Question

# A software engineer creates a LAN game where an $8$ digit code made up of $1,2,3,4,5,6,7,8$ has to be decided on universal code. There is a condition that each number has to be used and no number can be repeated. What is the probability that first $4$ digits of the code are even numbers ?

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Solution

## Solve for the required probability:There are $8$ numbers to be filled in the $8$ digit code without any repetition.We know that probability $P\left(E\right)=\frac{n\left(E\right)}{n\left(s\right)}$ where $n\left(E\right)$ and $n\left(s\right)$ are number of favorable outcomes and total number of possible outcomes respectively.The number of possible arrangements of this are $8!$.Hence, the total number of outcomes$n\left(s\right)=8!$ $...\left(i\right)$Out of the given $8$ numbers exactly $4$ numbers are even$\left(2,4,6,8\right)$.If the first $4$ digits are to be occupied by even numbers then the last $4$ digits must be occupied by odd numbers.Now , the ways of arranging $4$ even numbers in first $4$digits are $4!$The ways of arranging $4$ odd numbers in last $4$digits are $4!$Hence, the favorable number of outcomes$n\left(E\right)=4!×4!$ $...\left(ii\right)$$⇒$$P\left(E\right)=\frac{4!×4!}{8!}$ $=\frac{24}{8×7×6×5}$$⇒$$P\left(E\right)=\frac{1}{70}$Hence, the probability that first $4$ digits of the $8$ digit code being even is $\frac{1}{70}$.

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