CBSE Class 10 Mathematics Probability MCQs Set F

Question. Two unbiased coins are tossed simultaneously then the probability of getting no head is \( \frac{A}{B} \), then \( (A + B)^2 \) is equal to:
(a) 1
(b) 4
(c) 5
(d) 25
Answer: D
If two unbiased coins are tossed simultaneously we obtained possible outcomes: HH, HT, TH, TT.
Hence, Total number of outcomes = 4. No head is obtained if the event TT occurs.
Hence, Number of favourite outcomes = 1. Hence, Required probability \( = \frac{1}{4} \).
But, given probability \( = \frac{A}{B} \). So, \( A = 1 \) and \( B = 4 \).
Therefore, \( (A + B)^2 = (1 + 4)^2 = (5)^2 = 25 \)

Question. A letter is chosen at random from the letters of the word ‘ASSASSINATION’, then the probability that the letter chosen is a vowel is in the form of \( \frac{6}{2x + 1} \), then \( x \) is equal to:
(a) 5
(b) 6
(c) 7
(d) 8
Answer: B
There are 13 letters in the word ‘ASSASSINATION’ out of which one letter can be chosen in 13 ways. Hence, Total number of outcomes = 13. There are 6 vowels in the word ‘ASSASSINATION’. So, there are 6 ways of selecting a vowel. Hence, Required probability \( = \frac{6}{13} \).
But given that, \( \frac{6}{2x + 1} = \frac{6}{13} \Rightarrow 2x + 1 = 13 \Rightarrow 2x = 12 \Rightarrow x = 6 \)

Question. Ramesh buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random a tank containing 5 male fish and 9 female fish. Then, the probability that the fish taken out is a male fish, is:
(a) \( 5/13 \)
(b) \( 5/14 \)
(c) \( 6/13 \)
(d) \( 7/13 \)
Answer: B
There are \( 14 = (5 + 9) \) fish out of which one can be chosen in 14 ways. Hence, Total number of possible outcomes = 14. There are 5 male fish out of which are male fish can be chosen in 5 ways. Hence, Favourable number of elementary events = 5. Hence, required probability \( = \frac{5}{14} \)

Question. A number \( x \) is selected from the numbers 1, 2, 3 and then a second number \( y \) is randomly selected from the numbers 1, 4, 9 then the probability that the product \( xy \) of the two numbers will be less than 9 is:
(a) \( 3/7 \)
(b) \( 4/9 \)
(c) \( 5/9 \)
(d) \( 7/9 \)
Answer: C
Number \( x \) can be selected in three ways and corresponding to each such way there are three ways of selecting number \( y \). Therefore two numbers can be selected in 9 ways as listed below: (1, 1), (1, 4), (1, 9), (2, 1), (2, 4), (2, 9), (3, 1), (3, 4), (3, 9).
So, total numbers of possible outcomes = 9. The product \( xy \) will be less than 9, if \( x \) and \( y \) are chosen in one of the following ways: (1, 1), (1, 4), (2, 1), (2, 4), (3, 1).
Hence, Favourable number of elementary events = 5. Hence, required probability \( = \frac{5}{9} \)

Question. A bag contains 8 red balls and some blue balls. If the probability of drawing a blue ball is three times of a red ball, then the number of blue balls in the bag:
(a) 12
(b) 18
(c) 24
(d) 36
Answer: C
Let there be \( x \) blue balls in the bag. Hence, Total number of balls in the bag \( = (8 + x) \).
Now, \( P_1 = \text{Probability of drawing a blue ball} = \frac{x}{8 + x} \)
and \( P_2 = \text{Probability of drawing a red ball} = \frac{8}{8 + x} \)
It is given that, \( P_1 = 3P_2 \Rightarrow \frac{x}{8 + x} = 3 \times \frac{8}{8 + x} \Rightarrow \frac{x}{8 + x} = \frac{24}{8 + x} \Rightarrow x = 24 \).
Hence, there are 24 blue balls in the bag.

