If 4-digit numbers greater than 5,000 are ran | vedclass

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(B) The set of digits is $S = \{0, 1, 3, 5, 7\}$. We need to form $4$-digit numbers greater than $5,000$ without repetition.The thousands place can be

Vedclass · Accepted Answer

$\frac{3}{8}$

Vedclass · Answer

(B) The set of digits is $S = \{0, 1, 3, 5, 7\}$. We need to form $4$-digit numbers greater than $5,000$ without repetition.The thousands place can be filled by $5$ or $7$ ($2$ choices).The remaining $3$ places can be filled by the remaining $4$ digits in $P(4, 3) = 4 	imes 3 	imes 2 = 24$ ways.Total $4$-digit numbers greater than $5,000 = 2 	imes 24 = 48$.For a number to be divisible by $5$,the units place must be $0$ or $5$.Case $1$: Thousands place is $5$.Units place must be $0$ ($1$ choice). The remaining $2$ places can be filled by the remaining $3$ digits in $P(3, 2) = 3 	imes 2 = 6$ ways.Case $2$: Thousands place is $7$.Units place can be $0$ or $5$ ($2$ choices). The remaining $2$ places can be filled by the remaining $3$ digits in $P(3, 2) = 3 	imes 2 = 6$ ways. Total $= 2 	imes 6 = 12$ ways.Total favorable outcomes $= 6 + 12 = 18$.Probability $= \frac{18}{48} = \frac{3}{8}$.

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 the repetition of digits is not allowed?

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