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Answer :
In a single throw of a die, we can get any one of the six numbers marked on its six faces.
Therefore, the total number of outcomes = 6
(i) Let A denote the event “getting a prime number”. So,event A occurs if we obtain 2, 3, 5 as an outcome.
∴ Favourable number of outcomes = 3
Hence, P\(\left( A \right) =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
(ii) Let A denote the event “not getting a prime number”. So, event A occurs if we obtain 1, 4, 6 as an outcome.
∴ Favourable number of outcomes = 3
Hence, P\(\left( A \right) =\frac { 3 }{ 6 } =\frac { 1 }{ 2 }\)
(iii) The event “getting a number greater than 5” will occur if we obtain the number 6.
∴ Favourable number of outcomes = 1
Hence, the required probability = \(\frac { 1}{ 6 } \)
(iv) The event “getting a number not greater than 5” will occur if we obtain one of the numbers 1, 2, 3, 4, 5.
∴ Favourable number of outcomes = 5
Hence, required probability = \(\frac { 5 }{ 6 } \)