$\begin{array}{1 1}(A)\;\large\frac{1}{5}\\(B)\;\large\frac{4}{5}\\(C)\;\large\frac{1}{30}\\(D)\;\large\frac{5}{9}\end{array} $

Step 1:

Given four digit numbers are to be formed.

$\therefore$ The numbers can't be formed starting with zero.

$\therefore$ Total numbers formed

Starting with 2=3!=6

Starting with 3=3!=6

Starting with 5=3!=6

$\therefore$ total =6+6+6=18

Step 2:

The numbers being divisible with 5,have to end with 0 or 5

$\therefore$ Total such numbers =4+4+2=10

$\therefore$ Required probability =$\large\frac{10}{18}=\frac{5}{9}$

Hence (D) is the correct option.

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