Question

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 the given 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?

Answer 1

(i) Total number of possible outcomes = 8
$\text{Probability}\ \text{ of}\ \text{ getting}\ \ \text{8} =\frac{\text{Number}\ \text{ of} \ \text{favourable}\ \text{ outcomes}}{\text{Number}\ \text{ of }\ \text{total }\ \text{possible} \ \text{outcomes}}$ $=\frac{1}{8}$

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(ii) Total number of odd numbers on spinner = 4
$\text{Probability}\ \text{ of}\ \text{ getting}\ \text{ an }\ \text{odd}\ \text{ number}=\frac{\text{Number}\ \text{ of} \ \text{favourable}\ \text{ outcomes}}{\text{Number}\ \text{ of }\ \text{total }\ \text{possible} \ \text{outcomes}}$ $=\frac{4}{8} =\frac{1}{2}$

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(iii) The numbers greater than 2 are 3, 4, 5, 6, 7, and 8.

Therefore, total numbers greater than 2 = 6
$\text{Probability}\ \text{ of}\ \text{ getting}\ \text{ a }\ \text{number}\ \text{ greater}\ \text{than}\ \text{2}=\frac{\text{Number}\ \text{ of} \ \text{favourable}\ \text{ outcomes}}{\text{Number}\ \text{ of }\ \text{total }\ \text{possible} \ \text{outcomes}}$ $= \frac{6}{8}=\frac{3}{4}$

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(iv) The numbers less than 9 are 1, 2, 3, 4, 6, 7, and 8.
Therefore, total numbers less than 9 = 8
Probability of getting a number less than 9= $\frac{8}{8}=1.$