# Twelve tickets are numbered from 1 to 12. One tickets is drawn at random, then the probability of the number to be divisible by 2 or 3 is

$\begin {array} {1 1} (A)\;\large\frac{2}{3} & \quad (B)\;\large\frac{7}{12} \\ (C)\;\large\frac{5}{6} & \quad (D)\;\large\frac{3}{4} \end {array}$

## 1 Answer

Let E be the event of numbers to be divisible by 2 or 3
$\therefore E = \{2,3,4,6,8,9,10,12 \}$
$\Rightarrow n(E) = 8$
$n (S) = 12$
Hence required probability = $\large\frac{n(E)}{n(S)}$
$= \large\frac{8}{12}$
$\large\frac{2}{3}$
Ans : (A)
answered Jan 16, 2014

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