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?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

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) \( = \ \frac{6}{36} \),

P(B) \( = \ \frac{6}{36} \) and

P(not B) \( = \ \frac{5}{6} \)

So, P(not A) \( = \ 1 \ - \ \frac{6}{36} \ = \ \frac{5}{6} \)

The required probability \(= \ \frac{5}{6} \ × \ \frac{5}{6} \ = \ \frac{25}{36} \)

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

\(\therefore \) P(E) \( = \ \frac{25}{36} \)

(ii) Number of events when 5 comes at least once = 5 + 6 = 11

\(\therefore \) The required probability \( = \ \frac{11}{36} \)