A dice is thrown (2n +1) times. The probability that faces with even…

We have,

p= Probability of getting an even number in a throw

$⇒ p=\frac{3}{6}=\frac{1}{2}$

$∴ q= 1 -p=\frac{1}{2}$

Let X denote the number of times an even number is shown in (2n +1) throws of a dice. Then, X follows binomial distribution such that

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

∴ Required probability $= \sum\limits^{2n+1}_{r=1}P(X=r)= \sum\limits^{2n+1}_{r=1}{^{2n+1}C}_r \left(\frac{1}{2}\right)^{2n+1}$

$= \left(\frac{1}{2}\right)^{2n+1}\begin{Bmatrix}{^{2n+1}C}_1 + {^{2n+1}C}_3 + ...+{^{2n+1}C}_{2n+1}\end{Bmatrix}$

$= \frac{1}{2^{2n+1}}×2^{2n}=\frac{1}{2}$