A die used in a certain board game has eight faces, of which 3 are red, 3 are yellow, and 2 are blue. Each face is equally likely to land faceup when the die is tossed. In the game, a player tosses the die until blue lands faceup, and the number of tosses before blue lands faceup is counted. For example, a player who tosses the sequence shown in the following table has tossed the die 3 times before blue lands faceup. What is the probability that a player will toss the die at least 2 times before blue lands faceup? A 0.1406 B 0.4219 C 0.4375 D 0.5625 E 0.5781

Using the binomial distribution, it is found that the probability that a player will toss the die at least 2 times before blue lands face up is:

D 0.5625

For each toss, there are only two possible outcomes, either blue lands face up, or it does not. The probability of blue landing face up on a toss is independent of any other toss, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

2 out of the 8 faces are blue, hence [tex]p = \frac{2}{8} = 0.25[/tex]

The probability that a player will toss the die at least 2 times before blue lands face up is P(X = 0) when n = 2, hence:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{2,0}.(0.25)^{0}.(0.75)^{2} = 0.5625[/tex]

Hence, option D is correct.

A similar problem is given at https://brainly.com/question/24863377