The Mathematics of Maybe Introduction to Probability Class 9 MCQ Maths Chapter 7

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MCQ on The Mathematics of Maybe Introduction to Probability Class 9

The Mathematics of Maybe Introduction to Probability MCQ Class 9

Class 9 Maths The Mathematics of Maybe Introduction to Probability MCQ

Question 1.
If you roll a standard 6-sided die, how would you describe the probability of rolling a number less than 7?
(a) Impossible
(b) Equally likely
(c) Less likely
(d) Certain
Answer:
(d) Certain

Explanation:
Given, a standard 6-sided die is rolled.
Let S be the sample space of all possible outcomes.
S = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6
Let E be the event of rolling a number less than 7. Since, all numbers on the die (1, 2, 3, 4, 5, 6) are strictly less than 7.
E = {1, 2, 3, 4, 5, 6)
∴ n(E) = 6
Probability is calculated,
∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{6}{6}\) = 1
A probability of 1 indicates that the event is guaranteed to happen.

Question 2.
A coin is tossed 100 times and head appears 46 times. Now, if we toss a coin at random then what is the probability of getting a tail?
(a) \(\frac{23}{50}\)
(b) \(\frac{27}{50}\)
(c) \(\frac{7}{50}\)
(d) \(\frac{9}{50}\)
Answer:
(b) \(\frac{27}{50}\)

Explanation:
Given, a coin is tossed 100 times, therefore total number of trials = 100
Also, head appears 46 times.
Now, tail appears 100 – 46 = 54 times
∴ P(getting a tail) = \(\frac{54}{100}=\frac{27}{50}\)

Question 3.
A Mathematics book contains 250 pages. A page is selected at random. What is the probability that the number on the page selected is a perfect square?
(a) \(\frac{4}{25}\)
(b) \(\frac{3}{25}\)
(c) \(\frac{7}{25}\)
(d) \(\frac{9}{25}\)
Answer:
(b) \(\frac{3}{25}\)

Explanation:
∵ Total number of pages = 250 [given]
Now, numbers on the page, which are perfect square are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144,169, 196, 225.
So, the total number on the pages, which are perfect square = 15.
∴ Required probability = \(\frac{54}{100}=\frac{27}{50}\)

Question 4.
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes

The probability of getting two heads is
(a) 0.21
(b) 0.36
(c) 0.49
(d) 0.50
Answer:
(b) 0.36

Explanation:
Let us define the event A = Getting two heads
Total number of trials = 200
Number of trials in which we get two heads in
which event A happens = 72
Hence, P(A) = \(\frac{72}{200}\) = 0.36

Question 5.
In a cricket match, if a batsman hits a boundary 8 times out of 40 balls he plays. Then, the probability that he did not hit a boundary is
(a) 0.2
(b) 0.4
(c) 0.6
(d) 0.8
Answer:
(d) 0.8

Explanation:
Let A denote the event that the batsman did not hit a boundary.
We have, total number of trials = 40.
Number of trials in which the event A happened
= 40 – 8 = 32
∴ P(A) = \(\frac{32}{40}=\frac{4}{5}\) = 0.8

Question 6.
The record of a weather station shows that out of the past 250 consecutive days, its weather forecast were correct 175 times. Then, the probability that on a given day it was correct and it was not correct, is
(a) 0.1, 0.2
(b) 0.7,0.5
(c) 0.9, 0.1
(d) 0.7, 0.5
Answer:
(b) 0.7,0.5

Explanation:
We have, total number of days for which the weather forecast was made = 250
Number of days for which the weather forecast was correct = 175
Number of days for which the forecast was not correct = 250 – 175 = 75
Therefore, probability that the forecast was correct on a given day
= \(\frac{\text { Number of days for which the forecast was correct }}{\text { Number of days for which the forecast was made }}\)
= \(\frac{175}{250}\) = 0.7

and probability that the forecast was not correct on a given day
= \(\frac{\text { Number of days for which the forecast was correct }}{\text { Number of days for which the forecast was made }}\)
= \(\frac{75}{250}\) = 0.3

Question 7.
The maximum temperature in Celsius of some cities on a day are given as below.

