**Question:**

An ordinary dice is rolled for 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 for odd number of times is:

Correct Option: , 4

**Solution:**

${ }^{\mathrm{n}} \mathrm{C}_{2}\left(\frac{1}{2}\right)^{\mathrm{n}}={ }^{\mathrm{n}} \mathrm{C}_{3}\left(\frac{1}{2}\right)^{\mathrm{n}} \Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{2}={ }^{\mathrm{n}} \mathrm{C}_{3}$

$\Rightarrow \quad n=5$

Probability of getting an odd number for odd number of times is

${ }^{5} \mathrm{C}_{1}\left(\frac{1}{2}\right)^{5}+{ }^{5} \mathrm{C}_{3}\left(\frac{1}{2}\right)^{5}+{ }^{5} \mathrm{C}_{5}\left(\frac{1}{2}\right)^{5}=\frac{1}{2^{5}}(5+10+1)$

$=\frac{1}{2}$