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# If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when, (ii) the repetition of digits is not allowed?

9.(ii) If $\small 4$-digit numbers greater than  $\small 5,000$  are randomly formed from the digits $\small 0,1,3,5$ and $\small 7$, what is the probability of forming a number divisible by $\small 5$ when, (ii) the repetition of digits is not allowed?

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(ii)

Since 4-digit numbers greater than 5000 are to be formed,

The $1000's$ place digit can be filled up by either 7 or 5 in $^{2}\textrm{C}_{1}$ ways

Since repetition is not allowed,

The remaining 3 places can be filled by remaining 4 digits in $^{4}\textrm{C}_{3}\times3!$  ways.

$\therefore$ Total number of 4-digit numbers greater than 5000 = $^{2}\textrm{C}_{1}\times(^{4}\textrm{C}_{3}\times3!)= 2\times4\times6 = 48$

We know, a number is divisible by 5 if unit’s place digit is either 0 or 5.

Case 1. When digit at $1000's$ place is 5, the units place can be filled only with 0.

And the $100's$ & $10's$ places can be filled with any two of the remaining digits {1,3,7} in $^{3}\texttrm{C}_{2}\times2!$

$\therefore$ Number of 4-digit numbers starting with 5 and divisible by 5 = $1\times ^{3}\texttrm{C}_{2}\times2! = 1.3.2 = 6$

Case 2. When digit at $1000's$ place is 7, the units place can be filled by 0 or 5 in 2 ways.

And the $100's$ & $10's$ places can be filled with any two of the remaining 3 digits in $^{3}\texttrm{C}_{2}\times2!$

$\therefore$ Number of 4-digit numbers starting with 7 and divisible by 5 = $1\times2\times( ^{3}\texttrm{C}_{2}\times2! )= 1.2.3.2 = 12$

$\therefore$ Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 6 + 12 = 18

Therefore, the required probability =

$P(without\ repetition) = \frac{18}{48} = \frac{3}{8}$

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