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

$\frac{5}{11}$

Explanation for the correct option:

Step 1: Determining the probability of one in an even number of throw

The probability of getting one in the second throw = $P\left(2\right)$ = $\frac{5}{6}\times \frac{1}{6}$

The probability of getting one in the fourth row =$P\left(4\right)$= ${\left(\frac{5}{6}\right)}^3\left(\frac{1}{6}\right)$

The probability of getting one in the sixth row =$P\left(6\right)$= ${\left(\frac{5}{6}\right)}^5\left(\frac{1}{6}\right)$

Hence, the required probability = $P\left(2\right)+P\left(4\right)+P\left(6\right)+.......$

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

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

Step 2: Applying the concept of infinite geometric series

So, the required probability = $\left(\frac{1}{6}\right)\left(\frac{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}{1-{\left({\displaystyle \frac{5}{6}}\right)}^2}\right)=\frac{5}{11}$

Hence, the probability of getting one in an even number of throw $=\frac{5}{11}$

Therefore, the correct option is (B)