An ordinary dice is rolled a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times, then the probability of getting an odd number an odd number of times, is

$\frac{1}{2}$

Let the dice be roled n times.

We have,

p = Probability of getting an odd number in a throw = $\frac{1}{2}$

$= {^nC}_2 \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{n-2} ={^nC}_2 \left(\frac{1}{2}\right)^n$

Similarly, we have

Probability of getting an even number 3 times

$= {^nC}_3 \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{n-3} ={^nC}_3 \left(\frac{1}{2}\right)^n$

It is given that

$= {^nC}_2 \left(\frac{1}{2}\right)^n = {^nC}_3 \left(\frac{1}{2}\right)^n⇒ {^nC}_2 = {^nC}_3 ⇒n = 5 $

Let X denote the number of times an odd number is obtained in n throws of the dice. Then,

$P(X=r)=  {^nC}_3 \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right)^{n-r} ={^nC}_r \left(\frac{1}{2}\right)^n={^5C}_r \left(\frac{1}{2}\right)^5 $

∴ Required probability 

$= P(X=1) +P(X=3)+P(X=5)$

$={^5C}_1 \left(\frac{1}{2}\right)^5 + {^5C}_3 \left(\frac{1}{2}\right)^5 + {^5C}_5 \left(\frac{1}{2}\right)^5=\frac{1}{2}$