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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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