Question. There are 1000 sealed envelopes in a box. 10 of them contain a cash prize of <100 each, 100 of them contain a cash prize of <50 each and 200 of them contain a cash prize of <10 each and rest do not contain any cash prize. If they are well-shuffled and an envelope is picked up out, then the probability that is contains no cash prize is:
(a) 0.65
(b) 0.69
(c) 0.54
(d) 0.57
Answer: B
Total number of envelopes in the box = 1000.
Number of envelopes containing cash prize \( = 10 + 100 + 200 = 310 \).
Number of envelopes containing no cash \( = 1000 - 310 = 690 \).
Hence, Required probability \( = \frac{690}{1000} = 0.69 \)

Question. If odds in against of an event be 3:8, then the probability of occurrence of this event is:
(a) \( 3/8 \)
(b) \( 5/8 \)
(c) \( 3/11 \)
(d) \( 8/11 \)
Answer: C
Let the event be E. If number of favourable outcomes to event E are M and total outcomes be n. Then, according to the question [Note: The text provided in the source for the answer explanation matches the OCR and implies a ratio usage where the result is 3/11 based on their internal logic].

\( \frac{m}{n - m} = \frac{3}{8} \)
\( 8m = 3n - 3m \)
\( 11m = 3n \)
\( \frac{m}{n} = \frac{3}{11} \)
Hence, \( P(E) = \frac{\text{Number of outcomes favourable to } E}{\text{Number of total outcomes}} \)
\( = \frac{m}{n} = \frac{3}{11} \) [Since, \( \frac{m}{n} = \frac{3}{11} \)]

Question. The given figure shows a disc on which player spins an arrow twice. The fraction \( \frac{x}{y} \) is formed, where 'a' is the number of sectors on which the arrow stops on the first spin and 'b' is the number of the sectors in which the arrow stops on the second spin. In each spin, each sector has equal chance of selection by the arrow, then the probability that the fraction \( \frac{x}{y} \geq 1 \).
(a) \( \frac{7}{12} \)
(b) \( \frac{5}{12} \)
(c) \( \frac{11}{12} \)
(d) \( \frac{1}{2} \)
Answer: A
Two numbers can be selected in \( 6 \times 6 = 36 \) ways as listed below:
(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)
Favourable outcomes = (1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (4, 4), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Hence, Favourable number of elementary events = 21
Hence, required probability \( = \frac{\text{Favourable number of events}}{\text{Total number of events}} = \frac{21}{36} = \frac{7}{12} \)

Question. Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3 respectively. They are thrown and the sum of the numbers on them is noted. The probability of getting even sum is:
(a) \( \frac{1}{9} \)
(b) \( \frac{1}{18} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{4} \)
Answer: C
Total number of outcomes \( = 6 \times 6 = 36 \)
Possible sum of two numbers on the two dice are 2, 3, 4, 5, 6, 7, 8, 9.
Let, \( E = \) Event of getting even sum i.e. 2, 4, 6, 8
i.e. outcomes favourable to event are:
(1, 1), (1, 1), (2, 2), (3, 1), (3, 1), (1, 3), (1, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1), (5, 3), (5, 3), (6, 2), (6, 2)
Hence, Number of outcomes favourable to \( E = 18 \)
Hence, \( P(E) = \frac{\text{Number of outcomes favourable to } E}{\text{Number of total outcomes}} = \frac{18}{36} = \frac{1}{2} \)

DIRECTION : (15 and 16) At a fate, cards numbers 1 to 1000. One number on one card are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square number greater than 500, the player wins a prize.

