Problem 35
Question
A die is rolled four times. Find the probability of obtaining: Exactly three sixes.
Step-by-Step Solution
Verified Answer
The probability of obtaining exactly three sixes when a die is rolled four times is \(\frac{5}{54}\).
1Step 1: Calculate the number of combinations
In this step, we will calculate C(n, x), which is the number of combinations of rolling the die four times and getting exactly three sixes:
\(C(n, x) = \frac{n!}{x!(n-x)!} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!}\)
We know that,
\(4! = 4 \times 3 \times 2 \times 1 = 24\)
\(3! = 3 \times 2 \times 1 = 6\)
\(1! = 1\)
So, using these factorials:
\(C(n, x) = \frac{24}{6 \times 1} = 4\)
2Step 2: Calculate the probability of obtaining exactly three sixes
Now that we have the number of combinations, we can plug our values into the binomial probability formula:
\(P(x) = C(n, x) * p^x * (1-p)^{(n-x)}\)
Plugging in our values, we get:
\(P(3) = 4 * (\frac{1}{6})^3 * (\frac{5}{6})^{(4-3)}\)
3Step 3: Simplify and solve
Now, we just need to simplify and solve the equation to find the probability:
\(P(3) = 4 * (\frac{1}{6})^3 * (\frac{5}{6})^1\)
\(P(3) = 4 * \frac{1}{216} * \frac{5}{6}\)
\(P(3) = \frac{20}{216}\)
We can simplify the fraction by dividing both the numerator and denominator by the greatest common divisor (4):
\(P(3) = \frac{5}{54}\)
So, the probability of obtaining exactly three sixes when a die is rolled four times is \(\frac{5}{54}\).
Key Concepts
CombinatoricsBinomial DistributionFactorials
Combinatorics
Combinatorics is like a fancy word for counting different ways things can happen. Imagine you have a small basket with blocks of different colors. If you want to know how many unique ways you can pick a certain number of blocks, ''combinatorics'' helps you figure that out.
In the exercise, when a die is rolled four times, we're interested in how many ways we can get exactly three sixes. This is where ''combinations'' come in. Combinations are a way to count choices without worrying about the order. This is calculated using a special formula:
In the exercise, when a die is rolled four times, we're interested in how many ways we can get exactly three sixes. This is where ''combinations'' come in. Combinations are a way to count choices without worrying about the order. This is calculated using a special formula:
- The formula is: \( C(n, x) = \frac{n!}{x!(n-x)!} \)
- Here, ''n'' is the total rolls (4 in this case), and ''x'' is the number of times we want a six (3).
- The exclamation mark "!" is called a ''factorial'' and it means to multiply all the numbers from 1 up to that number.
Binomial Distribution
The ''binomial distribution'' is like a game plan for predicting how often a yes/no event will happen. It deals with scenarios where you have two outcomes: success and failure.
In our rolling dice exercise, getting a six can be considered as a success (yes), and not getting a six is a failure (no). For rolling a die four times, we use the binomial formula to calculate the probability of getting exactly three sixes:
In our rolling dice exercise, getting a six can be considered as a success (yes), and not getting a six is a failure (no). For rolling a die four times, we use the binomial formula to calculate the probability of getting exactly three sixes:
- The main formula is: \( P(x) = C(n, x) \cdot p^x \cdot (1-p)^{n-x} \)
- \( C(n, x) \) is our combinations from the first part.
- ''p'' is the probability of success on one roll (rolling a six, which is \( \frac{1}{6} \)).
- The probability of failure (not rolling a six) is \( \frac{5}{6} \).
Factorials
Factorials are like a shortcut for multiplying a series of descending whole numbers down to 1. This operation is super handy in calculating combinations and permutations, which are crucial in probability theory.
When you see a number with an exclamation mark, like 4! or 3!, it is saying to multiply that number by each whole number below it. Here's how it works:
When you see a number with an exclamation mark, like 4! or 3!, it is saying to multiply that number by each whole number below it. Here's how it works:
- \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
- \( 3! = 3 \times 2 \times 1 = 6 \)
- The concept of factorial is crucial for combinations in determining \( C(n, x) \). We used it to find how many different ways a specific outcome could happen when rolling a dice.
Other exercises in this chapter
Problem 34
Find the number of ways a committee of five can be formed from a group of five boys and four girls, if each committee must contain: At least one boy and at leas
View solution Problem 35
Evaluate each sum. $$ a\left(\begin{array}{l}{n} \\\ {0}\end{array}\right)+(a+d)\left(\begin{array}{l}{n} \\\ {1}\end{array}\right)+(a+2 d)\left(\begin{array}{l
View solution Problem 35
Find the number of ways 10 quarters can be distributed among three people \(-\) Aaron, Beena, and Cathy - so that both Aaron and Beena get at least one quarter,
View solution Problem 35
Let \(a_{n}\) denote the number of multiplications needed to compute \(D_{n}\) using the formula in Exercise 34. Show that \(a_{n}=\mathrm{O}\left(n^{2}\right)\
View solution