Question

Let $A$ be a set of all 4 -digit natural numbers whose exactly one digit is 7 . Then the probability that a randomly chosen element of A leaves remainder 2 when divided by 5 is :

• A. $\frac{2}{9}$
• B. $\frac{122}{297}$
• C. $\frac{97}{297}$
• D. $\frac{1}{5}$

Answer 1

Answer: (C)

$n(s)=n($ when 7 appears on thousands place)
$+n(7$ does not appear on thousands place)
$=9\times 9\times 9+8\times 9\times 9\times 3$
$=33\times 9\times 9$
$n(E)=n$( last digit $7\&7$ appears once)
$+n($ last digit 2 when 7 appears once)
$=8\times 9\times 9+(9\times 9+8\times 9\times 2)$
$\therefore P(E)=\frac{8\times 9\times 9+9\times 25}{33\times 9\times 9}=\frac{97}{297}$