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  • 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 : Option: 1

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 :
Option: 1 2/9
Option: 2 122/297
Option: 3 97/297
Option: 4 1/5

Answers (1)

best_answer

Sample space

n(s) = n(when 7 appears on thousands place) + n(7 doesn't appear on thousands    place) 

       = 1× 9× 9× 9 + 3C1 × 9 × 9 = 729 + 1944 = 2673

Events

n(E) = n(last digit 7 & 7 appears once) + n(last digit 2 when 7 appears once)

        = 8 × 9 × 9 × 1 + ((1 × 9 × 9 × 1) + (9 × 8 × 2C1 ×1)) = 873

\therefore \mathrm{P}(\mathrm{E})=\frac{873}{2673}=\frac{97}{297}

 

Posted by

himanshu.meshram

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