Probability: Rolling A 5 Exactly Three Times

Let's break down how to calculate the probability of rolling a 5 exactly three times in ten rolls of a standard six-sided die. This is a classic probability problem that involves understanding binomial probability. We'll explore the concepts, the formula, and how to apply it to this specific scenario.

Understanding Binomial Probability

Before diving into the solution, let's understand the underlying principles. Binomial probability deals with situations where there are a fixed number of independent trials, each with only two possible outcomes: success or failure. In our case, each roll of the die is a trial. A 'success' is rolling a 5, and a 'failure' is rolling any other number.

• Fixed Number of Trials (n): We have a set number of attempts, which is 10 rolls of the die.
• Independent Trials: Each roll doesn't affect the outcome of any other roll. The die has no memory!
• Two Possible Outcomes: Each roll results in either rolling a 5 (success) or not rolling a 5 (failure).
• Constant Probability of Success (p): The probability of rolling a 5 remains the same for each roll, which is 1/6.

The binomial probability formula helps us calculate the probability of getting exactly k successes in n trials:

$P(X = k) = {n ame{C} k} * p^k * (1-p)^{(n-k)}$

Where:

• $P(X = k)$ is the probability of getting exactly k successes.
• ${n ame{C} k}$ is the number of combinations of n items taken k at a time (also written as $\binom{n}{k}$), which tells us how many different ways we can get k successes in n trials. It is calculated as $n! / (k! * (n-k)!)$, where "!" denotes the factorial (e.g., 5! = 5 * 4 * 3 * 2 * 1).
• $p$ is the probability of success on a single trial.
• $(1-p)$ is the probability of failure on a single trial.
• $k$ is the number of successes we want to find the probability for.
• $n$ is the total number of trials.

Applying the Formula to Our Problem

In our problem, we want to find the probability of rolling a 5 exactly three times in ten rolls. Let's identify the values for our variables:

• $n = 10$ (number of rolls)
• $k = 3$ (number of times we want to roll a 5)
• $p = ame{1/6}$ (probability of rolling a 5 on a single roll)
• $(1-p) = ame{5/6}$ (probability of not rolling a 5 on a single roll)

Now, we plug these values into the binomial probability formula:

$P(X = 3) = {10 ame{C} 3} * ( ame{1/6})^3 * ( ame{5/6})^7$

Let's break down each part of this expression:

• ${10 ame{C} 3}$: This is the number of ways to choose 3 rolls out of 10 to be the ones where we roll a 5. It is calculated as $10! / (3! * 7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120$.
• $( ame{1/6})^3$: This is the probability of rolling a 5 on three specific rolls. Since each roll is independent, we multiply the probability of rolling a 5 on each of those three rolls together: $( ame{1/6}) * ( ame{1/6}) * ( ame{1/6}) = ame{1/216}$.
• $( ame{5/6})^7$: This is the probability of not rolling a 5 on the other seven rolls. Again, since each roll is independent, we multiply the probability of not rolling a 5 on each of those seven rolls together: $( ame{5/6}) * ( ame{5/6}) * ( ame{5/6}) * ( ame{5/6}) * ( ame{5/6}) * ( ame{5/6}) * ( ame{5/6}) = ame{78125/279936}$.

Putting it all together:

$P(X = 3) = 120 * ( ame{1/216}) * ( ame{78125/279936}) ≈ 0.039$

So, the probability of rolling a 5 exactly three times in ten rolls of a six-sided die is approximately 0.039, or 3.9%.

Calculating the Combination

The combination ${10 ame{C} 3}$ represents the number of ways to choose 3 successes (rolling a 5) from 10 trials (rolls). The formula for combinations is:

${n ame{C} k} = n! / (k! * (n-k)!)$

Where '!' denotes the factorial. So, in our case:

${10 ame{C} 3} = 10! / (3! * 7!) = (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * (7 * 6 * 5 * 4 * 3 * 2 * 1))$

We can simplify this by canceling out the 7! from both the numerator and denominator:

${10 ame{C} 3} = (10 * 9 * 8) / (3 * 2 * 1) = 720 / 6 = 120$

This means there are 120 different ways to get exactly three 5s in ten rolls.

The Importance of Independent Events

It's crucial to understand that each roll of the die is an independent event. This means the outcome of one roll has absolutely no influence on the outcome of any other roll. This independence is what allows us to use the binomial probability formula. If the events were dependent (for example, if the die changed after each roll), we would need a different approach to calculate the probability.

Common Mistakes to Avoid

• Forgetting the Combination: A common mistake is to calculate just the probability of getting three 5s in a specific order (e.g., 5, 5, 5, and then seven non-5s). However, we need to account for all the different orders in which we could get three 5s. This is why we use the combination ${n ame{C} k}$.
• Incorrectly Calculating Factorials: Make sure you correctly calculate the factorials when computing the combination. Remember that $n! = n * (n-1) * (n-2) * ... * 2 * 1$.
• Using the Wrong Probability: Ensure you're using the correct probability of success (p) and failure (1-p). In this case, the probability of rolling a 5 is 1/6, and the probability of not rolling a 5 is 5/6.

Conclusion

The binomial probability formula is a powerful tool for analyzing situations with a fixed number of independent trials and two possible outcomes. By understanding the formula and its components, you can accurately calculate the probability of a specific number of successes in a given number of trials. In the case of rolling a six-sided die ten times, the probability of getting exactly three 5s is calculated using the formula ${10 ame{C} 3} * ( ame{1/6})^3 * ( ame{5/6})^7$, which is approximately 3.9%.

For further learning on probability concepts, you can visit Khan Academy's Probability Section.