even or odd !!!!!!!!!!!!!

Let's consider numbers from 0-9 since the probability will remain same with any other case due to symmetry (You can check for inteval of any 10 , 20 ..etc. numbers and you'll get same results)

Now, total ways of selection = ${ 10 }^{ 4 }$

For product to be a prime number all the four selections should be prime (1,3,5,7,9)

So, total ways of selecting = $5*5*5*5$ = 625

$P \left( odd \right)$ = $\frac { 625 }{ { 10 }^{ 4 } }$

$P \left( even \right)$ = $1 - P \left( odd \right) \quad$ = 1 - $\frac { 625 }{ { 10 }^{ 4 } }$ = $\frac { 9375 }{ { 10 }^{ 4 } }$

$P\left( even \right) =\frac { 15*625 }{ 10000 } =15*P\left( odd \right)$

$X = \boxed{15}$

We have 5 odd numbers and 5 even numbers, so:

Prob. of picking an odd number = prob. of picking an even number = $\frac{1}{2}$

Notice that the product of the 4 numbers would be odd only if we pick 4 odd numbers

Prob. of picking 4 odd numbers = $(\frac{1}{2})^{4}$ = $\frac{1}{16}$

So, prob. of picking any other choice = 1 - $\frac{1}{16}$ = $\frac{15}{16}$

Therefore it is $\boxed{15}$ times more probable to have an even product.

P[odd]=\frac{5^4}{10^4}= \frac{1}{16} and P[even]=1-P[odd] =\frac{15}{16} So, simply \boxed{X=15}