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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?

Given:

A die is thrown twice.

To do:

We have to find the probability that

(i) 5 will not come up either time.

(ii) 5 will come up at least once.

Solution:

When a die is rolled twice, the total possible outcomes are $6\times6=36$.

This implies,

The total number of possible outcomes $n=36$

(i) Outcomes, where 5 will not come up either time, are $(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 6),$

$ (4, 1), (4, 2), (4, 3), (4, 4), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 6)$

Number of outcomes, where 5 will not come up either time $=25$

Total number of favourable outcomes $=25$

Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

Therefore,

Probability that 5 will not come up either time $=\frac{25}{36}$

The probability that 5 will not come up either time is $\frac{25}{36}$.

(ii) Outcomes, where 5 will come up at least once, are $(1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 5)$

Number of outcomes, where 5 will come up at least once $=11$

Total number of favourable outcomes $=11$

Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

Therefore,

Probability that 5 will come up at least once $=\frac{11}{36}$

The probability that 5 will come up at least once is $\frac{11}{36}$.

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