A game of chance consists of spinning an arro | vedclass

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(C) The total number of possible outcomes on the circular disc is $n = 12$ (since the numbers are $1$ to $12$).The event is that the arrow points at t

Vedclass · Accepted Answer

$\frac{1}{12}$

Vedclass · Answer

(C) The total number of possible outcomes on the circular disc is $n = 12$ (since the numbers are $1$ to $12$).The event is that the arrow points at the number $7$. There is only $1$ such outcome, so the number of favourable outcomes is $m = 1$.The probability of an event is given by the formula: $P(E) = \frac{	ext{Number of favourable outcomes}}{	ext{Total number of possible outcomes}}$.Therefore, the required probability is $P = \frac{m}{n} = \frac{1}{12}$.

$A$ game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers $1, 2, 3, \dots, 11, 12$ on a circular disc. All these outcomes are equally likely. Then, the probability that the arrow rests to point at $7$ is ..........

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