Two fair dice are thrown till outcome is $12$. The probability that one has to do $20$ throws for this is ______.
Step-by-Step Solution
Key Concept: This is a geometric distribution problem where we need exactly 19 failures before the first success, giving probability $(1-p)^{n-1} \cdot p$ with $p = \frac{1}{36}$ and $n = 20$.
We need exactly 19 failures (not getting 12) followed by 1 success (getting 12) on the 20th throw. The probability of getting 12 (sum of two dice) is $P(12) = \frac{1}{36}$ (only outcome: 6,6). The probability of not getting 12 is $P(\text{not 12}) = \frac{35}{36}$. For exactly 20 throws needed, we require 19 consecutive failures and then success on throw 20: $P = \left(\frac{35}{36}\right)^{19} \times \frac{1}{36}$. Computing: $\left(\frac{35}{36}\right)^{19} \approx 0.5897$ and $0.5897 \times \frac{1}{36} \approx 0.0130$.
Correct Answer: 0.0130