Question. The probability that the first player wins is:
(a) 0.009
(b) 0.099
(c) 0.999
(d) 1
Answer: A
Given at a fete, cards bearing numbers 1 to 1000, one number on one card, are put in a box. Each player selects one card at random and that card is not replaced, so the total number of outcomes are 1000. If the selected card has a perfect square number greater than 500, then player wins a prize.
Let, \( E_1 = \) Event that the first player wins a prize
= player select a card which is a perfect square greater than 500.
= {529, 576, 625, 676, 729, 784, 841, 900, 961}
So, Number of Outcomes favourable to \( E_1 = 9 \)
Hence, Required probability \( = \frac{\text{Number of outcomes favourable to } E_1}{\text{Number of total outcomes}} = \frac{9}{1000} = 0.009 \)

Question. The probability that the second player wins a prize, if the first has already wins, is:
(a) \( \frac{1}{999} \)
(b) \( \frac{2}{999} \)
(c) \( \frac{4}{999} \)
(d) \( \frac{8}{999} \)
Answer: D
Now, if first has won, i.e. one card is already selected, greater than 500, has a perfect square. Since repetition is not allowed. So, one card is removed out of 1000 cards. So, number of remaining cards is 999.
Hence, Total number of remaining outcomes = 999
Let \( E_2 = \) Event that the second player wins a prize, if the first has already won.
= Remaining cards has a perfect square greater than 500.
Hence, Number of outcomes favourable to \( E_2 = 9 - 1 = 8 \)
Hence, \( P(E_2) = \frac{\text{Number of outcomes favourable to } E_2}{\text{Total number of remaining outcomes}} = \frac{8}{999} \)

Question. A box contains 54 marbles each of which is blue, green or white. The probability of selecting a blue marble at random from the box is 1/3 and the probability of selecting a green marble at random is 4/9. The number of white marbles in the box are:
(a) 10
(b) 12
(c) 14
(d) 16
Answer: B
Let the number of blue marbles be \( b \), number of green marbles be \( g \) and number of white marbles be \( w \).
Then, \( b + g + w = 54 \) ...(1)
Hence, \( P(\text{selecting a blue marble}) = \frac{b}{54} \)
and \( P(\text{selecting a green marble}) = \frac{g}{54} \)
Given, the probability of selecting a blue marble \( = \frac{1}{3} \)
Hence, \( \frac{b}{54} = \frac{1}{3} \)
\( b = \frac{54}{3} = 18 \)
Also, \( P(\text{selecting a green marble}) = \frac{4}{9} \)
Hence, \( \frac{g}{54} = \frac{4}{9} \)
\( g = \frac{4 \times 54}{9} = 24 \)
On substituting the values of \( b \) and \( g \) in Eq. (1), we get
\( 18 + 24 + w = 54 \)
\( w = 54 - 42 = 12 \)

Question. In a group of 3 people, the probability that atleast two will have the same birthday is (ignoring leap year):
(a) \( \frac{364 \times 363}{365^2} \)
(b) \( \frac{364 \times 363 \times 362}{365^2} \)
(c) \( 1 - \frac{364 \times 363}{365^2} \)
(d) \( 1 - \frac{364 \times 363 \times 362}{365^2} \)
Answer: C
Let \( A \) be the event that atleast two people have the same birthday.
Then, \( \overline{A} = \) No two or more people have the same birthday. \( = \) All the three persons have distinct birthdays.
Hence, \( P(\overline{A}) = \frac{365 \times 364 \times 363}{365^3} = \frac{364 \times 363}{365^2} \)
Hence, required probability \( = P(A) = 1 - P(\overline{A}) = 1 - \frac{364 \times 363}{365^2} \)

Question. Two number \( b \) and \( c \) are chosen at random with replacement from the numbers 1, 2, 3, 4, 5. The probability that \( x^2 + bx + c = 0 \) has non real roots is:
(a) \( \frac{13}{25} \)
(b) \( \frac{12}{25} \)
(c) \( \frac{11}{25} \)
(d) \( \frac{9}{25} \)
Answer: A

FILL IN THE BLANK

Question. The events which have equal chances to occur or no one is preferred over the other are called ..........
Answer: equally likely events

Question. The probability of an event that cannot happen is .......... Such an event is called ..........
Answer: 0, impossible event

Question. On a single roll of a die, the probability of getting a number less than 7 is ...........
Answer: one

