Classical Approach to Probability - Free MCQ Practice Test with solutions,

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

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.

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)

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.

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

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 (1 is neither prime nor composite so it will not be considered)
Favourable number of events = 3
Probability that it will be a prime number = 3/6 = 1/2

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

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

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

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.

Given: P(win) = 0.3.

Assuming the only possible outcomes are win or lose, these are complementary events and their probabilities sum to 1.

Therefore, P(lose) = 1 − P(win) = 1 − 0.3 = 0.7.

Hence the probability of losing is 0.7, so option B is correct.

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

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

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

Probability =  1/4

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

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