What is the probability that an integer x \((1 \leq x \leq 25)\) chosen at random is divisible by both 2 and 3?

A.

\(\frac{1}{25}\)

B.

\(\frac{1}{5}\)

C.

\(\frac{4}{25}\)

D.

\(\frac{3}{4}\)

Correct answer is C

\((1 \leq x \leq 25)\) = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25}

Number N of x divisible by both 2 and 3 is 4.

n(\(\varepsilon\)) = 25

= \(\frac{N}{n(\varepsilon)}\)

= \(\frac{4}{25}\)