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 the probability that it will point to an odd number is

The total number of possible outcomes here is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

So, n(S) = 12

Let A be the event that the arrow will point to an odd number.

Thus, the favourable outcomes = 1, 3, 5, 7, 9, 11.

I.e., n(A) = 6

Hence, the probability that the arrow will point to an odd number, P(A) = n(A)/n(S)

P(A) = 6/12

P(A) = 1/2.

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