Probability of Exactly 2 Sixes in 4 Fair Die Rolls

This problem involves calculating the probability of a specific outcome (getting a six) occurring a fixed number of times (exactly 2) in a fixed number of independent trials (4 rolls of a fair die). This is a classic example of a binomial probability distribution problem.

Understanding Binomial Probability

A binomial probability distribution applies when there are:

• A fixed number of trials (n). Here, n = 4 (4 rolls).
• Each trial has only two possible outcomes: "success" or "failure". Here, "success" is rolling a six, and "failure" is not rolling a six.
• The probability of success (p) is constant for each trial. For a fair die, the probability of rolling a six is \(p = \frac{1}{6}\).
• The trials are independent. The outcome of one roll does not affect the outcome of others.

The probability of failure (q) is \(q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}\).

We want to find the probability of getting exactly \(k\) successes in \(n\) trials. The formula for binomial probability is:

\(P(X=k) = \binom{n}{k} p^k q^{n-k}\)

where \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).

Calculating the Probability

In this problem, we have:

• Number of trials, \(n = 4\)
• Number of successes (exactly 2 sixes), \(k = 2\)
• Probability of success (rolling a six), \(p = \frac{1}{6}\)
• Probability of failure (not rolling a six), \(q = \frac{5}{6}\)

We need to calculate \(P(X=2)\).

First, calculate the binomial coefficient \(\binom{4}{2}\):

\(\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{24}{4} = 6\)

Next, calculate the probabilities raised to their respective powers:

• \(p^k = \left(\frac{1}{6}\right)^2 = \frac{1^2}{6^2} = \frac{1}{36}\)
• \(q^{n-k} = \left(\frac{5}{6}\right)^{4-2} = \left(\frac{5}{6}\right)^2 = \frac{5^2}{6^2} = \frac{25}{36}\)

Now, multiply these values according to the binomial probability formula:

\(P(X=2) = \binom{4}{2} p^2 q^2 = 6 \times \left(\frac{1}{36}\right) \times \left(\frac{25}{36}\right)\)

Multiply the fractions:

\(P(X=2) = 6 \times \frac{1 \times 25}{36 \times 36} = 6 \times \frac{25}{1296}\)

Now, multiply 6 by the fraction:

\(P(X=2) = \frac{6 \times 25}{1296} = \frac{150}{1296}\)

Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor. Both are divisible by 6:

\(\frac{150 \div 6}{1296 \div 6} = \frac{25}{216}\)

So, the probability of getting exactly 2 sixes when a fair die is rolled 4 times is \(\frac{25}{216}\).

Summary of Calculation Steps

1. Identify \(n=4\), \(k=2\), \(p=1/6\), \(q=5/6\).
2. Calculate the binomial coefficient \(\binom{4}{2} = 6\).
3. Calculate \(p^k = (1/6)^2 = 1/36\).
4. Calculate \(q^{n-k} = (5/6)^2 = 25/36\).
5. Multiply the results: \(6 \times \frac{1}{36} \times \frac{25}{36}\).
6. Simplify the product: \(\frac{150}{1296} = \frac{25}{216}\).

The final probability is \(\frac{25}{216}\).