- Home
- Standard 11
- Mathematics
14.Probability
hard
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
$\frac{1}{5}$
(JEE MAIN-2021)
Solution
$n ( s )= n ($ when 7 appears on thousands place) $+ n (7$ does not appear on thousands place)
$=9 \times 9 \times 9+8 \times 9 \times 9 \times 3$
$=33 \times 9 \times 9$
$n ( E ) = n (\text { last digit } 7 \& 7 \text { appears once })$
$+ n ($ last digit 2 when 7 appears once)
$=8 \times 9 \times 9+(9 \times 9+8 \times 9 \times 2)$
$\therefore P ( E )=\frac{8 \times 9 \times 9+9 \times 25}{33 \times 9 \times 9}=\frac{97}{297}$
Standard 11
Mathematics
