Final answer:
To calculate the probability distribution for the number of rolls required to win when two spaces away in a board game, we use the geometric distribution formula P(X = n) = q^(n-1) × p, where 'p' is the probability of rolling snake eyes (0.0278) with two dice and 'q' is the probability of not rolling snake eyes (0.9722). The probabilities are calculated for each roll up to ten.
Step-by-step explanation:
To find the probability distribution for the number of rolls required to win the board game when a player is two spaces from the end, we need to consider that the only way to roll exactly two with two dice is to roll a 1 on one die and a 1 on the other (also known as rolling "snake eyes"). The probability of rolling a 1 on a fair six-sided die is ½, and since the two dice are independent, the probability of rolling a 1 on both dice is ½ × ½ = ⅛.
The probability of not rolling a total of two on two dice is 1 minus the probability of rolling snake eyes, which is 1 - (⅛) = ⅝. If we let the random event that we roll two snake eyes be a "success" and all other outcomes be "failures," the probability of each outcome does not change from roll to roll. Therefore, we can model this scenario using the geometric distribution to identify the probability of the first success (rolling snake eyes) on the nth roll.
Using the geometric probability formula, P(X = n) = q(n-1) × p, where n is the roll number, p is the success probability, q is the failure probability, and X is the random variable representing the number of rolls until the first success, we can calculate the probabilities up to the tenth roll.
- Roll 1: P(X = 1) = (⅝)(1-1) × (⅛) = 0.0278
- Roll 2: P(X = 2) = (⅝)(2-1) × (⅛) = 0.0259
- ... (continue calculating similarly for rolls 3 to 10) ...
- Roll 10: P(X = 10) = (⅝)(10-1) × (⅛) = 0.0146
Note: All probabilities should be rounded to four decimal places, as these are the standards for probability problems.