If a number is chosen at random from the four | vedclass

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(B) Total number of $4$-digit numbers formed using digits ${0, 1, 2, 3, 4, 6}$ without repetition: Thousands place can be filled in $5$ ways (excludin

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$\frac{17}{50}$

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(B) Total number of $4$-digit numbers formed using digits ${0, 1, 2, 3, 4, 6}$ without repetition: Thousands place can be filled in $5$ ways (excluding $0$). Hundreds place can be filled in $5$ ways. Tens place can be filled in $4$ ways. Units place can be filled in $3$ ways. Total $= 5 	imes 5 	imes 4 	imes 3 = 300$. $A$ number is divisible by $4$ if the number formed by its last two digits is divisible by $4$. The possible last two-digit combinations from ${0, 1, 2, 3, 4, 6}$ are: $04, 12, 16, 20, 24, 32, 36, 40, 60, 64$. Case $1$: Last two digits are $04, 20, 24, 40, 60, 64$ (contain $0$ or $4$). - If the last two digits are $04, 20, 40, 60$: Remaining $2$ places can be filled in $4 	imes 3 = 12$ ways each. Total $= 4 	imes 12 = 48$ ways. - If the last two digits are $24, 64$: Thousands place cannot be $0$ or the digits used. Thousands place has $3$ choices,Hundreds place has $3$ choices. Total $= 2 	imes (3 	imes 3) = 18$ ways. Case $2$: Last two digits are $12, 16, 32, 36$: Thousands place cannot be $0$ or the digits used. Thousands place has $3$ choices,Hundreds place has $3$ choices. Total $= 4 	imes (3 	imes 3) = 36$ ways. Total favorable outcomes $= 48 + 18 + 36 = 102$. Probability $= \frac{102}{300} = \frac{17}{50}$.

If a number is chosen at random from the four-digit numbers formed by using the digits $0, 1, 2, 3, 4, 6$ without repetition,then the probability that it is divisible by $4$ is:

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