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  4. 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 :

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Q. 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 :

JEE MainJEE Main 2021Probability - Part 2

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 $( last digit $ 7 \& 7 $ appears once)
$+ n ($ last digit 2 when 7 appears once)
$=8 \times 9 \times 9+(9 \times 9+8 \times 9 \times 2)$
$\therefore P(E)=\frac{8 \times 9 \times 9+9 \times 25}{33 \times 9 \times 9}=\frac{97}{297}$