Probability
Infinite Trials and Geometric Distributions
Grade 12
Question:
<p>A six faced fair die is thrown until 1 comes. Then, the probability that 1 comes in even number of trials, is</p>
<p>(a) \(\frac{5}{11}\)</p>
<p>(b) \(\frac{5}{6}\)</p>
<p>(c) \(\frac{6}{11}\)</p>
<p>(d) \(\frac{1}{6}\)</p>
Step-by-Step Solution
Key Concept: Sum the infinite geometric series representing all even outcomes: each even trial needs exactly an odd number of failures followed by success.
<p>The probability of getting 1 on a fair die is $\frac{1}{6}$ and not getting 1 is $\frac{5}{6}$. For 1 to come in an even number of trials, it must come in 2nd, 4th, 6th, ... trial. $P(\text{even}) = P(\text{no 1, then 1}) + P(\text{no 1, no 1, no 1, then 1}) + \ldots = \frac{5}{6} \cdot \frac{1}{6} + \left(\frac{5}{6}\right)^3 \cdot \frac{1}{6} + \ldots = \frac{1}{6} \cdot \frac{5}{6} \left(1 + \left(\frac{5}{6}\right)^2 + \left(\frac{5}{6}\right)^4 + \ldots\right) = \frac{5}{36} \cdot \frac{1}{1 - \frac{25}{36}} = \frac{5}{36} \cdot \frac{36}{11} = \frac{5}{11}$.</p>
Correct Answer: A