A six faced fair dice is thrown until $1$ comes. The probability that $1$ comes in even no. of trails is
- $\displaystyle \frac{5}{11}$
- $\displaystyle \frac{5}{6}$
- $\displaystyle \frac{6}{11}$
- $\displaystyle \frac{1}{6}$
The probability of rolling a 1 is p = 1/6, and not rolling a 1 is q = 5/6. The probability of getting the first 1 on an even trial is q*p + q^3*p + q^5*p + ... which is a geometric series with first term a = (5/6)*(1/6) = 5/36 and common ratio r = (5/6)^2 = 25/36. The sum is a/(1-r) = (5/36) / (1 - 25/36) = (5/36) / (11/36) = 5/11.
Let p be the probability of rolling a 1, which is 1/6, and let q be the probability of not rolling a 1, which is 5/6. The probability that the first 1 appears on an even-numbered throw is the infinite sum S = q*p + q^3*p + q^5*p + ..., which forms an infinite geometric series. Using the formula for an infinite geometric series, this sum equals p*q / (1 - q^2), so substituting the values gives (1/6)*(5/6) / (1 - 25/36) = (5/36) / (11/36) = 5 / 11.