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Aleksandr-060686 [28]
3 years ago
15

A spinner has equal regions numbered 1 through 20. What is the probability that the spinner will stop on an odd number or a mult

iple of 5? Don't forget to reduce fraction.
Mathematics
1 answer:
klio [65]3 years ago
7 0

Answer:

<em>Probability that the spinner will stop on an odd number or a multiple of 5 is </em><em>0.6</em>

Step-by-step explanation:

Probability = \frac{Required outcomes}{Total possible outcomes}

We are given the equal regions numbered from 1 through 20 which means that our total possible outcomes are 20

<em>Total possible outcomes: 20</em>


<em>Outcomes that spinner will stop on an odd number, n(Odd): 10</em>

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Probability of spinner stoping on Odd number:

P(Odd) = \frac{n(Odd)}{Total} = \frac{10}{20} = \frac{1}{2} = 0.5


Outcomes that spinner will stop on a multiple of 5, n(5): 4

5, 10, 15, 20

Probability of spinner stoping on multiple of 5:

P(5) = \frac{n(5)}{Total} = \frac{4}{20} = \frac{1}{5} = 0.2

Odd numbers which are a multiple of 5 are: 5 and 15

So,

P(Odd and 5) = \frac{2}{20}=\frac{1}{10}=0.1

Thus Probability of spinner stopping at odd number or a multiple of 5 becomes:

P(Odd or 5) = P(Odd) + P(5) - P(Odd and 5) = 0.5 + 0.2 - 0.1 = 0.6

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