Question

Getting to the Finish In a certain board game participants roll a standard six-sided die and need to hit a particular value to get to the finish line exactly. For example, if Carol is three spots from the finish, only a roll of 3 will let her win; anything else and she must wait another turn to roll again. The chance of getting the number she wants on any roll is \(p=1 / 6\) and the rolls are independent of each other. We let a random variable \(X\) count the number of turns until a player gets the number needed to win. The possible values of \(X\) are \(1,2,3, \ldots\) and the probability function for any particular count is given by the formula $$ P(X=k)=p(1-p)^{k-1} $$ (a) Find the probability a player finishes on the third turn. (b) Find the probability a player takes more than three turns to finish.

Short answer

(a) The probability of a player finishing on the third turn is calculated using the formula \(P(X=3)= (1/6)(5/6)^2\). (b) The probability of a player taking more than three turns to finish is found by taking the complement of the sum of the probabilities of finishing in one, two, or three turns, which is calculated as \(1 - P(X = 1) - P(X = 2) - P(X = 3)\).

Step-by-step solution

Calculate the probability of finishing on the third turn

First, plug in \(p = 1 / 6\) (the probability of winning in one turn), and \(k = 3\) (the third turn) into the formula: \(P(X=3)= (1/6)(1-1/6)^{3-1} = (1/6)(5/6)^2\). Calculate this expression to find the probability of finishing on the third turn.

Calculate the probability of finishing in 1, 2 or 3 turns

We need this for calculating the probability of finishing in more than three turns. For each turn up to the third, calculate \(P(X=k)\) with \(k = 1, 2, 3\) and sum up these probabilities:

Calculate the probability of finishing in more than three turns

The probability of finishing in more than three turns is the complement of finishing in three or fewer turns. So, it equals \(1 - P(X = 1) - P(X = 2) - P(X = 3)\)