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In a Game of Chance, the Spinning Arrow Rests at One of the Numbers 1, 2, 3, 4, 5, 6, 7, 8. All These Are Equally Likely Outcomes. Find the Probabilities of the Following Events. - Algebra

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Question

In a game of chance, the spinning arrow rests at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8. All these are equally likely outcomes. Find the probabilities of the following events.
(A) The arrow rests at an odd number.
(B) It rests at a prime
number.
(C) It rests at a multiple
of 2.

Solution

s = {1, 2, 3, 4, 5, 6, 7, 8}
∴ n(s) = 8
A = {1, 3, 5, 7}
∴ n(A) = 4
`∴ p(A) = (n(A))/(n(S))=4/8 = 1/2`
B = {2, 3, 5, 7}
∴  n(B) = 4
∴` p(B) = (n(B))/(n(S))=4/8=1/2`
C = {2, 4, 6, 8}
∴ n(C) = 4
`P(C) = (n(c))/(n(s))=4/8=1/2`

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In a Game of Chance, the Spinning Arrow Rests at One of the Numbers 1, 2, 3, 4, 5, 6, 7, 8. All These Are Equally Likely Outcomes. Find the Probabilities of the Following Events. Concept: Equally Likely Outcomes.
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