CBSE Class 10 Mathematics Probability MCQs Set E

Question. The probability of getting a number greater than 2 in throwing a die is
(a) \( 2/3 \)
(b) \( 1/3 \)
(c) \( 4/3 \)
(d) \( 1/4 \)
Answer: A
Required probability \( = \frac{4}{6} = \frac{2}{3} \)

Question. Out of one digit prime numbers, one number is selected at random. The probability of selecting an even number is
(a) \( 1/2 \)
(b) \( 1/4 \)
(c) \( 4/9 \)
(d) \( 2/5 \)
Answer: B
One digit prime numbers are 2, 3, 5, 7. Out of these numbers, only the number 2 is even.

Question. A bag contains 3 red and 2 blue marbles. If a marble is drawn at random, then the probability of drawing a blue marble is:
(a) \( 1/5 \)
(b) \( 2/5 \)
(c) \( 3/5 \)
(d) \( 4/5 \)
Answer: B
There 5 marbles in the bag. Out of these 5 marbles one can be choose in 5 ways. Hence, Total number of possible outcomes = 5. Since, the bag contains 2 blue marbles. Therefore, one blue marble can be drawn in 2 ways. Hence, Favourable number of elementary events = 2. Hence, \( P (\text{getting a blue marble}) = \frac{2}{5} \)

Question. A single letter is selected at random from the word “PROBABILITY”. The probability that the selected letter is a vowel is
(a) \( 2/11 \)
(b) \( 3/11 \)
(c) \( 4/11 \)
(d) \( 0 \)
Answer: C
Required probability \( = \frac{1 + 2 + 1}{11} = \frac{4}{11} \)

Question. A three digit number is to be formed using the digits 3, 4, 7, 8 and 2 without repetition. The probability that it is an odd number is
(a) \( 2/5 \)
(b) \( 1/5 \)
(c) \( 4/5 \)
(d) \( 3/5 \)
Answer: A

Question. Two coins are tossed simultaneously. The probability of getting at most one head is
(a) \( 1/4 \)
(b) \( 1/2 \)
(c) \( 3/4 \)
(d) \( 1 \)
Answer: C
Total outcomes = HH, HT, TH, TT
Favourable outcomes = HT, TH, TT
\( P(\text{at most one head}) = \frac{3}{4} \)

Question. A fair die is thrown once. The probability of getting a composite number less than 5 is
(a) \( 1/3 \)
(b) \( 1/6 \)
(c) \( 2/3 \)
(d) \( 0 \)
Answer: B
The outcomes are 1, 2, 3, 4, 5, 6. Out of these, 4 is the only composite number which is less than 5.

Question. If a letter is chosen at random from the letter of English alphabet, then the probability that it is a letter of the word ‘DELHI’ is
(a) \( 1/5 \)
(b) \( 1/26 \)
(c) \( 5/26 \)
(d) \( 21/26 \)
Answer: C
The English alphabet has 26 letters in all. The word ‘DELHI’ has 5 letter, so the number of favourable outcomes = 5.

Question. The probability that a two digit number selected at random will be a multiple of ‘3’ and not a multiple of ‘5’ is
(a) \( 2/15 \)
(b) \( 4/15 \)
(c) \( 1/15 \)
(d) \( 4/90 \)
Answer: B
24 out of the 90 two digit numbers are divisible by ‘3’ and not by ‘5’. The required probability is therefore, \( \frac{24}{90} = \frac{4}{15} \)

Question. If in a lottery, there are 5 prizes and 20 blanks, then the probability of getting a prize is
(a) \( 2/5 \)
(b) \( 4/5 \)
(c) \( 1/5 \)
(d) \( 1 \)
Answer: C
Required probability \( = \frac{5}{25} = \frac{1}{5} \)

Question. If a number \( x \) is chosen at random from the numbers \( -2, -1, 0, 1, 2 \). Then, the probability that \( x^2 < 2 \) is:
(a) \( 2/5 \)
(b) \( 4/5 \)
(c) \( 1/5 \)
(d) \( 3/5 \)
Answer: D
Clearly, number \( x \) can take anyone of the five given values. So, total number of possible outcomes = 5. We observe that \( x^2 < 2 \) when \( x \) takes anyone of the following three values \( -1, 0 \) and \( 1 \). So, favourable number of elementary events = 3. Hence, \( P(x^2 < 2) = \frac{3}{5} \)

Question. Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random, then the probability that the ticket has a number which is a multiple of 3 or 7, is:
(a) \( 2/5 \)
(b) \( 3/5 \)
(c) \( 4/5 \)
(d) \( 1/5 \)
Answer: A
Out of 20 tickets numbered from 1 to 20, one can be chosen in 20 ways. So, total number of possible outcomes associated with the given random experiment is 20. Out of 20 tickets numbered 1 to 20, tickets bearing numbers which are multiple of 3 or 7 bear numbers 3, 6, 7, 9, 12, 14, 15 and 18. Hence, Favourable number of elementary events = 8. Hence, required probability \( = \frac{8}{20} = \frac{2}{5} \)

Question. Which of the following relationship is the correct?
(a) \( P(E) + P(\bar{E}) = 1 \)
(b) \( P(E) - P(\bar{E}) = 1 \)
(c) \( P(E) = 1 + P(\bar{E}) \)
(d) None of these
Answer: A
\( P(E) + P(\bar{E}) = 1 \)

Question. A dice is thrown twice. The probability of getting 4, 5 or 6 in the first throw and 1, 2, 3 or 4 in the second throw is
(a) 1/3
(b) 2/3
(c) 1/2
(d) 1/4
Answer: A 
Let \( P(A) \) and \( P(B) \) be the probability of the events, then \( P(A \text{ and } B) = P(A) \cdot P(B) = \frac{3}{6} \times \frac{4}{6} = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \)