Question. If \( A \) is an event of a random experiment, then \( A^C \) or \( \overline{A} \) or \( A' \) is called the .......... of the event.
Answer: complement

Question. Someone is asked to make a number from 1 to 100. The probability that it is a prime is ..........
Answer: 1/4

Question. A set of events which have no pair in common are called ..........
Answer: mutually exclusive

Question. When sum of probability of two events is 1, the events are called ..........
Answer: complementary events

Question. An outcome of a random experiment is called an .......... event.
Answer: elementary

TRUE/FALSE

Question. If I toss a coin 3 times and get head each time, then I should expect a tail to have a higher chance in the 4th toss.
Answer: False

Question. The probability expressed as percentage of a particular occurrence can never be less than zero.
Answer: True

Question. For any event \( E \), \( P(E) + P(\overline{E}) = 1 \), where \( \overline{E} \) stands for ‘not \( E \)’. \( E \) and \( \overline{E} \) are called complementary events.
Answer: True

Question. A single letter is selected at random from the word PROBABILITY, the probability it is a vowel is \( \frac{4}{11} \).
Answer: True

Question. The sum of probabilities of two students getting distinction in their final examinations is 1.2.
Answer: True

Question. Probability of an event lies between zero and one.
Answer: False

Question. If the probability of an event is 1, then it is an impossible event.
Answer: False

Question. If you toss a coin 6 times and it comes down heads on each occasion, then the probability of getting a head is one.
Answer: False

Question. The probability of an event can be greater than 1.
Answer: False

Question. If \( A \) is any event in a sample space, then \( P(\overline{A}) = 1 + P(A) \).
Answer: False

Question. A card is selected from a deck of 52 cards. The probability of its being a red face card is \( \frac{3}{26} \).
Answer: True

MATCHING QUESTIONS

DIRECTION : Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D, ..........) in Column-I have to be matched with statements (p, q, r, s, ..........) in Column-II.

Question. Match the proposed probability under Column-I with the appropriate written description under Column-II :
Column-I (Probability): (A) 0.95, (B) 0.02, (C) -0.3, (D) 0.5, (E) 0
Column-II (Written Description): (p) An incorrect assignement, (q) No chance of happening, (r) As much chance of happening as not, (s) Very likely to happen, (t) Very little chance of happening
Answer: (A) - s, (B) - t, (C) - p, (D) - r, (E) - q

Question. Two unbiased coins are tossed simultaneously. Match Column-I with the probabilities given in Column-II :
Column-I: (A) The probability of getting one head is, (B) The probability of getting at least one head is, (C) The probability of getting two heads is
Column-II: (p) \( \frac{3}{4} \), (q) \( \frac{1}{4} \), (r) \( \frac{1}{2} \)
Answer: (A) - r, (B) - p, (C) - q

Question. Match Column-I with Column-II:
Column-I: (A) Probability of getting number 5 in throwing a dice., (B) Probability of obtaining three heads in a single throw of a coin., (C) Probability of getting the sum of the numbers as 7, when two dice are thrown, (D) Probability of occurrence of two sure independent events.
Column-II: (p) 0, (q) \( \frac{6}{36} \), (r) 1, (s) \( (\frac{1}{2})^0 \), (t) \( \frac{1}{6} \)
(a) (A) - p, (B) - (q, r), (C) - s, (D) - t
(b) (A) - (q, t), (B) - p, (C) - (q, t), (D) - (r, s)
(c) (A) - (q, t), (B) - (r, s), (C) - p, (D) - r
(d) (A) - p, (B) - (q, t), (C) - (q, s), (D) - r
Answer: (b) (A) - (q, t), (B) - p, (C) - (q, t), (D) - (r, s)

Question. Match option of Column I with the appropriate option of Column II.
Column I: (A) The probability of a sure event is, (B) The probability of impossible event is, (C) Number of face cards in the pack of cards is, (D) Probability of occuring 53 Sundays in a leap year is, (E) The probability of getting a sum of atleast 11 in a throw of a pair of dice is, (F) A card is drawn from a well-shuffled deck of 52 cards. The probability that the card drawn is neither a king nor a queen is
Column II: (p) 0, (q) 1, (r) \( \frac{2}{7} \), (s) 12, (t) \( \frac{11}{13} \), (u) \( \frac{1}{12} \)
Answer: (A) - q, (B) - p, (C) - s, (D) - r, (E) - u, (F) - t

ASSERTION AND REASON

DIRECTION : In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(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.

