Problem 37
Question
A die is rolled four times. Find the probability of obtaining: Not more than two sixes.
Step-by-Step Solution
Verified Answer
The probability of obtaining not more than two sixes when rolling a die four times is: \(P(X \leq 2) = (\frac{5}{6})^{4} + 4 * \frac{1}{6} * (\frac{5}{6})^3 + 6 * \frac{1}{36} * (\frac{5}{6})^2\).
1Step 1: Calculate probability of obtaining zero sixes
To calculate the probability of obtaining zero sixes, we need to find P(X=0). Following the binomial probability formula:
P(X=0) = C(4, 0) * \((\frac{1}{6})^0\) * \((\frac{5}{6})^{4-0}\)
First, calculate the number of combinations C(4,0):
C(4,0) = 1 (since there's only one way to get zero sixes)
Next, the probability term is calculated as
\((\frac{1}{6})^0\)= 1 and \((\frac{5}{6})^{4-0}\) = \((\frac{5}{6})^{4}\)
Now, put all the numbers back into the equation:
P(X=0) = 1 * 1 * \((\frac{5}{6})^{4}\) = \((\frac{5}{6})^{4}\)
2Step 2: Calculate probability of obtaining one six
To calculate the probability of obtaining one six, we need to find P(X=1). Applying the binomial probability formula:
P(X=1) = C(4, 1) * \((\frac{1}{6})^1\) * \((\frac{5}{6})^{4-1}\)
First, calculate the number of combinations C(4,1):
C(4,1) = 4 (since there are 4 possible trials where a single six can be rolled)
Next, calculate the probability terms:
\((\frac{1}{6})^1\)= \(\frac{1}{6}\) and \((\frac{5}{6})^{4-1}\) = \((\frac{5}{6})^{3}\)
Finally, put all the numbers back into the equation:
P(X=1) = 4 * \(\frac{1}{6}\) * \((\frac{5}{6})^{3}\)
3Step 3: Calculate probability of obtaining two sixes
To calculate the probability of obtaining two sixes, we need to find P(X=2). Using the binomial probability formula:
P(X=2) = C(4, 2) * \((\frac{1}{6})^2\) * \((\frac{5}{6})^{4-2}\)
First, calculate the number of combinations C(4,2):
C(4,2) = 6 (since there are 6 possible pairs where two sixes can be rolled)
Next, calculate the probability terms:
\((\frac{1}{6})^2\)= \(\frac{1}{36}\) and \((\frac{5}{6})^{4-2}\) = \((\frac{5}{6})^{2}\)
Finally, put all the numbers back into the equation:
P(X=2) = 6 * \(\frac{1}{36}\) * \((\frac{5}{6})^{2}\)
4Step 4: Sum the probabilities of obtaining 0, 1, and 2 sixes
Final step is to sum the calculated probabilities to find the probability of obtaining not more than two sixes:
P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2)
Calculate the sum of the probabilities:
P(X ≤ 2) = \((\frac{5}{6})^{4}\) + 4 * \(\frac{1}{6}\) * \((\frac{5}{6})^3\) + 6 * \(\frac{1}{36}\) * \((\frac{5}{6})^{2}\)
Now you have the probability of obtaining not more than two sixes when rolling a die four times.
Key Concepts
Probability of Obtaining SixesCombinatoricsBinomial Distribution
Probability of Obtaining Sixes
Understanding the probability of obtaining sixes when rolling a die is a fundamental concept in basic probability. Let's explore a scenario where a die is rolled multiple times - in our case, four times. The interest is in finding the likelihood of rolling a six no more than two times within those four rolls. The six faces on a die are all equally probable, so rolling any specific number, like a six, has a probability of \( \frac{1}{6} \).
We approach this problem using the binomial probability formula, which requires both the number of trials (four rolls) and the success probability for each trial (rolling a six). The intricate part comes when we consider different outcomes that fit our 'not more than two sixes' condition. We must consider all the possible ways we could roll zero sixes, one six, or two sixes, calculate the probability for each scenario, and then sum these probabilities up. This methodical process ensures a comprehensive solution and a deeper understanding of the probabilities involved.
We approach this problem using the binomial probability formula, which requires both the number of trials (four rolls) and the success probability for each trial (rolling a six). The intricate part comes when we consider different outcomes that fit our 'not more than two sixes' condition. We must consider all the possible ways we could roll zero sixes, one six, or two sixes, calculate the probability for each scenario, and then sum these probabilities up. This methodical process ensures a comprehensive solution and a deeper understanding of the probabilities involved.
Combinatorics
The heart of solving such probability problems lies in combinatorics, the branch of mathematics dealing with combinations and permutations of objects. In our problem, we define 'success' as rolling a six and we want to know the number of ways we can achieve zero, one, or two successes in four trials.
The formula for calculating combinations, denoted as \( C(n, k) \), tells us how many ways we can choose \( k \)-items from a larger set of \( n \)-items without regard to the order. For instance, the number of ways where no sixes appear, \( C(4, 0) \), is only one, since this is equivalent to none of the four dice showing a six. Similarly, \( C(4, 1) \), corresponds to the scenario where we achieve one success—a single six. Small numbers here are easy to compute, but understanding the combinatorics concept is key when dealing with larger, more complex scenarios.
The formula for calculating combinations, denoted as \( C(n, k) \), tells us how many ways we can choose \( k \)-items from a larger set of \( n \)-items without regard to the order. For instance, the number of ways where no sixes appear, \( C(4, 0) \), is only one, since this is equivalent to none of the four dice showing a six. Similarly, \( C(4, 1) \), corresponds to the scenario where we achieve one success—a single six. Small numbers here are easy to compute, but understanding the combinatorics concept is key when dealing with larger, more complex scenarios.
Binomial Distribution
Lastly, binomial distribution is a probability distribution that summarizes the likelihood that a value will take on one of two independent values under a given number of observations or trials. It's defined by two parameters: the number of trials \( n \), and the probability of success \( p \).
In practical terms for our problem, each roll of the die is an independent trial; the result of one roll doesn't affect the others. Since the outcome we consider a 'success' (rolling a six) has a probability of \( \frac{1}{6} \) and each of the four rolls is independent, we can use the binomial distribution formula to calculate the probability for zero, one, or two sixes.
When we use the binomial distribution, we deal with a fixed number of trials (four dice rolls), two possible outcomes (a six or not a six), and a constant probability of the 'success' occurring on each trial. The power of binomial distribution allows for precise calculations and predictions for a wide range of similar probabilistic scenarios.
In practical terms for our problem, each roll of the die is an independent trial; the result of one roll doesn't affect the others. Since the outcome we consider a 'success' (rolling a six) has a probability of \( \frac{1}{6} \) and each of the four rolls is independent, we can use the binomial distribution formula to calculate the probability for zero, one, or two sixes.
When we use the binomial distribution, we deal with a fixed number of trials (four dice rolls), two possible outcomes (a six or not a six), and a constant probability of the 'success' occurring on each trial. The power of binomial distribution allows for precise calculations and predictions for a wide range of similar probabilistic scenarios.
Other exercises in this chapter
Problem 36
A die is rolled four times. Find the probability of obtaining: At least one six.
View solution Problem 36
Evaluate each sum. Show that \(C(n, r-1)
View solution Problem 37
Using the recursive definition of \(P(n, r),\) evaluate each. $$P(5,4)$$
View solution Problem 37
Verify that \(\left(\begin{array}{l}n \\\ r\end{array}\right)=\frac{n}{r}\left(\begin{array}{l}n-1 \\\ r-1\end{array}\right),\) where \(n \geq r \geq 1\)
View solution