A die is tossed twice. The probability of having a number greater than 4 on each toss is
Answer & explanation
Correct answer: option 2
We have,
Total number of elementary events = 6 x 6 = 36
A number greater than 4 can be obtained in each toss in one of the following ways
(5,5), (5, 6), (6, 5), (6, 6)
∴ Favourable number of elementary events =4
Hence, required probability =$\frac{4}{36}=\frac{1}{9}$