Suppose we throw a die once. $(i)$ What is the probability of getting a number greater than $4$? $(ii)$ What is the probability of getting a number less than or equal to $4$?
(N/A) $(i)$ Here,let $E$ be the event 'getting a number greater than $4$'. The total number of possible outcomes is $6$ $(1, 2, 3, 4, 5, 6)$. The outcomes favourable to $E$ are $5$ and $6$. Therefore,the number of outcomes favourable to $E$ is $2$.
$P(E) = P(\text{number greater than } 4) = \frac{2}{6} = \frac{1}{3}$.
$(ii)$ Let $F$ be the event 'getting a number less than or equal to $4$'.
The total number of possible outcomes is $6$.
The outcomes favourable to the event $F$ are $1, 2, 3, 4$.
Therefore,the number of outcomes favourable to $F$ is $4$.
$P(F) = P(\text{number less than or equal to } 4) = \frac{4}{6} = \frac{2}{3}$.
$P(E) = P(\text{number greater than } 4) = \frac{2}{6} = \frac{1}{3}$.
$(ii)$ Let $F$ be the event 'getting a number less than or equal to $4$'.
The total number of possible outcomes is $6$.
The outcomes favourable to the event $F$ are $1, 2, 3, 4$.
Therefore,the number of outcomes favourable to $F$ is $4$.
$P(F) = P(\text{number less than or equal to } 4) = \frac{4}{6} = \frac{2}{3}$.
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