Question

Four digit numbers are formed using each of the digits $1,2,.......,8$ only once. One number from them is picked at random then the probability that the selected number contains unity is

• A. $\frac{1}{2}$
• B. $\frac{1}{8}$
• C. $\frac{1}{4}$
• D. $\frac{1}{3}$

Answer 1

The correct option is A $\frac{1}{2}$

Total numbers ${=}^8{C}_4\ast 4!$
Numbers with unity ${=}^7{C}_3\ast 4!$
$P{=}^7{C}_3\ast 4!{/}^8{C}_4\ast 4!=1/2$