If a number is chosen at random from the set {x: 4 $\leq x \leq 15$ }. Find the probability that it is a multiple of 3 or a multiple of 4

$\frac{1}{12}$

$\frac{5}{12}$

$\frac{7}{12}$

$\frac{11}{12}$

Correct answer is C

set = 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

multiple of 3 = 6, 9, 12, 15

multiple of 4 = 4, 8, 12

Prob(multiple of 3 or 4) = Prob(multiple of 3) + pro(multiple of 4)

= $\frac{4}{12} + \frac{3}{12} = \frac{7}{12}$

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