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 \(\frac{4}{10} = \frac{2}{5}\).

Key Concepts

Geometric Series

The concept of a geometric series is a fundamental part of probability theory and plays a crucial role in solving this problem involving dice. A geometric series is a series of terms where each term is a constant ratio of the previous term. In simpler terms, it's a sequence of numbers where each number is a fixed multiple of the number before it. This fixed number is called the 'common ratio'.
When we're rolling the dice in our exercise, we're dealing with an infinite geometric series because there is a small chance that it might take an infinite number of rolls before getting a sum of 5 or 7. The probability of each roll without a sum of 5 or 7 behaves like a geometric sequence. The first term in the sequence is the probability of getting a sum of 5 on the first roll, and each subsequent term is reduced by the probability that neither 5 nor 7 appeared on previous rolls.
By calculating the sum of this infinite series, we can find the probability of rolling a sum of 5 first. The formula for the sum of an infinite geometric series is \[ S = \frac{a}{1 - r} \] where \( a \) is the first term of the series, and \( r \) is the common ratio.
Here, \( a = \frac{4}{36} \) (probability of getting a sum of 5 on the first roll), and \( r = \frac{26}{36} \), the probability of rolling neither a 5 nor a 7. By plugging into the formula, we compute the probability that the sum of 5 occurs first.

Dice Probabilities

Understanding dice probabilities is essential for solving problems involving dice, like the one in this exercise. When rolling two dice, each die has 6 faces, so there are a total of 36 possible outcomes (6 times 6). Different sums can be obtained depending on the combination of numbers from the two dice.
To find the probability of any specific outcome, like the sum of 5 or 7, we need to count the possible combinations that result in that sum. For instance:

• A sum of 5 can happen through the combinations (1,4), (2,3), (3,2), and (4,1) - giving us 4 possibilities.
• A sum of 7 can occur through the combinations (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1) - providing us with 6 possibilities.

Altogether, there are 10 ways to get a sum of 5 or 7, thus the probability of getting either is \( \frac{10}{36} \).
By understanding these probabilities, we can set the foundation for calculating probabilities of more complex events like what occurs in multiple dice rolls.

Event Probability

In probability theory, understanding the concept of event probability is fundamental. An event is a specific outcome or group of outcomes from a random situation. When rolling dice multiple times, events can be complex sequences. In our exercise, the event \( E_n \) is a particular case of interest.
Event \( E_n \) represents the scenario where the first \( n-1 \) rolls do not result in a sum of 5 or 7, and a sum of 5 is achieved on the \( n \)-th roll.

• This means that for the first \( n-1 \) rolls, the outcomes are any of the 26 outcomes that are not a sum of 5 or 7.
• The probability of no 5 or 7 in \( n-1 \) rolls is \( \left( \frac{26}{36} \right)^{n-1} \).
• On the \( n \)-th roll, we need a sum of 5, which occurs with probability \( \frac{4}{36} \).

Therefore, the probability \( P(E_n) \) is the product of these probabilities: \( \left( \frac{26}{36} \right)^{n-1} \times \frac{4}{36} \).
By considering all such possible values of \( n \), we can determine the event probability that a 5 occurs before a 7 is rolled, helping us solve problems where outcomes over multiple trials are of interest.