What is the probability that a number selected from the numbers 1, 2, 3,..., 20 is a prime number, where each of the given numbers is equally likely to be selected?

\frac{7}{10}

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\frac{2}{15}

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\frac{2}{5}

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\frac{3}{5}

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Solution

The correct option is C $\frac{2}{5}$
$E = {2, 3, 5, 7, 11, 13, 17, 19} \Rightarrow n (E) = 8 S = {1, 2, 3, 4, \dots 20} \Rightarrow n (S) = 20 ∴ P (E) = \frac{n (E)}{n (S)} = \frac{8}{20} = \frac{2}{5}$

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What is the probability that a number selected from the numbers 1, 2, 3,..., 20 is a prime number, where each of the given numbers is equally likely to be selected?

The correct option is C 25E={2,3,5,7,11,13,17,19}⇒n(E)=8S={1,2,3,4,⋯20}⇒n(S)=20∴P(E)=n(E)n(S)=820=25

The correct option is C $\frac{2}{5}$
$E = {2, 3, 5, 7, 11, 13, 17, 19} \Rightarrow n (E) = 8 S = {1, 2, 3, 4, \dots 20} \Rightarrow n (S) = 20 ∴ P (E) = \frac{n (E)}{n (S)} = \frac{8}{20} = \frac{2}{5}$