Explore numerous Class 9 Maths MCQ and Ganita Manjari Class 9 Maths Chapter 7 The Mathematics of Maybe Introduction to Probability MCQ Questions Online Test with Answers provided with detailed solutions by looking below.
MCQ on The Mathematics of Maybe Introduction to Probability Class 9
Class 9 Maths The Mathematics of Maybe Introduction to Probability MCQ
Choose the correct option from the given options:
Question 1.
The probability of an impossible event is:
(a) 1
(b) 0
(c) \(\frac{1}{2}\)
(d) \(\frac{2}{3}\)
Solution:
(b) 0
Explanation:
Impossible event is which can not occur. So, its probability will be zero.
Question 2.
In an experiment there are exactly two outcomes. The probability of one of the events is:
(a) \(\frac{1}{2}\)
(b) 1
(c) 0
(d) cannot be determined
Solution:
(a) \(\frac{1}{2}\)
Explanation:
Here, n(S) = 2, n(E) = 1
Now, P(E) = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{1}{2}\)
Question 3.
A coin is tossed once. Its sample space is:
(a) {H,T}
(b) {H}
(c) {T}
(d) none of these
Solution:
(a) {H,T}
Explanation:
Possible outcomes are H and T.
∴S = {H, T}.
Question 4.
A die is rolled once. The value of n(S) is:
(a) 1
(b) 2
(c) 6
(d) 4
Solution:
(c) 6
Explanation:
When a die is rolled, there are 6 possible outcomes, namely 1, 2, 3, 4, 5, 6.
So, n(S) = 6.
Question 5.
A coin is tossed 500 times and number of times heads turn up was 255. Relative frequency of head is:
(a) 0.51
(b) 0.255
(c) \(\frac{51}{10}\)
Solution:
(a) 0.51
Explanation:
Here, n(S) = 500, n(H) = 255
So, Relative frequency of head = \(\frac{\mathrm{n}(\mathrm{H})}{\mathrm{n}(\mathrm{~S})} \frac{255}{500}\)
= \(\frac{51}{100}\)
= 0.51
The Mathematics of Maybe Introduction to Probability MCQ Class 9
Question 6.
If the probability of happening of an event is \(\frac{2}{3}\), then probability of its not happening is:
(a) \(\frac{2}{3}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{1}{6}\)
(d) \(\frac{1}{2}\)
Solution:
(b) \(\frac{1}{3}\)
Explanation:
Sum of probabilities of an event taking place and probability of its not taking place = 1.
So, probability of not taking place of an event = 1 – probability of its taking place
Hence, required probability = 1 – \(\frac{2}{3}=\frac{1}{3}\)
Question 7.
A die is rolled. The probability of getting 4, is:
(a) \(\frac{1}{2}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{1}{6}\)
(d) \(\frac{1}{3}\)
Solution:
(c) \(\frac{1}{6}\)
Explanation:
Let E:getting a 4.
Here, n(s) = 6; n(E) = 1
Required probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{1}{6}\)
Question 8.
A letter is picked at random from the word ‘VANDE MATARAM’. The probability of picking the letter ‘M’ is:
(a) \(\frac{1}{2}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{1}{6}\)
(d) \(\frac{1}{3}\)
Solution:
(c) \(\frac{1}{6}\)
Explanation:
Let E:picking the letter ‘M’.
Here, n(S) = 12; n(E) = 2
Required probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{2}{12}=\frac{1}{6}\)
Question 9.
A number is selected from first 10 natural numbers. What is the probability of it being a prime number?
(a) \(\frac{1}{2}\)
(b) \(\frac{2}{5}\)
(c) \(\frac{1}{5}\)
(d) \(\frac{1}{3}\)
Solution:
(b) \(\frac{2}{5}\)
Explanation:
There are 4 prime numbers in first 10 natural numbers, namely 2, 3, 5 and 7.
Let E:getting a prime number.
