Probability: Rolling A 4 Twice In 7 Rolls

What's up, math enthusiasts! Today, we're diving deep into the fascinating world of probability, and specifically, we're going to break down a common problem that might trip you up: figuring out the chances of a specific event happening a certain number of times within a set number of trials. Think of it like this: you're playing a game, and you want to know the odds of landing on a particular square exactly twice after spinning a spinner seven times. It sounds complicated, but with the right tools, it's totally manageable. We're going to use a super handy formula, the binomial probability formula, to solve this. This formula is your best friend when you're dealing with situations where there are only two possible outcomes for each trial (like rolling a 4 or not rolling a 4), the probability of success stays the same for each trial, and all the trials are independent. So, get ready, guys, because we're about to unlock the secrets to calculating these kinds of probabilities with confidence!

Understanding the Binomial Probability Formula

The core of solving this kind of problem lies in the binomial probability formula. You'll often see it written as $P(k ext{ successes}) = {}_n C_k imes p^k imes (1-p)^{n-k}$. Let's break down what each part of this beast means, shall we? First up, we have '$n$', which represents the total number of trials. In our specific problem, Thuy is rolling a number cube 7 times, so $n=7$. Easy peasy, right? Next, '$k$' is the number of times we want a specific outcome (which we call a 'success') to happen. In this case, Thuy wants to roll a 4 exactly 2 times, so $k=2$. Now, '$p$' is the probability of 'success' on a single trial. For a standard number cube, there are six faces (1, 2, 3, 4, 5, 6). The probability of rolling a 4 on any single roll is 1 out of 6, so p = rac{1}{6}. The term $(1-p)$ then represents the probability of not getting a success on a single trial. If the probability of rolling a 4 is rac{1}{6}, then the probability of not rolling a 4 is 1 - rac{1}{6} = rac{5}{6}. Finally, we have the term ${}_n C_k$. This is the combination formula, which tells us how many different ways we can choose $k$ successes from $n$ trials, without regard to the order in which they occur. It's calculated as {}_n C_k = rac{n!}{(n-k)!k!}, where '!' denotes the factorial (e.g., $5! = 5 imes 4 imes 3 imes 2 imes 1$). So, ${}_7 C_2$ means the number of ways to choose exactly 2 rolls out of 7 that result in a 4. Putting it all together, the formula helps us calculate the probability of getting exactly $k$ successes in $n$ independent trials, each with a probability of success $p$. It accounts for both the probability of the specific successes happening and the number of different ways those successes can be arranged within the trials.

Applying the Formula: Step-by-Step

Alright guys, let's get our hands dirty and plug our specific numbers into the binomial probability formula. Remember, we want to find the probability of Thuy rolling a 4 exactly 2 times in 7 rolls. We've already identified our variables: $n=7$ (total rolls), $k=2$ (number of times rolling a 4), p=rac{1}{6} (probability of rolling a 4 on one roll), and (1-p) = rac{5}{6} (probability of not rolling a 4 on one roll). The first part we need to calculate is ${}_n C_k$, which is ${}_7 C_2$. Using the combination formula, {}_n C_k = rac{n!}{(n-k)!k!}, we get:

{}_7 C_2 = rac{7!}{(7-2)!2!} = rac{7!}{5!2!} = rac{7 imes 6 imes 5 imes 4 imes 3 imes 2 imes 1}{(5 imes 4 imes 3 imes 2 imes 1) imes (2 imes 1)}

We can cancel out the $5!$ from the numerator and denominator:

{}_7 C_2 = rac{7 imes 6}{2 imes 1} = rac{42}{2} = 21

This means there are 21 different ways Thuy could roll a 4 exactly two times in seven rolls. For example, she could roll a 4 on the first two rolls and not on the rest, or roll it on the first and last roll, and so on. There are 21 unique sequences for this to happen.

Now, let's tackle the probability part. We need to calculate $p^k$ and $(1-p)^{n-k}$.

p^k = (rac{1}{6})^2 = rac{1^2}{6^2} = rac{1}{36}

This is the probability of getting two 4s in two specific rolls.

(1-p)^{n-k} = (rac{5}{6})^{7-2} = (rac{5}{6})^5 = rac{5^5}{6^5} = rac{3125}{7776}

This is the probability of not getting a 4 on the remaining five rolls.

Finally, we multiply all these parts together to get our final probability:

P( ext{exactly 2 fours}) = {}_7 C_2 imes (rac{1}{6})^2 imes (rac{5}{6})^5

P( ext{exactly 2 fours}) = 21 imes rac{1}{36} imes rac{3125}{7776}

To simplify, we can multiply the numerators and the denominators:

P( ext{exactly 2 fours}) = rac{21 imes 1 imes 3125}{36 imes 7776} = rac{65625}{279936}

We can also simplify this fraction by dividing both the numerator and denominator by their greatest common divisor, which is 3.

