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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.A) The arrow rests at an odd number.B) It rests at a prime number.C) It rests at a multiple of $2$.

Last updated date: 18th Jun 2024
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Hint: In this question, we have to find the required probabilities. For that we have to write the sample spaces and then we have to find the number of elements in it. Also we should find the number of elements in every event. Finally we should substitute it in the probability formula to find the required result.

Formula used: Probability is a type of ratio where we compare how many times an outcome can occur compared to all possible outcomes.
$Probability = \dfrac{\text{Number of possible outcomes}}{\text{Total number of outcomes}}$

It is given that, in a game of chance, the spinning arrow rests at one of the numbers $1,2,3,4,5,6,7,8$
Also all these are equally likely outcomes.
We need to find out 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 $2$.
Let, S=$\left\{ {1,2,3,4,5,6,7,8} \right\}$ be the sample space of the event happening.
Total number =$8$
That is $n\left( S \right) = 8$.
Let A be the event such that the arrow rests at an odd number.
Odd numbers that can be the outcome that is A=$\left\{ {1,3,5,7} \right\}$
$n\left( A \right)$=number of possible outcomes for event A $= 4$
The probabilities of the events that, the arrow rests at an odd number
$\Rightarrow P\left( A \right) =$$\dfrac{{n\left( A \right)}}{{n\left( S \right)}} = \dfrac{4}{8} = \dfrac{1}{2}$
Let B be the event that it rests at a prime number.
Prime numbers that can be the outcome that is B =$\left\{ {2,3,5,7} \right\}$
$n\left( B \right)$ =number of possible outcomes for event B $= 4$.
The probabilities of the events that, the arrow rests at a prime number
$\Rightarrow P\left( B \right) =$$\dfrac{{n\left( B \right)}}{{n\left( S \right)}} = \dfrac{4}{8} = \dfrac{1}{2}$ .
Let C be the event that it rests at a multiple of $2$.
The numbers multiple of $2$ that can be the outcome that is C=$\left\{ {2,4,6,8} \right\}$
$n\left( C \right)$ =number of possible outcomes for event C $= 4$.
The probabilities of the events that, the arrow rests at a multiple of $2$
$\Rightarrow P\left( C \right) =$$\dfrac{{n\left( C \right)}}{{n\left( S \right)}} = \dfrac{4}{8} = \dfrac{1}{2}$
Hence,
(A)The probability of the event, the arrow rests at an odd number is $\dfrac{1}{2}$.
(B) The probability of the event resting at a prime number is $\dfrac{1}{2}$.
(C) The probability of the event, it rests at a multiple of $2$ is $\dfrac{1}{2}$.

A prime number is a natural number greater than $1$ that is not a product of two smaller natural numbers. Prime numbers are numbers that have only two factors $1$ and themselves.