Question. Two dice are thrown together. The probability that sum of the two numbers will be a multiple of 4, is:
(a) \( 1/2 \)
(b) \( 1/3 \)
(c) \( 1/8 \)
(d) \( 1/4 \)
Answer: D
Here, \( S = \{(3, 1), (2, 2), (1, 3), (6, 2), (5, 3), (4, 4), (3, 5), (2, 6), (6, 6)\} \)
Total number of outcomes = 36
Number of favourable outcomes = 9
Hence, \( P(\text{sum of two numbers will be multiple of 4}) = \frac{9}{36} = \frac{1}{4} \)

FILL IN THE BLANK

Question. A die is thrown once, the probability of getting a prime number is ..........
Answer: 1/2

Question. The probability of an event that is certain to happen is .......... Such an event is called ..........
Answer: 1, sure or certain event

Question. The sum of the probabilities of all the elementary events of an experiment is ..........
Answer: 1

Question. On a single roll of a die, the probability of getting a number 8 is ..........
Answer: zero

Question. The probability of an event is greater than or equal to .......... and less than or equal to ..........
Answer: 0, 1

Question. If \( P(E) = 0.05 \), the probability of ‘not \( E \)’ is ..........
Answer: .95

Question. Probability of an event \( E \) + Probability of the event ‘not \( E \)’ = ..........
Answer: 1

TRUE/FALSE

Question. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears a two-digit number is 0.9.
Answer: True

Question. A bag contains slips numbered from 1 to 100. If Fatima chooses a slip at random from the bag, it will either be an odd number or an even number. Since, this situation has only two possible outcomes, so the probability of each is \( \frac{1}{2} \).
Answer: True

Question. A card is selected at random from a well shuffled deck of 52 playing cards. The probability of its being a face card is \( \frac{3}{13} \).
Answer: True

Question. An event associated to a random experiment is a compound event if it is obtained by combining two or more elementary events associated to the random experiment.
Answer: True

Question. In a family, having three children, there may be no girl, one girl, two girls or three girls. So, the probability of each is \( \frac{1}{4} \).
Answer: False

Question. I toss three coins together. The possible outcomes are no head, 1 head, 2 heads and 3 heads. So, I say that probability of no head is \( \frac{1}{4} \).
Answer: False

Question. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, the number of blue balls in the bag is 10.
Answer: True

Question. The sum of probabilities of all the outcomes of an experiment is greater than one.
Answer: False

Question. The sum of the probabilities of all the elementary events of an experiment is 1.
Answer: True

Question. An event A associated to a random experiment is said to occur if any one of the elementary events associated to the event A is an outcomes.
Answer: True

Question. In every situation that has only two possible outcomes, each outcome will have probability \( \frac{1}{2} \).
Answer: False

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 box contains 5 white, 2 red and 4 black marbles, then the probability of not drawing a white marble from the box is \( \frac{5}{11} \).
Reason : \( P(\overline{E}) = 1 - P(E) \), where \( E \) is any 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: D
Assertion (A) is false but reason (R) is true. Assertion is not correct, but reason is correct. \( P(\text{white marble}) = \frac{5}{5+2+4} = \frac{5}{11} \). \( P(\text{not white marble}) = 1 - \frac{5}{11} = \frac{6}{11} \).

Question. Let \( A \) and \( B \) be two independent events. Assertion : If \( P(A) = 0.3 \) and \( P(A \cup B) = 0.8 \), then \( P(B) \) is \( \frac{2}{7} \).
Reason : \( P(\overline{E}) = 1 - P(E) \), where \( E \) is any 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).

Question. Assertion : If \( P(A) = 0.25 \), \( P(B) = 0.50 \) and \( P(A \cap B) = 0.14 \), then the probability that neither \( A \) nor \( B \) occurs is 0.39.
Reason : \( \overline{A \cup B} = \overline{A} \cup \overline{B} \).
(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.

Question. Assertion : When two coins are tossed simultaneously then the probability of getting no tail is \( \frac{1}{4} \).
Reason : The probability of getting a head (i.e., no tail) in one toss of a coin is \( \frac{1}{2} \).
(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). Probability of getting no tail when two coins tossed simultaneously i.e., both are head. Probability of both head = \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \).

Question. Assertion : An event is very unlikely to happen. Its probability is 0.0001
Reason : If \( P(A) \) denote the probability of an event \( A \), then \( 0 \leq P(A) \leq 1 \).
(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: B
Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). Assertion and Reason is correct but Reason is not correct explanation for Assertion.

Question. Assertion : If the probability of an event is \( P \) then probability of its complementary event will be \( 1 - P \).
Reason : When \( E \) and \( \overline{E} \) are complementary events, then \( P(E) + P(\overline{E}) = 1 \).
(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). Both statements are correct and Reason is the correct for Assertion.

Question. Assertion : If a die is thrown, the probability of getting a number less than 3 and greater than 2 is zero.
Reason : Probability of an impossible event is zero.
(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 statements are correct. Event given in Assertion is an impossible event.

Question. Assertion : In a simultaneously throw of a pair of dice. The probability of getting a double is \( \frac{1}{6} \).
Reason : Probability of an event may be negative.
(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. When two dice are tossed. Total possible outcomes = 36. \( n(S) = 36 \) and total favourable outcomes (doublet) = \( \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\} \). \( n(E) = 6 \). Probability = \( \frac{6}{36} = \frac{1}{6} \).