Here, n(S) = 10; n(E) = 4
Required probability =\(\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{4}{10}=\frac{2}{5}\)
Question 10.
A bag contains 4 red and 5 green balls. The probability of drawing a green ball from the bag is:
(a) \(\frac{1}{9}\)
(b) \(\frac{2}{9}\)
(c) \(\frac{5}{9}\)
(d) \(\frac{4}{9}\)
Solution:
(d) \(\frac{4}{9}\)
Explanation:
There are 4 red and 5 green balls.
Let E:getting a green ball.
Here, n(S) = 10; n(E) = 4
Required probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{4}{10}=\frac{2}{5}\)
Question 11.
The set of all possible outcomes of a random experiment is called __________.
(a) Sample Space
(b) Favourable Space
(c) Likely Space
(d) Happening Space
Solution:
(a) Sample Space
Explanation:
The set of all possible outcomes are called Sample Space.
Question 12.
The probability of an event that is sure to happen is:
(a) 1
(b) \(\frac{2}{3}\)
(c) \(\frac{1}{4}\)
(d) \(\frac{1}{2}\)
Solution:
(a) 1
Explanation:
The probability of sure event is 1.
Question 13.
The probability of a sure event is:
(a) 1
(b) 0
(c) \(\frac{1}{2}\)
(d) \(\frac{2}{3}\)
Solution:
(a) 1
Question 14.
In an experiment there are exactly three equally likely outcomes. The probability of one of the events is:
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{3}\)
(c) 0
(d) cannot be determined
Solution:
(b) \(\frac{1}{3}\)
Question 15.
Result of a match is predicted for a team as W: win, L: loose, D: drawn, A: abandoned. Its sample space is:
(a) {W, L}
(b) {L,A}
(c) {W, A, L}
(d) {W,A,L,D}
Solution:
(d) {W,A,L,D}
Question 16.
A coin is tossed twice. The value of n(S) is:
(a) 1
(b) 2
(c) 6
(d) 4
Solution:
(d) 4
Question 17.
A coin is tossed 500 times and number of times heads turn up was 255. Relative frequency of tail is:
(a) 0.51
(b) 0.255
(c) \(\frac{51}{10}\)
(d) 0.49
Solution:
(d) 0.49
Question 18.
If the probability of happening of an . 4 event is \(\frac{4}{5}\), then probability of its not 5 happening is:
(a) \(\frac{2}{5}\)
(b) \(\frac{3}{5}\)
(c) \(\frac{1}{5}\)
(d) \(\frac{4}{5}\)
Solution:
(c) \(\frac{1}{5}\)
Question 19.
A die is rolled. The probability of getting 3 or 5, is:
(a) \(\frac{1}{2}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{1}{6}\)
(d) \(\frac{1}{3}\)
Solution:
(d) \(\frac{1}{3}\)
Question 20.
A letter is picked at random from the word ‘VANDE MATARAM’. The probability of picking the letter ‘A’ is:
(a) \(\frac{1}{2}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{1}{6}\)
(d) \(\frac{1}{3}\)
Solution:
(d) \(\frac{1}{3}\)
Question 21.
A number is selected from first 10 natural numbers. What is the probability of it being a multiple of 2?
(a) \(\frac{1}{2}\)
(b) \(\frac{2}{5}\)
(c) \(\frac{1}{5}\)
(d) \(\frac{1}{3}\)
Solution:
(a) \(\frac{1}{2}\)
Question 22.
A bag contains 4 red and 5 green balls. The probability of drawing a red ball from the bag is:
(a) \(\frac{1}{9}\)
(b) \(\frac{2}{9}\)
(c) \(\frac{5}{9}\)
(d) \(\frac{4}{9}\)
Solution:
(d) \(\frac{4}{9}\)
The Mathematics of Maybe Introduction to Probability Class 9 Assertion and Reason Questions
Direction: A statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option from the following options.
(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 1.
