Question

In a throw of a dice the probability of getting one in even number of throw is

• A. $\frac{5}{36}$
• B. $\frac{5}{11}$
• C. $\frac{6}{11}$
• D. $\frac{1}{6}$

Answer 1

The correct option is B $\frac{5}{11}$

Probability of getting 1 on $2^{nd}$ throw,

$P\left(2\right)=\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)$

Probability of getting 1 on $4^{th}$ throw,

$P\left(4\right)={\left(\frac{5}{6}\right)}^3\left(\frac{1}{6}\right)$

Probability of getting 1 on $6^{th}$ throw,

$P\left(6\right)={\left(\frac{5}{6}\right)}^5\left(\frac{1}{6}\right)$

Therefore total probability,

$P=P\left(2\right)+P\left(4\right)+P\left(6\right)+$...…

$P=\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)+{\left(\frac{5}{6}\right)}^3\left(\frac{1}{6}\right)+{\left(\frac{5}{6}\right)}^3\left(\frac{1}{6}\right)+$...….

$P=\frac{1}{6}[\left(\frac{5}{6}\right)+{\left(\frac{5}{6}\right)}^3+{\left(\frac{5}{6}\right)}^5+...\dots ]$

By sum of an infinite geometric series,

$P=\frac{1}{6}\left[\frac{\frac{5}{6}}{1-{\left(\frac{5}{6}\right)}^2}\right]$

$\therefore P=\frac{5}{11}$.