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A game of chance consists of spinning an arrow on a circular board, divided into 8 equal parts, which comes to rest pointing at one of the numbers 1, 2, 3, ..., 8 (Fig. 9), which are equally likely outcomes. What is the probability that the arrow will point at (i) an odd number (ii) a number greater than 3 (iii) a number less than 9.

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#### Solution

Arrow can come to rest at any of the numbers 1, 2, 3, 4, 5, 6, 7 and 8.

Total number of events = 8

(i) There are four odd numbers 1, 3, 5 and 7.

Probability that the arrow will point at an odd number is given by P (Arrow point at odd number)= `4/8=1/2`

(ii) There are five numbers greater than 3, that is, 4, 5, 6, 7 and 8.

Probability that the arrow will point at a number greater than 3 is given by P (Arrow point at a number greater than 3) = `5/8`

(iii) All the numbers are less than 9.

Probability that the arrow will point at a number less than 9 is given by P (Arrow point at a number less than 9)= `8/8=1`

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