A die has two faces each with number '1', three faces each with number '2' and one face with number '3'. If die is rolled once determine
(i) P(2)
 (ii) P(1 or 3)
 (iii) P(not 3)

Hint: To solve this problem we will have to know about the favourable sample space and total sample space. Then find the favourable sample space and total sample spaces in each case then use the formula of probability and find the required answers.

Complete step-by-step answer:
Here in this question it given that a 6 faces dies is rolled
So the,
Total number of sample spaces \[ = 6\]
Now coming to part (i)
The total number of favourable sample spaces or number faces with number ′2′ \[ = 3\]
Now we know the probability of an event is
Probability (P) = total number of favourable sample space/ total number of sample spaces
So
∴ \[\;P\left( 2 \right) = \dfrac{3}{6} = \dfrac{1}{2}\]
So, the correct answer is “ $ \dfrac{1}{2} $ ”.

similarly for part (ii)
The total number of favourable sample spaces or number faces with number 1 or 2 = The total number of favourable sample spaces or number faces with number not 2
Now applying the simple property of probability that sum of happening and not happening event is always equal to 1
We get,
 \[P\left( {1{\text{ }}or{\text{ }}3} \right) = P\left( {not{\text{ }}2} \right) = 1 - P\left( 2 \right)\]
 \[\; \Rightarrow P\left( {1{\text{ }}or{\text{ }}3} \right) = 1 - \dfrac{{1}}{2} = \dfrac{1}{2}\]
So, the correct answer is “ $ \dfrac{5}{6} $ ”.

Similarly for part (iii)
The total number of favourable sample spaces or number faces with number ′′ =1
By using the probability
We get,
 \[P\left( 3 \right) = \dfrac{1}{6}\]
Thus, \[\] \[P\left( {not{\text{ }}3} \right) = 1 - P\left( 3 \right) = 1 - \dfrac{1}{6} = \dfrac{5}{6}\]
So, the correct answer is “ $ \dfrac{5}{6} $ ”.

Note: Students always should be aware of finding the probability of not happening an event. Sometimes finding not happening events are more convenient than finding the happening events.