Problem 25 - A First Course in Probability

Question

A pair of dice is rolled until a sum of either 5 or 7 appears. Find the probability that a 5 occurs first. Hint: Let \(E_{n}\) denote the event that a 5 occurs on the \(n\) th roll and no 5 or 7 occurs on the first \(n-1\) rolls. Compute \(P\left(E_{n}\right)\) and argue that \(\sum_{n=1}^{\infty} P\left(E_{n}\right)\) is the desired probability.

Step-by-Step Solution

Answer

The probability that a sum of 5 occurs first when rolling a pair of dice is \(\frac{2}{5}\).

Step 1: Probability of getting a sum of 5 or 7 on a single roll

First, let's calculate the probability of getting a sum of 5 or 7 when rolling a pair of dice. There are 6 x 6 = 36 equally likely outcomes when rolling two dice. Four of these outcomes result in a sum of 5 (1,4), (2,3), (3,2), and (4,1), and six outcomes result in a sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). Therefore, the probability of getting either a sum of 5 or 7 on a single roll is \(\frac{4 + 6}{36} = \frac{10}{36}\).

Step 2: Probability of not getting a sum of 5 or 7 on a single roll

Since there are 10 outcomes resulting in a sum of 5 or 7, there are 36 - 10 = 26 outcomes that do not result in a sum of 5 or 7. So, the probability of not getting either a sum of 5 or 7 on a single roll is \(\frac{26}{36}\).

Step 3: Calculate the probability of event \(E_n\)

Event \(E_n\) denotes getting a sum of 5 on the nth roll and not getting a sum of 5 or 7 in previous n-1 rolls. To calculate this probability \(P(E_n)\), we first consider the probability of not getting a sum of 5 or 7 in the first n-1 rolls, which is \((\frac{26}{36})^{n-1}\). Then, on the nth roll, we want to get a sum of 5, which has a probability of \(\frac{4}{36}\). So the probability of event \(E_n\) can be written as: \(P(E_n) = \left(\frac{26}{36}\right)^{n-1} \cdot \frac{4}{36}\).

Step 4: Calculate the desired probability

We are interested in finding the probability that a 5 occurs first, which means we need to calculate the sum of the probabilities of all events \(E_n\) for n ranging from 1 to infinity. This can be expressed as: \(P(\text{5 occurs first}) = \sum_{n=1}^{\infty} P(E_n) = \sum_{n=1}^{\infty} \left(\frac{26}{36}\right)^{n-1} \cdot \frac{4}{36}\). This is an infinite geometric series with the first term \(a = \frac{4}{36}\) and the common ratio \(r = \frac{26}{36}\). The sum of an infinite geometric series is \(\frac{a}{1 - r}\). Applying this formula, we get: \(P(\text{5 occurs first}) = \frac{\frac{4}{36}}{1 - \frac{26}{36}} = \frac{4}{10}\). So, the probability that a sum of 5 occurs first when rolling a pair of dice is \(\frac{4}{10} = \frac{2}{5}\).