Test: Classical Approach to Probability - Class 10 MCQ

# Test: Classical Approach to Probability - Class 10 MCQ

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## 15 Questions MCQ Test Mathematics (Maths) Class 10 - Test: Classical Approach to Probability

Test: Classical Approach to Probability for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation. The Test: Classical Approach to Probability questions and answers have been prepared according to the Class 10 exam syllabus.The Test: Classical Approach to Probability MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Classical Approach to Probability below.
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Test: Classical Approach to Probability - Question 1

### If the probability of winning a game is 0.995, then the probability of losing is

Detailed Solution for Test: Classical Approach to Probability - Question 1

Let P(winning the game)=0.995

Since the probability of two complimentary event sums to 1 so-

P(not winning the game) + P(winning the game )=1

=P(losing it )=1-P(winning the game )

=1-0.995=0.005

Test: Classical Approach to Probability - Question 2

### What is the probability of a sure event?​

Detailed Solution for Test: Classical Approach to Probability - Question 2

A sure event is an event, which always happens. For example, it's a sure event to obtain a number between “1”“1” and “6”“6” when rolling an ordinary die.

The probability of a sure event has the value of 11.

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Test: Classical Approach to Probability - Question 3

### The probability of an event that is certain to happen is

Detailed Solution for Test: Classical Approach to Probability - Question 3

The probability of an event is a number describing the chance that the event will happen. An event that is certain to happen has a probability of 1. An event that cannot possibly happen has a probability of zero. If there is a chance that an event will happen, then its probability is between zero and 1.

Examples of Events:

• tossing a coin and it landing on heads
• tossing a coin and it landing on tails
• rolling a '3' on a die
• rolling a number > 4 on a die
• it rains two days in a row
• drawing a card from the suit of clubs
• guessing a certain number between 000 and 999 (lottery)
Test: Classical Approach to Probability - Question 4

Which of the following cannot be the probability of an event?​

Detailed Solution for Test: Classical Approach to Probability - Question 4

Probability of an event (E) is always greater than or equal to 0.

Also, it is always less than or equal to one.

This implies that the probability of an event cannot be negative or greater than 1.

Therefore, out of these alternatives, −1.5 cannot be a probability of an event.

Test: Classical Approach to Probability - Question 5

The probability of getting a prime number in single throw of a dice is:​

Detailed Solution for Test: Classical Approach to Probability - Question 5

Given, A dice is thrown once. So,

Total number of outcomes (n) = 6

Number of prime numbers = {2, 3,5}

So, Favorable number of outcomes (m) = 3

Thus, probability of getting a prime number = m/n = 3/6 = 1/2

Test: Classical Approach to Probability - Question 6

If a die is thrown once, the probability of getting a prime number is​

Detailed Solution for Test: Classical Approach to Probability - Question 6

Possible numbers of events on throwing a dice = 6
Numbers on dice = 1,2,3,4,5 and 6

Prime numbers = 2, 3 and 5
Favourable number of events = 3
Probability that it will be a prime number = 3/6 = 1/2

Test: Classical Approach to Probability - Question 7

The probability that a non-leap year selected at random will have 53 Mondays is

Detailed Solution for Test: Classical Approach to Probability - Question 7

We know that there are 52 weeks in a year.

There are 7 days in a week

52 weeks will be  to 7*52=364 days.

The remaining 1 day can be any day among Monday, Tuesday,..., Sunday.
Sample space has seven days as options.
So probability of getting 53 Sundays in a non-leap year is 1/7

Test: Classical Approach to Probability - Question 8

A card is drawn at random from a pack of 52 playing cards. The probability of getting a face card is

Detailed Solution for Test: Classical Approach to Probability - Question 8

Clearly, there are 52 cards, out of which there are 12 face cards.

P (getting a face card) = 12/52 = 3/13.

Test: Classical Approach to Probability - Question 9

Which one of the following cannot be the probability of an event?​

Detailed Solution for Test: Classical Approach to Probability - Question 9

If the probability is in percentage we divide it by 100
So d) option is 0.05.And we know If an event is impossible its probability is zero. Similarly, if an event is certain to occur, its probability is one. The probability of any event lies in between these values. It is called the range of probability and is denoted as 0 ≤ P (E) ≤ 1.And probability more than one means that favourable outcomes are more than total outcomes which is wrong.

Test: Classical Approach to Probability - Question 10

If the probability of winning a game is 0.3, then the probability of losing it is:​

Detailed Solution for Test: Classical Approach to Probability - Question 10

Let the probability of losing the game be x
We know that sum of the the sum of probabilities is equal to 1
So
0.3+x=1
x=0.7

Test: Classical Approach to Probability - Question 11

If a letter of English alphabet is chosen at random, then the probability that the letter is a consonant is:

Detailed Solution for Test: Classical Approach to Probability - Question 11

The probability that the letter is a consonant = (26 - 5)/26 = 21/26.

Test: Classical Approach to Probability - Question 12

If a digit is chosen at random from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 then the probability that it is odd is​

Detailed Solution for Test: Classical Approach to Probability - Question 12

Sample Space=1,2,3,4,5,6,7,8,9
Total number of outcomes=9
No of favourable outcomes(odd)=1,3,5,7,9=5
Probability of odd number=No.of favourable outcomes/total number of outcomes=5/9

Test: Classical Approach to Probability - Question 13

Two coins are tossed together. The probability of getting head on both the coins is​

Detailed Solution for Test: Classical Approach to Probability - Question 13

The sample space for the event is :    (H,H)  (T,H)  (H,T)  (T,T)
Therefore total outcomes= 4

Probability =  1/4

Test: Classical Approach to Probability - Question 14

The probability of getting an even number, when a die is thrown once, is:​

Detailed Solution for Test: Classical Approach to Probability - Question 14

If we throw a die once, then possible outcomes (s), are
S = { 1, 2, 3, 4, 5, 6 }
⇒    n(E) = 6
(i) Let E be the favourable outcomes of getting an even number, then
E = { 2, 4, 6 }
⇒ n(S) = 3

Test: Classical Approach to Probability - Question 15

The probability that a randomly chosen number from one to twelve is a divisor of twelve is​

Detailed Solution for Test: Classical Approach to Probability - Question 15

We have the numbers from 1 to 12. Total possible outcomes = 12 Now, divisors of 12 : 1, 2, 3, 4, 6 and 12 Number of divisors of 12 = 6 Number of favourable outcomes = 6
Required probability

## Mathematics (Maths) Class 10

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## Mathematics (Maths) Class 10

116 videos|420 docs|77 tests