The probability that maximum temperature lies between 16°C and 30°C, is
(a) \(\frac{41}{79}\)
(b) \(\frac{56}{81}\)
(c) \(\frac{21}{53}\)
(d) \(\frac{1}{10}\)
Answer:
(a) \(\frac{41}{79}\)

Explanation:
Total number of cities
= 1 + 2 + 18 + 21 + 19 + 18 = 79
Let E be the event, which show maximum temperature of cities.
Number of cities, whose maximum temperature lies between 16°C and 30°C = 2 + 18 + 21 = 41
∴ Probabilities that the temperature of cities lies between 16°C and 30°C = \(\frac{41}{79}\).

Question 8.
On one page of a telephone directory, there were 200 telephone number. The frequency distribution of their unit place digit (e.g. In the number 25828573 the unit’s place digit is 3) is given in the following table.

Without looking at the page, the pencil is placed on one of these numbers i.e. the number is chosen at random. What is the probability that the digit in its unit place is 9?
(a) 0.1
(b) 0.03
(c) 0.05
(d) 0.07
Answer:
(a) 0.1

Explanation:
Here, 200 telephone numbers are given on one page of a telephone directory.
∴ Total number of selected telephone numbers = 200 The frequency of digit 9 at unit place is 20 i.e. in 20 telephone numbers 9 is at unit place.
∴ Required probability
= \(\frac{\text { Frequency of } 9}{\text { Total number of selected telephone numbers }}\)
= \(\frac{20}{200}\)
= 0.1

Question 9.
A die is thrown 250 times and the outcomes are noted as given below.

If a die is thrown at random, then the probabilities of getting 1 and 5 are respectively
(a) 0.26, 0.132
(b) 0.21, 0.139
(c) 0.20, 0.139
(d) 0.11, 0.112
Answer:
(a) 0.26, 0.132

Explanation:
Here, a die is thrown 250 times.
So, total number of trials = 250
In a random throw of a die, let E₁ and E₂ be the events of getting 1 and 5, respectively.
Here, frequency of outcomes 1 is 65 i.e. 1 occurs on a die 65 times.
∴ P (getting 1) = P(E₁) = \(\frac{\text { Number of times } 1 \text { occur }}{\text { Total number of trials }}\)
= \(\frac{65}{250}\)
= 0.26
Here, 5 occurs on a die 33 times.
∴ P (getting 5) = P(E₅) = \(\frac{\text { Number of times } 5 \text { occurs }}{\text { Total number of trials }}\)
= \(\frac{33}{250}\) = 0.132

The Mathematics of Maybe Introduction to Probability Class 9 Assertion and Reason Questions

Direction (Q. Nos. 1-2) In the questions given below, there are two statements marked as Assertion (A) and Reason (R). Read the statements and choose the correct option.

Question 1.
Assertion (A): In a cricket match, a batsman hits a boundary 5 times out of 45 balls he plays. The probability that in a given ball, he does not hit the boundary is \(\frac{8}{9}\).
Reason (R) P(E) + P (not E) = 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.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Explanation:
(A) A bastman hits a boundary 9 times out of 45 balls he plays.
P(A) = P (he hits a boundary) = \(\frac{5}{45}=\frac{1}{9}\)
Since, P(A) + P(Ā) = 1
P(Ā) = 1 – P(A)
= 1 – \(\frac{1}{9}=\frac{8}{9}\)
∴ Probability that he does not hit the boundary is \(\frac{4}{5}\).
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Question 2.
Assertion (A): If E is an event such that
P(E) = \(\frac{1}{999}\) then P(Ē) = 0.001.
Reason (R): P(E) + P(Ē) = 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.
Answer:
(d) A is false but R is true.

Explanation:
Given, P(E) = \(\frac{1}{999}\) = 0.001
We know that P(E) + P(E) = 1
∴ P(E) = 1 – P(Ē) = 1 – 0.001 = 0.999
Hence, Assertion is false but Reason is true.