The probability that a randomly chosen 5-digi | vedclass

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(C) Total number of $5-digit$ numbers is $9 	imes 10^{4}$.Case $1$: The two digits chosen are both non-zero.Number of ways to choose $2$ digits from

Vedclass · Accepted Answer

$\frac{135}{10^{4}}$

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(C) Total number of $5-digit$ numbers is $9 	imes 10^{4}$.Case $1$: The two digits chosen are both non-zero.Number of ways to choose $2$ digits from ${1, 2, \dots, 9}$ is $^{9}C_{2} = 36$.For each selection,the number of $5-digit$ numbers using both digits is $2^{5} - 2 = 30$.Total for Case $1 = 36 	imes 30 = 1080$.Case $2$: One digit is zero and the other is non-zero.Number of ways to choose $1$ non-zero digit from ${1, 2, \dots, 9}$ is $^{9}C_{1} = 9$.The first digit must be non-zero (fixed in $1$ way),and the remaining $4$ positions can be filled by either $0$ or the chosen non-zero digit,excluding the case where all $4$ are non-zero. This gives $2^{4} - 1 = 15$ ways.Total for Case $2 = 9 	imes 15 = 135$.Total favorable outcomes $= 1080 + 135 = 1215$.Probability $= \frac{1215}{9 	imes 10^{4}} = \frac{135}{10^{4}}$.

The probability that a randomly chosen $5-digit$ number is made from exactly two digits is

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