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

Question

An unbiased die is thrown $n$ times. The probability that the product of numbers would be even is

1. $1/(2n)$
2. $1/[(6n)!]$
3. $1−6^{−n}$
4. $6^{−n}$
5. None of the above.

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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

$= 1 - \left(\cfrac{1}{2}\right)^n$

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

Answer 1

Your answer is correct.

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

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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 $1-\left(\frac36\right)^n$