Question

A game of chance consists of spinning an arrow , which comes to rest pointing at one of the numbers $1,2,3,4,5,6,7,8$ and these are equally likely outcomes , Find the probability that the arrows will point at any factor of $8$.

Answer 1

Hint: In this question we see that the number of events in sample space (or total number of outcomes ) is $8$. Now it is given that the favorable outcomes is the factor of 8 means that by which the number $8$ is divisible After that use simple formula i.e. Probability of getting a event
P(E) = $\dfrac{{Number\;of\;favorable\;outcomes\;}}{{Total\;number\;of\;outcomes\;}}$

Complete step-by-step answer:
Now we see that there are $8$ numbers that the arrow came to rest that numbers are either $1,2,3,4,5,6,7,8$
So total Number of events are : $8$
And the sample space looks like : $\left\{ {1,2,3,4,5,6,7,8} \right\}$
Now we have to find favorable outcomes that are nothing but the numbers of factor 8 present in the sample space .
Factor of number is a number that divides the other number exactly and without leaving a remainder.
So the Number that divides the $8$ exactly without leaving a remainder is $1,2,4,8$ .
Therefore the favorable outcomes are $4$ that is $ = \left\{ {1,2,4,8} \right\}$
Now we use the probability formula that is
P(E) = $\dfrac{{Number\;of\;favorable\;outcomes\;}}{{Total\;number\;of\;outcomes\;}}$
Number of Favorable outcomes is $ = 4$
Total Number of event is $ = 8$
Hence Probability of event is $ = \dfrac{4}{8}$
After dividing numerator and denominator by $4$
Probability of event is $ = \dfrac{1}{2}$
In decimal form = $0.5$

So the probability that the arrows will point at any factor of $8$ is = $0.5$.

Note: Always remember that the Prime Number only has two factors one is $1$ and the other one is the number itself . Prime numbers are like $2,3,5,7,11........$.
$1$ is not the prime number.
Probability of any event will always be in between $0$ and $1$.