A fair six-sided die has one blue face and five non-blue faces. What is the probability that a player will need to toss the die at least $2$ times before blue lands faceup for the first time?

• A. $\frac{25}{36}$
• B. $\frac{1}{6}$
• C. $\frac{1}{36}$
• D. $\frac{5}{6}$

Answer: (D)

Verified step by step guidance

1. Identify the probability of the event of interest: the die landing on the blue face. Since there is 1 blue face out of 6, the probability of blue on a single toss is \(\frac{1}{6}\).
2. Determine the complementary event: the die does NOT land on blue. This probability is \$1 - \frac{1}{6} = \frac{5}{6}$.
3. Understand that "needing at least 2 tosses before blue appears" means the first toss is NOT blue. So, the player must fail to get blue on the first toss.
4. Since each toss is independent, the probability that the first toss is not blue (and thus the first blue appears on the 2nd toss or later) is simply the probability of not getting blue on the first toss, which is \(\frac{5}{6}\).
5. Therefore, the probability that the player needs at least 2 tosses before blue lands faceup for the first time is \(\frac{5}{6}\).