If five-digit numbers are formed from the dig | vedclass

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(A) The total number of five-digit numbers formed using the digits $0, 1, 2, 3, 4$ exactly once is $4 	imes 4! = 4 	imes 24 = 96$. For a number to b

Vedclass · Accepted Answer

$\frac{5}{16}$

Vedclass · Answer

(A) The total number of five-digit numbers formed using the digits $0, 1, 2, 3, 4$ exactly once is $4 	imes 4! = 4 	imes 24 = 96$. For a number to be divisible by $4$,the last two digits must form a number divisible by $4$. The possible pairs for the last two digits are: $04, 20, 40$ (where $0$ is used): Each gives $3! = 6$ numbers. Total $= 3 	imes 6 = 18$. $12, 24, 32$ (where $0$ is not used): For each,the first digit cannot be $0$,so there are $3$ choices for the first digit and $2!$ for the remaining two. Total $= 3 	imes (3 	imes 2!) = 3 	imes 6 = 18$. Wait,let us re-evaluate: If last two digits are $04, 20, 40$: The remaining $3$ digits can be arranged in $3! = 6$ ways. $3 	imes 6 = 18$. If last two digits are $12, 24, 32$: The first digit cannot be $0$ and cannot be one of the two digits used. So $4 - 2 = 2$ choices for the first digit,then $2!$ for the remaining. $3 	imes (2 	imes 2!) = 3 	imes 4 = 12$. Total favorable numbers $= 18 + 12 = 30$. Probability $= \frac{30}{96} = \frac{5}{16}$.

List-$I$	List-$II$
$A$. Probability of getting $1$ red,$1$ white and $1$ blue ball	$(i)$ $\frac{3}{44}$
$B$. Probability of getting $2$ white and $1$ blue ball	$(ii)$ $\frac{21}{55}$
$C$. Probability of getting $2$ red and $1$ white ball	$(iii)$ $\frac{38}{55}$
$D$. Probability that none of the balls is white	$(iv)$ $\frac{3}{11}$
	$(v)$ $\frac{9}{110}$

If five-digit numbers are formed from the digits $0, 1, 2, 3, 4$ using every digit exactly once,then the probability that a randomly chosen number from those numbers is divisible by $4$ is:

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