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 :
Solution
n(s) = n(when 7 appears on thousands place)<br><br>+ n (7 does not appear on thousands place)<br><br>= 9 $\times$ 9 $\times$ 9 + 8 $\times$ 9 $\times$ 9 $\times$ 3<br><br>= 33 $\times$ 9 $\times$ 9<br><br>n(E) = n(last digit 7 & 7 appears once)<br><br>+n(last digit 2 when 7 appears once)<br><br>= 8 $\times$ 9 $\times$ 9 + (3 $\times$ 9 $\times$ 9 - 2 $\times$ 9)<br><br>$\therefore$ $$P(E) = {{8 \times 9 \times 9 + 9 \times 25} \over {33 \times 9 \times 9}} = {{97} \over {297}}$$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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