A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number $1,\ 2,\ 3……12$ ,then find the probability that it will point to an odd number.
Given: A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number $1,\ 2,\ 3……12$.
To do: To find the probability that it will point to an odd number.
Solution:
Let $F$ be event of pointing to an odd number in the game of chance with numbers $1$ to $12$
Odd no. up to $12=1,\ 3,\ 5,\ 7,\ 9,\ 11$
No. of favorable outcomes$=6$
Total no. of possible outcomes $=12$
As known that, Probability $P( F) =\frac{( No.\ of\ favorable\ outcomes)}{( Total\ no.\ of\ possible\ outcomes)}$
$=\frac{6}{12}$
$=\frac{1}{2}$
Therefore, the probability that the arrow points to an odd number is $\frac{1}{2}$.
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