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# 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 $$= \ \frac{11}{36}$$