A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
Outcomes are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
So, the total number of outcome = 6×6 = 36
(i) Method 1:
Consider the following events.
A = 5 comes in first throw,
B = 5 comes in second throw
P(A) = 6/36
P(B) = 6/36 and
P(not B) = 5/6
P(not A) = 1 - (6/36) = 5/6
∴ The required probability = (5/6)×(5/6) = 25/36
Method 2:
Let E be the event in which 5 does not come up either time.
So, the favourable outcomes are [36 – (5 + 6)] = 25
∴ P(E) = 25/36
(ii) Number of events when 5 comes at least once = 11(5+6)
∴ The required probability = 11/36