A positive integer is selected at random and | vedclass

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Question

(C) When a positive integer is divided by $9$,the possible remainders are the set $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8\}$.There are $9$ possible outcomes

Vedclass · Accepted Answer

$\frac{8}{9}$

Vedclass · Answer

(C) When a positive integer is divided by $9$,the possible remainders are the set $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8\}$.There are $9$ possible outcomes in total,so $n(S) = 9$.We want the probability that the remainder is not $1$.The set of favorable outcomes is $E = \{0, 2, 3, 4, 5, 6, 7, 8\}$.The number of favorable outcomes is $n(E) = 8$.The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of outcomes.$P(E) = \frac{n(E)}{n(S)} = \frac{8}{9}$.

$A$ positive integer is selected at random and is divided by $9$. What is the probability that the remainder is not $1$?

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