Problem 34 - Discrete Mathematics with Applications

Question

A die is rolled four times. Find the probability of obtaining: Exactly two sixes.

Step-by-Step Solution

Answer

The probability of obtaining exactly two sixes when rolling a die four times is \(\frac{25}{216}\).

Step 1: Identify the parameters of the binomial distribution

We are given a die is rolled four times (n = 4) and we want to find the probability of obtaining exactly two sixes (k = 2). There is a 1/6 probability of rolling a six and a 5/6 probability of not rolling a six in each trial.

Step 2: Calculate the number of successful outcomes (combinations)

We will use the binomial coefficient formula to calculate the number of successful outcomes, which is given by: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] where n is the number of trials and k is the number of successful trials. So, plugging in n = 4 and k = 2, we have: \[ C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4\times3}{2} = 6 \] There are 6 combinations of rolling exactly two sixes in four trials.

Step 3: Calculate the probability of each outcome

To find the probability of each outcome, we can use the binomial probability formula: \[ P(X=k) = C(n, k) \times p^k \times (1-p)^{(n-k)} \] where P(X=k) is the probability of obtaining exactly k sixes, C(n, k) is the number of combinations, p is the probability of success (rolling a six), and (1-p) is the probability of failure (not rolling a six). Plugging in the values, we have: \[ P(X=2) = 6 \times \left(\frac{1}{6}\right)^2 \times \left(\frac{5}{6}\right)^{(4-2)} \]

Step 4: Simplify the expression

Now, let's simplify the expression: \[ P(X=2) = 6 \times \frac{1}{36} \times \frac{25}{36} = 6 \times \frac{25}{1296} = \frac{150}{1296} \]

Step 5: Obtain the final probability

To obtain the final probability, simplify the fraction: \[ P(X=2) = \frac{150}{1296} = \frac{25}{216} \] Therefore, the probability of obtaining exactly two sixes when rolling a die four times is \(\frac{25}{216}\).

Key Concepts

Binomial Coefficient

Understanding the binomial coefficient is crucial when solving problems involving probability and combinations. The binomial coefficient, denoted as \( C(n, k) \), represents the number of ways to choose \( k \) successful outcomes from \( n \) trials, without considering the order of selection. This is often referred to as 'n choose k'.

The generic formula for the binomial coefficient is
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]
where \( n! \) (n factorial) is the product of all positive integers up to \( n \), \( k! \) is the product of all positive integers up to \( k \), and \( (n-k)! \) is the factorial of the difference between \( n \) and \( k \).

A practical example: If you roll a die four times, there are multiple outcomes where you can get exactly two sixes. By calculating the binomial coefficient for \( n=4 \) trials and \( k=2 \) successes, you'll find there are 6 possible combinations that result in exactly two sixes. Therefore, \( C(4, 2) = 6 \).

Probability of an Event

The probability of an event is a measure of how likely that event is to occur. It is a fundamental concept in statistics, described as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

The probability \( P \) of an event happening can be calculated if the number of likely outcomes and the total number of possible outcomes are known. The basic formula is:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \]

In the context of rolling a die, getting any specific number (like a six) has a probability of \( \frac{1}{6} \) because there are six possible outcomes, and only one is the desired six. If you toss the die four times, looking for exactly two sixes, the probability is not merely \( \frac{1}{6} \) but influenced by both the binomial coefficient (how the outcomes are combined) and the individual probabilities of success and failure, paving the way for using the binomial probability formula.

Factorial in Permutations

The factorial is an essential concept in both permutations and combinations. Represented by an exclamation mark \( ! \), a factorial signifies the product of an integer and all the non-zero integers below it. For instance, \( 5! \) is equal to \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

In permutations where the order matters, factorials are used to calculate the total number of arrangements. The formula for permutations of \( n \) items taken \( r \) at a time is given by:
\[ P(n, r) = \frac{n!}{(n-r)!} \]

For example, if you want to know how many two-letter arrangements can be made from four unique letters, you would calculate \( \frac{4!}{(4-2)!} \), which simplifies to \( 4 \times 3 = 12 \) arrangements. This emphasizes how factorial plays a pivotal role in determining permutations and, indirectly, in calculating probabilities involving permutations, such as in our dice-rolling scenario.