Question

Evaluate these probabilities: a. \(\frac{C_{1}^{3} C_{1}^{2}}{C_{2}^{5}}\) b. \(\frac{C_{2}^{4} C_{1}^{3}}{C_{3}^{7}}\) c. \(\frac{C_{4}^{5} C_{0}^{3}}{C_{4}^{8}}\)

Short answer

Question: Evaluate the following probability expressions: a. \(\frac{C_{1}^{3} C_{1}^{2}}{C_{2}^{5}}\) b. \(\frac{C_{2}^{4} C_{1}^{3}}{C_{3}^{7}}\) c. \(\frac{C_{4}^{5} C_{0}^{3}}{C_{4}^{8}}\) Answer: a. \(\frac{3}{5}\) b. \(\frac{18}{35}\) c. \(\frac{1}{14}\)

Step-by-step solution

Identify the combination values in the expression

We notice that in this expression, there are three combinations involved: \(C_{1}^{3}\), \(C_{1}^{2}\), and \(C_{2}^{5}\).

Calculate the value of each combination

Using the formula \(C_{n}^{r} = \frac{n!}{r!(n-r)!}\), we calculate the value for each combination: - \(C_{1}^{3} = \frac{3!}{1!(3-1)!} = \frac{3!}{1!2!} = \frac{6}{1\cdot 2} = 3\) - \(C_{1}^{2} = \frac{2!}{1!(2-1)!} = \frac{2!}{1!1!} = \frac{2}{1\cdot 1} = 2\) - \(C_{2}^{5} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{120}{2\cdot 6} = 10\)

Substitute the combination values and simplify the expression

Now that we have calculated the individual combination values, we can substitute them back into the original expression and simplify: \(\frac{C_{1}^{3} C_{1}^{2}}{C_{2}^{5}} = \frac{3 \cdot 2}{10} = \frac{6}{10}\)

Simplify the fraction

We can further simplify the fraction by dividing both the numerator and denominator by their greatest common divisor (GCD): \(\frac{6}{10} = \frac{3}{5}\) #a. The final probability is \(\frac{3}{5}\). #b. Evaluate \(\frac{C_{2}^{4} C_{1}^{3}}{C_{3}^{7}}\)#

Identify the combination values in the expression

We notice that in this expression, there are three combinations involved: \(C_{2}^{4}\), \(C_{1}^{3}\), and \(C_{3}^{7}\).

Calculate the value of each combination

Using the formula \(C_{n}^{r} = \frac{n!}{r!(n-r)!}\), we calculate the value for each combination: - \(C_{2}^{4} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{24}{2\cdot 2} = 6\) - \(C_{1}^{3} = \frac{3!}{1!(3-1)!} = \frac{3!}{1!2!} = \frac{6}{1\cdot 2} = 3\) - \(C_{3}^{7} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = \frac{5040}{6\cdot 24} = 35\)

Substitute the combination values and simplify the expression

Now that we have calculated the individual combination values, we can substitute them back into the original expression and simplify: \(\frac{C_{2}^{4} C_{1}^{3}}{C_{3}^{7}} = \frac{6 \cdot 3}{35} = \frac{18}{35}\) #b. The final probability is \(\frac{18}{35}\). #c. Evaluate \(\frac{C_{4}^{5} C_{0}^{3}}{C_{4}^{8}}\)#

Identify the combination values in the expression

We notice that in this expression, there are three combinations involved: \(C_{4}^{5}\), \(C_{0}^{3}\), and \(C_{4}^{8}\).

Calculate the value of each combination

Using the formula \(C_{n}^{r} = \frac{n!}{r!(n-r)!}\), we calculate the value for each combination: - \(C_{4}^{5} = \frac{5!}{4!(5-4)!} = \frac{5!}{4!1!} = \frac{120}{24\cdot 1} = 5\) - \(C_{0}^{3} = \frac{3!}{0!(3-0)!} = \frac{3!}{0!3!} = \frac{6}{1\cdot 6} = 1\) (Note: \(0! = 1\)) - \(C_{4}^{8} = \frac{8!}{4!(8-4)!} = \frac{8!}{4!4!} = \frac{40320}{24\cdot 24} = 70\)

Substitute the combination values and simplify the expression

Now that we have calculated the individual combination values, we can substitute them back into the original expression and simplify: \(\frac{C_{4}^{5} C_{0}^{3}}{C_{4}^{8}} = \frac{5 \cdot 1}{70} = \frac{5}{70}\)

Simplify the fraction

We can further simplify the fraction by dividing both the numerator and denominator by their greatest common divisor (GCD): \(\frac{5}{70} = \frac{1}{14}\) #c. The final probability is \(\frac{1}{14}\).

Key concepts

Combinations

Combinations are a crucial concept in probability and statistics, as they help us determine how many different groups or sets can be formed from a larger pool of items, where the order of selection does not matter. For combinations, we use the formula:\[ C_{n}^{r} = \frac{n!}{r!(n-r)!} \]This formula calculates the number of ways to choose \( r \) items from a total of \( n \), without regard to the order of the items.
- **\( n \)** refers to the total number of items.- **\( r \)** is the number of items to select from the total.- **\( ! \)** (factorial) means to multiply the number by all of its preceding whole numbers down to 1.This formula is vital in evaluating the probability exercises where you need to know how many different ways a selection can be made. When using combinations, ensure you carefully evaluate each part of the formula, especially the factorials, to determine the correct count.

Factorials

Factorials are a mathematical operation represented by an exclamation mark \( ! \) placed after a number. They are crucial when dealing with combinations and probabilities. The factorial of a number \( n \) is the product of all positive integers up to \( n \). For example, the factorial of 5 (written as \(5!\)) is:\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]Factorials are important because they are used in the combination formula to calculate the number of ways to pick a group of items from a larger set. Here are some points to remember:

• The factorial of zero, \(0!\), is always 1. This is a special rule that is important in combinations.
• Factorials grow very fast, becoming large numbers even with relatively small \( n \) values.

Understanding how and when to apply factorials is critical in solving mathematical problems involving permutations and combinations. Always ensure to simplify factorials where possible to make calculations more manageable.

Simplification of Fractions

Simplifying fractions is an essential skill in mathematics, especially when dealing with complex expressions in probability and statistics. Simplification involves reducing a fraction to its simplest form, where the numerator and denominator have no common factors other than 1.
To simplify a fraction, you can:

• Identify the greatest common divisor (GCD) of the numerator and the denominator.
• Divide both the numerator and the denominator by their GCD.

For instance, consider the fraction \( \frac{6}{10} \). The GCD of 6 and 10 is 2. Dividing both by 2 gives us \( \frac{3}{5} \), the simplified form.In probability exercises, fractions often arise from calculations involving combinations and other formulas. Simplification helps in understanding and interpreting the results more clearly. Always simplify your results for clearer and more concise answers.