Question

Find the probabilities. \(\frac{C_{2}^{4} C_{1}^{3}}{C_{3}^{7}}\)

Short answer

Answer: The probability is \(\frac{18}{35}\).

Step-by-step solution

Understand the formula

We are given the formula \(\frac{C_{2}^{4} C_{1}^{3}}{C_{3}^{7}}\). In this formula, \(C_{2}^{4}\) denotes the number of combinations of selecting 2 items from a set of 4 items. Similarly, \(C_{1}^{3}\) denotes the number of combinations of selecting 1 item from a set of 3 items, and \(C_{3}^{7}\) denotes the number of combinations of selecting 3 items from a set of 7 items. We need to calculate these values and simplify the fraction to find the probability.

Calculate \(C_{2}^{4}\)

To calculate \(C_{2}^{4}\), we can use the formula for combinations: \(C_n^r = \frac{n!}{r!(n-r)!}\) , where '!' denotes the factorial. In this case, we have \(n = 4\) and \(r = 2\). So, we need to find the values of \(4!\), \(2!\), and \((4-2)!\). \( 4! = 4 \times 3 \times 2 \times 1 = 24\) \( 2! = 2 \times 1 = 2\) \((4-2)! = 2! = 2 \times 1 = 2\) Now, plugging these values into the formula: \(C_{2}^{4} = \frac{4!}{2!(4-2)!} = \frac{24}{(2)(2)} = 6\)

Calculate \(C_{1}^{3}\)

Similarly, we can calculate \(C_{1}^{3}\) using the formula. In this case, we have \(n = 3\) and \(r = 1\). So, we need to find the values of \(3!\), \(1!\), and \((3-1)!\). \( 3! = 3 \times 2 \times 1 = 6\) \( 1! = 1\) \((3-1)! = 2! = 2 \times 1 = 2\) Now, plugging these values into the formula: \(C_{1}^{3} = \frac{3!}{1!(3-1)!} = \frac{6}{(1)(2)} = 3\)

Calculate \(C_{3}^{7}\)

Finally, we can calculate \(C_{3}^{7}\) using the formula. In this case, we have \(n = 7\) and \(r = 3\). So, we need to find the values of \(7!\), \(3!\), and \((7-3)!\). \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040\) \( 3! = 3 \times 2 \times 1 = 6\) \((7-3)! = 4! = 4 \times 3 \times 2 \times 1 = 24\) Now, plugging these values into the formula: \(C_{3}^{7} = \frac{7!}{3!(7-3)!} = \frac{5040}{(6)(24)} = 35\)

Calculate the probability

Now that we have the values of \(C_{2}^{4}\), \(C_{1}^{3}\), and \(C_{3}^{7}\), we can calculate the probability: \(P = \frac{C_{2}^{4} C_{1}^{3}}{C_{3}^{7}} = \frac{(6)(3)}{35} = \frac{18}{35}\) So, the probability is \(\frac{18}{35}\).

Key concepts

Factorial Notation

Understanding factorial notation is crucial for solving many problems in probability and combinatorics. A factorial, denoted by an exclamation point (!), is the product of all positive integers up to a given number. Specifically, for any non-negative integer 'n', the factorial \( n! \) is the product \( n \times (n-1) \times (n-2) \times ... \times 3 \times 2 \times 1 \). For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
There is a unique case for zero, which is defined as \( 0! = 1 \) by convention. Factorial notation simplifies the representation and calculation of permutations and combinations. The factorial terms in the aforementioned exercise—\( 4! \), \( 2! \), and so on—are foundational for evaluating combination formulas efficiently.

Combinatorial Probability

The term 'combinatorial probability' refers to the likelihood of an event occurring, based on the combinatorial selections possible within a set. It blends the concepts of probability with combinatorics, a branch of mathematics concerning the counting, arrangement, and combination of objects.
To calculate combinatorial probability, we need to determine the number of favorable combinations (those that comprise the event of interest) and the total number of possible combinations. The probability is then the ratio of these two numbers. In our exercise, we calculate the probability of selecting certain items from different sets, using the combinations formula. This approach is frequently used in games, genetics, decision making, and more.

Combinations Formula

When we want to know how many ways we can choose 'r' items from a larger set of 'n' items without regard to the order of selection, we use the combinations formula: \(C_n^r = \frac{n!}{r!(n-r)!}\).

Breaking Down the Formula

Let's dissect the components of this formula: \({n!}\) accounts for all the permutations of 'n' items. By dividing by \({r!}\), we disregard the order of 'r' items, and by dividing by \({(n-r)!}\), we exclude the permutations of the remaining \(n-r\) items not chosen. This results in the number of unique combinations.
In the exercise, the combinations formula allows for a straightforward calculation of how many ways we can select different numbers of items from distinct groups. For comprehension and learning, it's invaluable to understand the logic behind the formula and how it's derived from the fundamental principles of counting.