Question

A die is thrown twice. What is the probability that

(i) 3 will not come up either time?
(ii) 3 will come up at least once?

Answer 1

Let E be the event that 3 will come up atleast once when a die is thrown twice. Then outcomes favourable to E are
{(1, 3), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 3), (5, 3), (6, 3)}
n(S) = 36
(i) P (3 will not come either time) = 1 – P (3 will come atleast)
=1-n(E)/n(S) = 1-11/36 = 25/36
(ii) P (3 will come atleast once) = n(E)/n(S) = 11/36