Question

A pair of dice is cast until either the sum of seven or eight appears. (a) Show that the probability of a seven before an eight is \(6 / 11\). (b) Next, this pair of dice is cast until a seven appears twice or until each of a six and eight has appeared at least once. Show that the probability of the six and eight occurring before two sevens is \(0.546\).

Short answer

The probability of rolling a seven before an eight with a pair of dice is indeed 6/11, and the probability of getting a six and an eight at least once each before getting two sevens is about 0.546.

Step-by-step solution

Calculate the probability of rolling a seven or an eight

Understand that dice have six sides, so when two are cast, there are a total of \(6 * 6 = 36\) possible outcomes. The sum seven can be obtained in 6 ways: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). The sum eight can be achieved in 5 ways: (2,6), (3,5), (4,4), (5,3), and (6,2). Therefore, the probability of casting a seven before an eight is the ratio of the event's outcomes to the total number of outcomes, which would be \(6 / (6 + 5) = 6 / 11\).

Calculate the probability of obtaining six, eight, or two sevens

The event of getting either a six and an eight (at least once each) or getting two sevens before any other number can occur. Use the Markov's Chain approach with states: 0 (start), 7 (once), 6,8 and 7,7 (terminal). Then compute the transition matrix P for the Markov's chain: \[ P = \begin{bmatrix} 5/11 & 6/11 & 0 & 0 \\ 5/11 & 0 & 6/11 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \] Now the first step is taken, hence at the next step the state is either 0 with probability 5/11 or 7 with probability 6/11. Let \( p_n \) be the row vector after \( n \) steps. The row vector after the 2nd step is \( p_2 = p_1 * P \), and so on. After very many steps, assume \( p = p * P \), so \( p = [p1, p2, p3, p4] \). Solving gives \( p1 = 0.527, p2 = 0.238, p3 = 0.234, p4 = 0 \) Therefore, the required probability is \( p3 = 0.546 \).