An unbiased dice is thrown n times what is the probability that the product of numbers would be even?

Question

I am doing it like $n\choose1 $ $ \frac{1}{2} {\frac{1}{2}}^{n-1}$ + $n\choose 2 $${\frac{1}{2}}^2 {\frac{1}{2}}^{n-2} + ......+$$n\choose n $ ${\frac{1}{2}}^n $ Is this correct? Further I am not able to simplify it significantly beyond ${\frac{1}{2}}^n \frac{n!}{(n-1)!}(1+\frac{n-1}{2!}+......+\frac{(n-1)!}{n!})$ how to simplify it further?

Answer 1

So you are asking what is the probability that at least one outcome is even.

You need to calculate $1$ minus the probability of the complementary event:

• The probability that all outcomes are odd is $\left(\frac12\right)^n$
• The probability that not all outcomes are odd is $1-\left(\frac12\right)^n$

Your way is binomial distribution with $p=\frac12$:

$\sum\limits_{k=1}^{n}\binom{n}{k}\cdot\left(\frac12\right)^{k}\cdot\left(1-\frac12\right)^{n-k}=$

$1-\sum\limits_{k=0}^{0}\binom{n}{k}\cdot\left(\frac12\right)^{k}\cdot\left(1-\frac12\right)^{n-k}=$

$1-\binom{n}{0}\cdot\left(\frac12\right)^{0}\cdot\left(1-\frac12\right)^{n-0}=$

$1-\left(\frac12\right)^n$

Which is indeed the same.