A five-digit number is formed using the digit | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) number is divisible by $4$ if the number formed by its last two digits is divisible by $4$. Using the digits ${1, 2, 3, 4, 5}$, the possible two-d

Vedclass · Accepted Answer

$\frac{1}{5}$

Vedclass · Answer

(A) number is divisible by $4$ if the number formed by its last two digits is divisible by $4$. Using the digits ${1, 2, 3, 4, 5}$, the possible two-digit combinations divisible by $4$ are $12, 24, 32, 52$. There are $4$ such favorable combinations for the last two digits. The remaining $3$ positions can be filled by the remaining $3$ digits in $3!$ ways. Total number of favorable outcomes $= 4 	imes 3!$. Total number of possible five-digit numbers $= 5!$. Probability $= \frac{4 	imes 3!}{5!} = \frac{4 	imes 3!}{5 	imes 4 	imes 3!} = \frac{1}{5}$.

$A$ five-digit number is formed using the digits $1, 2, 3, 4, 5$ without repetition. What is the probability that the number is divisible by $4$?

Similar Questions

If a die is rolled three times,then the probability of getting a larger number on its face than the previous number each time is:

$A$ bag contains $5$ white,$7$ red,and $8$ black balls. If four balls are drawn one by one without replacement,what is the probability that all are white?

There are $4$ hotels in a town. If $3$ men check into the hotels in a day,then the probability that each checks into a different hotel is

Two numbers $b$ and $c$ are chosen at random in succession without replacement from the set $\{1, 2, 3, \ldots, 9\}$. Then the probability that $x^2 + bx + c > 0, \forall x \in R$ is

An urn contains $6$ yellow balls and $x$ black balls. When two balls are drawn at random,the probability that both are yellow is $\frac{5}{26}$,then the value of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module