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# 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 (see figure.), and these are equally likely outcomes. What is the probability that it will point at

**(i)** 8?

**(ii)** an odd number?

**(iii)** a number greater than 2?

**(iv)** a number less than 9?

"

Given:

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.

To do:

We have to find the probability that it will point at

(i) 8

(ii) an odd number

(iii) a number greater than 2

(iv) a number less than 9

Solution:

Numbers \( 1,2,3, \ldots, 8 \) are given.

This implies,

The total number of possible outcomes $n=8$.

(i) Total number of favourable outcomes(pointing at 8) $=1$.

We know that,

Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

Therefore,

Probability that it will point at 8 $=\frac{1}{8}$

The probability that it will point at 8 is $\frac{1}{8}$.

(ii) Total number of favourable outcomes(odd numbers) $=4$.

We know that,

Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

Therefore,

Probability that it will point at an odd number $=\frac{4}{8}$

$=\frac{1}{2}$

The probability that it will point at an odd number is $\frac{1}{2}$.

(iii) Total number of favourable outcomes(numbers greater than 2) $=6$.

We know that,

Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

Therefore,

Probability that it will point at a number greater than 2 $=\frac{6}{8}$

$=\frac{3}{4}$

The probability that it will point at a number greater than 2 is $\frac{3}{4}$.

(iv) Total number of favourable outcomes(numbers less than 9) $=8$.

We know that,

Therefore,

Probability that it will point at a number less than 9 $=\frac{8}{8}$

$=1$

The probability that it will point at a number less than 9 is $1$.

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