$A$ 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?
$(i)$ $5$ will not come up either time?
$(ii)$ $5$ will come up at least once?
(A) Total number of outcomes $= 6 \times 6 = 36$.
$(i)$ Let $E$ be the event that $5$ comes up at least once. The outcomes where $5$ appears are $(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)$.
Number of favourable outcomes for $E = 11$.
Probability of $5$ coming up at least once,$P(E) = \frac{11}{36}$.
The probability that $5$ will not come up either time is $P(\text{not } E) = 1 - P(E) = 1 - \frac{11}{36} = \frac{25}{36}$.
$(ii)$ The probability that $5$ will come up at least once is $P(E) = \frac{11}{36}$.
$(i)$ Let $E$ be the event that $5$ comes up at least once. The outcomes where $5$ appears are $(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)$.
Number of favourable outcomes for $E = 11$.
Probability of $5$ coming up at least once,$P(E) = \frac{11}{36}$.
The probability that $5$ will not come up either time is $P(\text{not } E) = 1 - P(E) = 1 - \frac{11}{36} = \frac{25}{36}$.
$(ii)$ The probability that $5$ will come up at least once is $P(E) = \frac{11}{36}$.
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