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

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(B) The total number of four-digit numbers formed using the digits $\{0, 1, 2, 3, 4, 6\}$ without repetition is calculated as follows: The first digit

Vedclass · Accepted Answer

$\frac{17}{50}$

Vedclass · Answer

(B) The total number of four-digit numbers formed using the digits $\{0, 1, 2, 3, 4, 6\}$ without repetition is calculated as follows: The first digit cannot be $0$,so there are $5$ choices. The remaining three positions can be filled in $5 	imes 4 	imes 3 = 60$ ways. Thus,total numbers $= 5 	imes 60 = 300$.$A$ number is divisible by $4$ if its last two digits form a number divisible by $4$. The possible pairs for the last two digits are $\{04, 12, 16, 20, 24, 32, 36, 40, 60, 64\}$.Case $1$: Pairs containing $0$ $(\{04, 20, 40, 60\})$: There are $4$ such pairs. For each,the remaining $2$ positions can be filled by the remaining $4$ digits in $4 	imes 3 = 12$ ways. Total $= 4 	imes 12 = 48$.Case $2$: Pairs not containing $0$ $(\{12, 16, 24, 32, 36, 64\})$: There are $6$ such pairs. For each,the first digit cannot be $0$ or the two digits already used,leaving $3$ choices for the first digit and $3$ choices for the second digit. Total $= 6 	imes (3 	imes 3) = 6 	imes 9 = 54$.Total favourable outcomes $= 48 + 54 = 102$.Required 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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