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The probability that a two-digit number selected at random will be a multiple of ' $3$ ' and a multiple of ' $5$ ' is:

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\frac{2}{15}

\frac{1}{30}

\frac{1}{15}

\frac{4}{90}

answer is B.

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Detailed Solution

Given a two-digit number that is multiple of 3 and multiple of 5.

\Rightarrow

Total number of two digit number

= 90

\Rightarrow

Total number of two digit number which is divisible by

3 = 30

Out of this there are

6

numbers divisible by

15 (15, 30, 45, 60, 75, 90)

which are also divisible by

5 .

\Rightarrow

Total two digit number which are divisible by

3

and 5 = 6

\Rightarrow

Required probability

= \frac{6}{90} = \frac{1}{30}

Hence option 2 correct.

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