Problem 35 - Algebra 2

Question

In a multi-state lottery, the player must guess which five of forty-nine white balls numbered from 1 to 49 will be drawn. The order in which the balls are drawn does not matter. The player must also guess which one of forty-two red balls numbered from 1 to 42 will be drawn. How many ways can the player fill out a lottery ticket?

Step-by-Step Solution

Answer

80,089,128

Step 1: Understanding the White Ball Combinations

The player needs to guess 5 out of 49 white balls. Since the order does not matter, we need to use combinations. The number of ways to choose 5 balls from 49 can be calculated using the formula for combinations: \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \). Here, \( n = 49 \) and \( r = 5 \).

Step 2: Calculating the White Ball Combinations

Substitute the values into the combination formula to get: \[ \binom{49}{5} = \frac{49!}{5!(49-5)!} = \frac{49 \times 48 \times 47 \times 46 \times 45}{5 \times 4 \times 3 \times 2 \times 1} \] Calculating this gives 1,906,884 ways.

Step 3: Understanding the Red Ball Choices

The player also needs to guess 1 out of 42 red balls. Since there is only one red ball choice, we use the combination formula for one item: \( \binom{n}{1} \), where \( n = 42 \).

Step 4: Calculating the Red Ball Choices

The number of ways to select 1 ball out of 42 using the combination formula is straightforward: \[ \binom{42}{1} = 42 \] So there are 42 ways to choose the red ball.

Step 5: Calculating the Total Number of Ways to Fill Out the Ticket

To find the total number of ways to fill out the lottery ticket, multiply the number of ways to choose the white balls by the number of ways to choose the red ball: \[ 1,906,884 \times 42 = 80,089,128 \] Thus, there are 80,089,128 different ways to fill out the lottery ticket.

Key Concepts

Probability

Probability is a numerical measure that describes the likelihood of an event occurring. It's often expressed as a fraction or percentage. In our lottery ticket scenario, we're interested in the probability of picking the winning combination of white and red balls.

The probability of a specific event happening can be calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this exercise, the total number of ways to fill out the lottery ticket is 80,089,128, representing all possible outcomes.

If we assume there's only one winning combination, then the probability of selecting this winning combination is \( \frac{1}{80,089,128} \).

• The formula for probability is: \( P(A) = \frac{\text{favorable outcomes}}{\text{total outcomes}} \).
• Probability values range from 0 (impossible event) to 1 (certain event).

Remember, probabilities are often used in real-world applications such as games of chance, risk assessment, and even weather forecasting.

Permutations

Permutations are arrangements of objects where the order does matter. However, in the lottery problem we're dealing with combinations, because the order of the numbers does not matter. But let's explore permutations briefly.

When working with permutations, you'd use the formula: \( P(n, r) = \frac{n!}{(n-r)!} \).
This formula gives you all the possible arrangements for a set of numbers where order matters. For example, picking a first, second, and third winner from 10 contestants would be a permutation problem.

• Suppose you have 3 digits (1, 2, 3) and you want to arrange them in different orders, this would be a permutation problem resulting in 6 different arrangements: 123, 132, 213, 231, 312, 321.
• Permutations are critical in situations like scheduling, ranking, and codes where sequence makes a difference.

Though permutations are not directly applicable in this lottery exercise, understanding them helps in comprehending when and how order plays a role in other types of problems.

Factorials

Factorial, denoted by an exclamation mark (!), is the product of an integer and all the integers below it. It's a key concept in both permutations and combinations.

For example, 5! (read as 'five factorial') is \(5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorials grow rapidly with larger numbers, and they're used to calculate permutations and combinations.

In our lottery problem, we used factorial in combination formula: \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\), which computes how many ways you can choose a subset of items from a larger set where order doesn't matter.

• Factorials are foundational in probability and statistics, and they simplify complex counting questions by finding the total number of permutations of a set.
• Important to remember: \(0! = 1\).

Factorials can seem daunting due to their rapid growth, but understanding their role will vastly improve your ability to solve problems involving complex arrangements.