Product of numbers is even , when an unbiased die rolled? An unbiased die is thrown $n$ times. The probability that the product of numbers would be even is

*

*$1/(2n)$

*$1/[(6n)!]$

*$1−6^{−n}$

*$6^{−n}$

*None of the above.


My attempt :
we have $3$ even number and $3$ odd number in the an unbiased dice , i.e. $2,4,6$ and $1,3,5$ respectively .
Probability to occur odd number , i.e. P(odd) $= \cfrac{1}{2}$
As we know , if all numbers are odd then product of these number will be odd , else even , in other words , if at least one even number occur then product of these number will be even .
Therefore , required probability is ,
$=$ probability for at least one even number occur
$= 1 -$ probability for all odd number occur
$= 1 - \left(\cfrac{1}{2}\right)^n$


Product of numbers is even , when an unbiased die rolled ?


Your answer is correct.
There's not much left to do with it, except perhaps writing it down slightly more "tidy".

Calculate $1$ minus the probability of the complementary event:


*

*The probability of getting an odd number in a roll is $\frac36$

*The probability of getting an odd product in $n$ rolls is $\left(\frac36\right)^n$

*The probability of getting an even product in $n$ rolls is $1-\left(\frac36\right)^n$

