A box contains a number of marbles with serial number $18$ to $38$ . A marble is picked at random. Find the probability that it is a multiple of $3$ .

\frac{3}{5}

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\frac{7}{20}

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\frac{3}{4}

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\frac{1}{3}

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Solution

The correct option is D $\frac{1}{3}$
A box contains a number of marbles with serial number 18 to 38. $N = 21$
the multiple of 3 are $18, 21, 24, 27, 30, 33, 36, M = 7$

$P (E) = \frac{M}{N} = \frac{7}{21} = \frac{1}{3}$

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A box contains a number of marbles with serial number 18 to 38. A marble is picked at random. Find the probability that it is a multiple of 3.

The correct option is D 13A box contains a number of marbles with serial number 18 to 38. N=21the multiple of 3 are 18,21,24,27,30,33,36,M=7P(E)=MN=721=13

A box contains a number of marbles with serial number $18$ to $38$ . A marble is picked at random. Find the probability that it is a multiple of $3$ .

The correct option is D $\frac{1}{3}$
A box contains a number of marbles with serial number 18 to 38. $N = 21$
the multiple of 3 are $18, 21, 24, 27, 30, 33, 36, M = 7$

$P (E) = \frac{M}{N} = \frac{7}{21} = \frac{1}{3}$