In a single toss of a fair die,the odds again | vedclass

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Question

(A) The sample space $S$ for a single toss of a fair die is $\{1, 2, 3, 4, 5, 6\}$,so the total number of outcomes is $n(S) = 6$.Let $E$ be the event

Vedclass · Accepted Answer

$2 : 1$

Vedclass · Answer

(A) The sample space $S$ for a single toss of a fair die is $\{1, 2, 3, 4, 5, 6\}$,so the total number of outcomes is $n(S) = 6$.Let $E$ be the event that the number $4$ or $5$ turns up. Then $E = \{4, 5\}$,so $n(E) = 2$.The probability of event $E$ is $P(E) = \frac{n(E)}{n(S)} = \frac{2}{6} = \frac{1}{3}$.The probability of the complement event $E^c$ (the event that $4$ or $5$ does not turn up) is $P(E^c) = 1 - P(E) = 1 - \frac{1}{3} = \frac{2}{3}$.The odds against an event $E$ are given by the ratio $P(E^c) : P(E)$.Thus,the odds against $E$ are $\frac{2}{3} : \frac{1}{3} = 2 : 1$.

In a single toss of a fair die,the odds against the event that number $4$ or $5$ turns up is

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