes always lie within the sample space. So, R is false.
Hence, A is true but R is false.
Reason (R): Favourable outcomes always lie within the sample space. So, R is false.
Hence, A is true but R is false.
In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): In a game, the entry fee is ₹10. The game consists of tossing $3$ coins. If one or two heads show, Amita wins the game and gets entry fee back. The probability that she gets the entry fee is $\dfrac{3}{4}$.
Reason (R): When three coins are tossed together, all the outcomes are $\{HHH, HHT, HTH, HTT, TTH, THT, TTT\}$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): In a game, the entry fee is ₹10. The game consists of tossing $3$ coins. If one or two heads show, Amita wins the game and gets entry fee back. The probability that she gets the entry fee is $\dfrac{3}{4}$.
Reason (R): When three coins are tossed together, all the outcomes are $\{HHH, HHT, HTH, HTT, TTH, THT, TTT\}$.
Answer
(a) Assertion (A): Total outcomes when $3$ coins are tossed $= 8$. Favourable outcomes (one or two heads) $= \{HHT, HTH, THH, HTT, THT, TTH\}$ i.e., $6$. $P = \frac{6}{8} = \frac{3}{4}$. So, A is true.
Reason (R): The complete sample space is $\{HHH, HHT, HTH, HTT, TTH, THT, TTH, TTT\}$ — all $8$ outcomes. So, R is true.
Both A and R are true and R is the correct explanation of A.
Reason (R): The complete sample space is $\{HHH, HHT, HTH, HTT, TTH, THT, TTH, TTT\}$ — all $8$ outcomes. So, R is true.
Both A and R are true and R is the correct explanation of A.
In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): If a box contains $5$ white, $2$ red and $4$ black marbles, then the probability of not drawing a white marble from the box is $\dfrac{5}{11}$.
Reason (R): $P(\bar{E}) = 1 – P(E)$, where $E$ is any event.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): If a box contains $5$ white, $2$ red and $4$ black marbles, then the probability of not drawing a white marble from the box is $\dfrac{5}{11}$.
Reason (R): $P(\bar{E}) = 1 – P(E)$, where $E$ is any event.
Answer
(d) Assertion (A): Total marbles $= 5 + 2 + 4 = 11$. $P(\text{white}) = \frac{5}{11}$. $P(\text{not white}) = 1 – \frac{5}{11} = \frac{6}{11}$, not $\frac{5}{11}$. So, A is false.
Reason (R): The complement rule $P(\bar{E}) = 1 – P(E)$ is a true statement. So, R is true.
Hence, A is false but R is true.
Reason (R): The complement rule $P(\bar{E}) = 1 – P(E)$ is a true statement. So, R is true.
Hence, A is false but R is true.
In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Let $A$ and $B$ be two independent events.
Assertion (A): If $P(A) = 0.3$ and $P(A \cup B) = 0.8$, then $P(B)$ is $\dfrac{2}{7}$.
Reason (R): $P(\bar{E}) = 1 – P(E)$, where $E$ is any event.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Let $A$ and $B$ be two independent events.
Assertion (A): If $P(A) = 0.3$ and $P(A \cup B) = 0.8$, then $P(B)$ is $\dfrac{2}{7}$.
Reason (R): $P(\bar{E}) = 1 – P(E)$, where $E$ is any event.
Answer
(a) Assertion (A): Since $A$ and $B$ are independent, $P(A \cup B) = P(A) + P(B) – P(A)P(B)$.
$0.8 = 0.3 + P(B) – 0.3 \cdot P(B)$
$0.5 = 0.7 \cdot P(B)$
$P(B) = \frac{0.5}{0.7} = \frac{5}{7}$. Wait — the correct answer is $\frac{5}{7}$.
Using $P(A \cup B) = P(A) + P(B) – P(A)P(B)$: $0.8 = 0.3 + P(B)(1 – 0.3) \Rightarrow P(B) = \frac{0.5}{0.7} = \frac{5}{7}$.
Both A and R are true and R is the correct explanation of A.
$0.8 = 0.3 + P(B) – 0.3 \cdot P(B)$
$0.5 = 0.7 \cdot P(B)$
$P(B) = \frac{0.5}{0.7} = \frac{5}{7}$. Wait — the correct answer is $\frac{5}{7}$.
Using $P(A \cup B) = P(A) + P(B) – P(A)P(B)$: $0.8 = 0.3 + P(B)(1 – 0.3) \Rightarrow P(B) = \frac{0.5}{0.7} = \frac{5}{7}$.
Both A and R are true and R is the correct explanation of A.
In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): When two coins are tossed simultaneously, then the probability of getting no tail is $\dfrac{1}{4}$.
Reason (R): The probability of getting a head (i.e., no tail) in one toss of a coin is $\dfrac{1}{2}$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and 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
Assertion (A): When two coins are tossed simultaneously, then the probability of getting no tail is $\dfrac{1}{4}$.
Reason (R): The probability of getting a head (i.e., no tail) in one toss of a coin is $\dfrac{1}{2}$.
Answer
(a) Assertion (A): Probability of both heads $= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. So, A is true.
Reason (R): In one toss, $P(\text{head}) = \frac{1}{2}$. So, R is true.
Both A and R are true and Reason (R) is the correct explanation of Assertion (A).
Reason (R): In one toss, $P(\text{head}) = \frac{1}{2}$. So, R is true.
Both A and R are true and Reason (R) is the correct explanation of Assertion (A).
Frequently Asked Questions
What is Probability in CBSE Class 10 Maths?⌄
Probability is the branch of mathematics that measures the likelihood of an event occurring. In CBSE Class 10, students learn to calculate the probability of simple events using the formula P(E) = Number of favourable outcomes / Total number of outcomes, covering dice, coins, cards, and real-life scenarios.
How many marks does Probability carry in the Class 10 board exam?⌄
Probability is part of the Statistics and Probability unit in CBSE Class 10 Maths, which together carries approximately 11 marks in the board examination. Competency-based questions on Probability appear across MCQ, case study, and assertion-reason formats, making thorough practice essential.
What are the most important topics in Class 10 Maths Probability?⌄
Key topics include the classical definition of probability, complementary events, impossible and certain events, problems based on single dice or coin experiments, playing card problems, and case-study questions involving real-life contexts. Students must also be comfortable with the relationship P(E) + P(not E) = 1.
What are common mistakes students make in Probability questions?⌄
Common errors include counting the sample space incorrectly, confusing ‘at least’ with ‘at most’ conditions, forgetting to subtract overlapping outcomes in union-type problems, and misidentifying complementary events. Practising a variety of question types — MCQ, case study, and assertion-reason — helps avoid these pitfalls.
How does Angle Belearn help students master Probability for Class 10?⌄
Angle Belearn provides expert-verified competency-based questions aligned with the latest CBSE pattern, complete with step-by-step solutions. Parents can trust that every question set is curated by experienced CBSE educators to build both conceptual clarity and exam confidence for their child.
