Problem 50 - College Algebra

Question

For the following exercises, use this scenario: a bag of M\&Ms contains 12 blue, 6 brown, 10 orange, 8 yellow, 8 red, and 4 green M\&Ms. Reaching into the bag, a person grabs 5 M\&Ms. What is the probability of getting no brown M\&Ms?

Step-by-Step Solution

Answer

The probability of getting no brown M&Ms is approximately 0.497.

Step 1: Calculate Total M&Ms

First, we need to find the total number of M&Ms in the bag by adding all the different colors together. 12 blue + 6 brown + 10 orange + 8 yellow + 8 red + 4 green = 48 M&Ms in total.

Step 2: Determine Total Ways to Choose 5 M&Ms

The total number of ways to choose 5 M&Ms from the bag of 48 is given by the combination formula: \[ \binom{48}{5} = \frac{48 \times 47 \times 46 \times 45 \times 44}{5 \times 4 \times 3 \times 2 \times 1} = 1,712,304 \]

Step 3: Calculate Non-Brown M&Ms

Next, calculate the total number of M&Ms that are not brown. 12 blue + 10 orange + 8 yellow + 8 red + 4 green = 42 non-brown M&Ms.

Step 4: Determine Ways to Choose 5 Non-Brown M&Ms

Now, find the number of ways to choose 5 M&Ms from the non-brown M&Ms:\[ \binom{42}{5} = \frac{42 \times 41 \times 40 \times 39 \times 38}{5 \times 4 \times 3 \times 2 \times 1} = 850,668 \]

Step 5: Calculate Probability of No Brown M&Ms

To find the probability of drawing no brown M&Ms, divide the number of ways to choose 5 non-brown M&Ms by the total number of ways to choose 5 M&Ms:\[ \text{Probability} = \frac{850,668}{1,712,304} \approx 0.4968 \]

Key Concepts

Combination Formula

The combination formula is a mathematical tool used to find the number of ways to choose a subset of items from a larger set, where the order of selection does not matter. This is quite useful in probabilistic scenarios where we deal with large sets or groups. In our M&Ms example, when we wanted to find out how many different ways we could choose 5 M&Ms from a total of 48, we used the combination formula:\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]Here, \( n \) represents the total number of items, while \( r \) is the number of items to choose. The symbol \( ! \) denotes factorial, which means multiplying a series of descending natural numbers. For example, \( 5! \) is \( 5 \times 4 \times 3 \times 2 \times 1 \). The lack of significance to the order of the items chosen differentiates it from permutations, where the order does matter.

Counting Techniques

Counting techniques are essential mathematical methods used to determine the number of possible outcomes in a given problem. Particularly in probability, it helps us calculate total favourable outcomes versus total possible outcomes. In the context of our bag of M&Ms, the counting technique assisted us by computing all potential ways to select subsets of M&Ms. We determined that there were 1,712,304 total ways to choose any 5 M&Ms from 48. Then, we computed the number of ways to select 5 M&Ms without a single brown one, which amounted to 850,668 different combinations. These calculations were crucial in establishing our probability solutions. Whenever counting large sets, remember to properly define your set limits and conditions to ensure all possible outcomes are correctly considered.

Probability of Compound Events

Probability deals with the likelihood of an event occurring. When we transition to compound events, these involve calculating the probability of two or more separate events happening at the same time. In scenarios like drawing M&Ms from a bag, compound events are common.To determine the probability of not drawing any brown M&Ms, we used the division of successful outcomes for the event by the total possible outcomes. This gives us the probability value, which is usually expressed in decimal form fraction or percentage. In our case, the favorable event was selecting 5 M&Ms that were not brown out of the total possible selections.\[ \text{Probability} = \frac{\text{Number of successful outcomes}}{\text{Total number of possible outcomes}} \]Using this formula, we found that the probability of choosing 5 non-brown M&Ms was approximately 0.4968, or 49.68% chance. Understanding how to properly set up these probability problems is essential for accurate calculations in compound and other complex probability scenarios.