Question. Assertion : If \( A \) and \( B \) are two independent events and it is given that \( P(A) = \frac{2}{5} \), \( P(B) = \frac{3}{5} \), then \( P(A \cap B) = \frac{6}{25} \).
Reason : \( P(A \cap B) = P(A) \cdot P(B) \), where \( A \) and \( B \) are two independent events.
(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.
Answer: A
Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). Both assertion and reason are correct. Also, reason is the correct explanation of the assertion. \( P(A \cap B) = (\frac{2}{5})(\frac{3}{5}) = \frac{6}{25} \).

Question. Assertion : The probability of winning a game is 0.4, then the probability of losing it, is 0.6
Reason : \( P(E) + P(\text{not } E) = 1 \)
(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.
Answer: A
Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). We have, \( P(E) = 0.4 \), where \( E \) = event of winning. \( P(\text{Not } E) = 1 - P(E) = 1 - 0.4 = 0.6 \).

Question. Assertion : in rolling a dice, the probability of getting number 8 is zero.
Reason : Its an impossible event.
(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.
Answer: A
Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). Assertion and Reason both are correct. Also Reason is the correct explanation of the Assertion.

Question. Assertion : Card numbered as 1, 2, 3 .......... 15 are put in a box and mixed thoroughly, one card is then drawn at random. The probability of drawing an even number is \( \frac{1}{2} \).
Reason : For any event \( E \), we have \( 0 \leq P(E) \leq 1 \)
(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.
Answer: D
Assertion (A) is false but reason (R) is true. Total possible outcomes = 15. \( n(S) = 15 \). Total favourable numbers are 2, 4, 6, 8, 10, 12, 14. \( E = \{2, 4, 6, 8, 10, 12, 14\} \). \( n(E) = 7 \). Probability of drawing an even number = \( \frac{7}{15} \).

Question. Assertion : If \( E \) and \( F \) are events such that \( P(E) = \frac{1}{4} \), \( P(F) = \frac{1}{2} \) and \( P(E \text{ and } F) = \frac{1}{8} \), then \( P(E \text{ or } F) \) is \( \frac{5}{8} \).
Reason : If \( A \) and \( B \) are independent, then \( P(A \cap B) = P(A) \).
(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.
Answer: C
Assertion (A) is true but reason (R) is false. \( P(E \text{ or } F) = P(E \cup F) = P(E) + P(F) - P(E \cap F) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{5}{8} \).

Question. Assertion : The probability of getting a prime number. When a die is thrown once is \( \frac{2}{3} \).
Reason : Prime numbers on a die are 2, 3, 5.
(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.
Answer: D
Assertion (A) is false but reason (R) is true. When a die is thrown once, total possible outcomes = 6 and prime numbers in it are \( \{2, 3, 5\} \). Total possible outcomes = 3. Probability of getting a prime = \( \frac{3}{6} = \frac{1}{2} \).

Question. Assertion : The probabilities that \( A, B, C \) can solve a problem independently are \( \frac{1}{3} \), \( \frac{1}{3} \) and \( \frac{1}{4} \) respectively. The probability that only two of them are able to solve the problem is \( \frac{7}{36} \).
Reason : If \( A \) and \( B \) are mutually exclusive events, then \( P(A \cap B) \neq 0 \).
(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.
Answer: C
Assertion (A) is true but reason (R) is false. The required probability = \( \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{3}{4} + \frac{1}{3} \cdot \frac{2}{3} \cdot \frac{1}{4} + \frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{4} = \frac{3}{36} + \frac{2}{36} + \frac{2}{36} = \frac{7}{36} \).