Practicing Success
A dice is thrown (2n +1) times. The probability that faces with even numbers show odd number of times is |
$\frac{1}{2}$ $<\frac{1}{2}$ $>\frac{1}{2}$ none of these |
$\frac{1}{2}$ |
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 $= \sum\limits^{2n+1}_{r=1}P(X=r)= \sum\limits^{2n+1}_{r=1}{^{2n+1}C}_r \left(\frac{1}{2}\right)^{2n+1}$ $= \left(\frac{1}{2}\right)^{2n+1}\begin{Bmatrix}{^{2n+1}C}_1 + {^{2n+1}C}_3 + ...+{^{2n+1}C}_{2n+1}\end{Bmatrix}$ $= \frac{1}{2^{2n+1}}×2^{2n}=\frac{1}{2}$ |