P( ext{exactly 2 fours}) = rac{65625 em 3}{279936 em 3} = rac{21875}{93312}

So, the probability of rolling a 4 exactly 2 times in 7 rolls is rac{21875}{93312}. Pretty neat, huh?

Evaluating the Options

Now that we've crunched the numbers and found the exact probability, let's look at the typical multiple-choice options you might see for a question like this. The question asks which expression represents the probability. This means we don't necessarily have to calculate the final numerical value, but rather identify the correct combination of the formula's components. The options usually look something like this:

• A. { }_7 C_2 imes (rac{1}{6})^2 imes (rac{5}{6})^{7-2}
• B. { }_7 C_2 imes (rac{1}{6})^{7-2} imes (rac{5}{6})^2
• C. { }_7 P_2 imes (rac{1}{6})^2 imes (rac{5}{6})^{7-2}
• D. { }_7 C_2 imes (rac{1}{6})^2 + (rac{5}{6})^{7-2}

Let's analyze each one based on our understanding of the binomial probability formula. Remember, the formula is $P(k ext{ successes}) = {}_n C_k imes p^k imes (1-p)^{n-k}$.

• Option A: { }_7 C_2 imes (rac{1}{6})^2 imes (rac{5}{6})^{7-2}. This option perfectly matches our formula. We have ${}_7 C_2$ for the number of combinations, (rac{1}{6})^2 for the probability of getting two successes (rolling a 4), and (rac{5}{6})^{7-2} which simplifies to (rac{5}{6})^5 for the probability of not getting a success on the remaining five rolls. This is exactly what we calculated!

• Option B: { }_7 C_2 imes (rac{1}{6})^{7-2} imes (rac{5}{6})^2. This option mixes up the exponents. It's using the probability of not rolling a 4 twice and the probability of rolling a 4 five times, which is the opposite of what we want.

• Option C: { }_7 P_2 imes (rac{1}{6})^2 imes (rac{5}{6})^{7-2}. This option uses ${ }_7 P_2$, which is the permutation formula. Permutations consider the order of events, but in probability like this, the order doesn't matter – we just care about the total number of successful outcomes. Combinations (${}_n C_k$) are the correct choice here because they count the number of ways to choose the successes, not arrange them.

• Option D: { }_7 C_2 imes (rac{1}{6})^2 + (rac{5}{6})^{7-2}. This option uses addition instead of multiplication for the probability terms. The binomial probability formula requires multiplying these probabilities together, not adding them.

Therefore, the correct expression that represents the probability of rolling a 4 exactly 2 times in 7 rolls is option A: { }_7 C_2 imes (rac{1}{6})^2 imes (rac{5}{6})^{7-2}. It's crucial to get each component of the binomial formula right, from the combinations to the correct probabilities and exponents, to arrive at the accurate answer. Keep practicing, and you'll master these in no time!

Key Takeaways for Probability Problems

So, there you have it, math whizzes! We've walked through how to tackle a probability problem using the binomial probability formula. The key takeaway here is to always break down the problem into its core components. Identify your $n$ (total trials), $k$ (desired successes), $p$ (probability of success), and $(1-p)$ (probability of failure). Once you have those, applying the formula becomes much more straightforward. Remember that ${}_n C_k$ (combinations) is used because the order in which the successes occur doesn't matter – we're just interested in the total count of successes within the given trials. Using permutations (${}_n P_k$) would be incorrect here because it implies that the order is significant, which it isn't for this type of probability calculation.

It's also vital to ensure you're raising the correct probabilities to the correct powers. The probability of success ($p$) should be raised to the power of the number of successes ($k$), and the probability of failure ($(1-p)$) should be raised to the power of the number of failures ($(n-k)$). Getting these exponents mixed up, as seen in option B, is a common mistake but easily avoidable with careful attention. Lastly, remember that the binomial probability formula involves multiplication of these components, not addition. Option D illustrates this pitfall; adding probabilities in this context doesn't yield the correct combined likelihood of the event occurring.

By carefully substituting the values into the formula $P(k ext{ successes}) = {}_n C_k imes p^k imes (1-p)^{n-k}$, we correctly identified the expression representing the probability of rolling a 4 exactly twice in seven rolls. It's about understanding why each part of the formula is there and what it represents in the real-world scenario. So next time you face a similar problem, remember these steps: identify variables, calculate combinations, apply probabilities with correct exponents, and multiply them all together. You've got this, guys! Keep practicing, stay curious, and happy calculating!