Assertion (A): The probability of getting an even number when a fair die is rolled is \(\frac{1}{2}\) .
Reason (R): There are 3 even numbers amongst 6 possible outcomes when a fair die is rolled.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Explanation:
There are three even numbers, namely 2, 4, 6, when a die is rolled. Total number of outcomes = 6.
Required probability = \(\frac{3}{6}\) = \(\frac{1}{2}\)
Hence, Assertion (A) is true.
There are 3 even numbers amongst 6 possible outcomes when a fair die is rolled.
Thus, Reason (R) is true and Reason (R) is the correct explanation of Assertion (A).
Question 2.
Assertion (A): Probability of getting a vowel if a letter is selected from the word ‘SCHOOL’
Reason (R): There are 4 consonants in the word ‘SCHOOL’.
Solution:
(d) Assertion (A) is false but Reason (R) is true.
Explanation:
There are 2 vowels O in the word SCHOOL. Total number of letters in the word SCHOOL is 6. So, probability of getting a vowel, when a letter is selected from the word SCHOOL = \(\frac{2}{6}\) = \(\frac{1}{3}\).
So, Assertion (A) is false.
There are 4 consonants namely S, C, H, L in the word SCHOOL.
So, Reason (R) is true.
Question 3.
Assertion (A): In a bag there are 4 red balls and 6 black balls. Probability of drawing a black ball from the bag is \(\frac{3}{5}\).
Reason (R): Probability of getting a red ball = \(\frac{1}{2}\)
Solution:
(c) Assertion (A) is true but Reason (R) is false.
Explanation:
There are 6 favourable outcomes and 10 total outcomes.
Probability of drawing a black ball = \(\frac{6}{10}\) = \(\frac{3}{5}\)
So, Assertion (A) is true.
Probability of drawing a red ball = \(\frac{4}{10}=\frac{2}{5}\)
So, Reason (R) is false.
Question 4.
Assertion (A): A number is selected at random from the numbers 1, 4, 9, 25 and 36. The probability that the number has odd number of factors is \(\frac{1}{4}\).
Reason (R): Perfect square numbers have an odd number of factors.
Solution:
(d) Assertion (A) is false but Reason (R) is true.
Explanation:
Since, given numbers are perfect squares.
So, any selected number will have an odd number of factors.
Thus, the probability = 1.
So, Assertion (A) is false.
Perfect square numbers have an odd number of factors.
So, Reason (R) is true.
Question 5.
Assertion (A): A number is selected from first 100 natural numbers. The probability of getting a prime number is \(\frac{1}{4}\).
Reason (R): There are 25 prime numbers in first hundred natural numbers.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Explanation:
Probability of getting a prime number from first 100 natural numbers is = \(\frac{25}{100}=\frac{1}{4}\)
Therefore, Assertion (A) is true.
There are 25 prime numbers in first 100 natural numbers.
So, Reason (R) is true and Reason (R) explains the existence of Assertion (A).
Question 6.
Assertion (A): The numerical value of probability of an event cannot be more than 1.
Reason (R): Number of favourable outcomes * cannot be greater than the total number of outcomes.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Question 7.
Assertion (A): If probability of occurrence of an event is \(\frac{2}{3}\), then probability of its non-occurrence is \(\frac{1}{3}\).
Reason (R): Sum of probability of occurrence of an even and its non-occurrence is 1.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Question 8.
Assertion (A): If we roll a die, then probability of getting a 6 is the least.
Reason (R): Getting any number from 1 to 6, when a fair die is rolled is equally likely events.
Solution:
(d) Assertion (A) is false but Reason (R) is true.
Question 9.
Assertion (A): We can not predict exactly whether there will be rain on a particular day
Reason (R): Event of rain is a random event.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Question 10.
Assertion (A): A number is randomly picked from the numbers 90 to 96. Probability of getting a prime number is zero.
Reason (R): There are 25 prime numbers from 1 to 100.
Solution:
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).