Section 6.1 Use Counting Principles

Vocabulary The Multiplication Counting Principle: 1 event can occur in m ways another event can occur in n ways both events can occur together is m*n ways The Addition Counting Principle: If possibilities in each aren’t in common, add possibilities in each group to get the total possibilities.

Example 1 U se the Multiplication Counting Principle At a shoe store, shoes are available in 6 different styles. Each style is available in 3 different colors. How many different choices does the shoe store offer? Yo u can use the multiplication counting principle to find the number of shoe choices. Multiply the number of shoe styles 6 by the number of color variations 3. Th e store offers 6 3 = 18 shoe choices.

Example 2 U se the Addition Counting Principle Suppose that each student at a certain school is assigned an identification card which contains a unique 4 character (letter and digit) barcode. Each barcode contains at most 1 digit. How many unique identification cards are possible? 0-digits: There are no digits and there are 26 choices for each letter. So, there are = 456,976 letter-letter-letter- letter possibilities 1-digit: There are = 175,760 digit-letter-letter-letter possibilities. The digit can be in any of four positions, so there are 4 175,760 = 703,040 possibilities. So, in total there are 456, ,040 = 1,160,016 possible identification cards.

Example 3 F ind a probability You are looking at action and strategy computer games. The action game is available in 3 versions: easy, medium, and hard. The strategy game is available in 2 versions: easy and medium. If you randomly choose one action game and one strategy game, what is the probability your choice includes an easy action game? Because there are 3 action game choices and 2 strategy game choices, the total number of possible choices is 3 2 = 6. If you limit yourself to only easy action games, the number of choices that includes an easy action game is 1 2 = 2. P(easy action game) = Easy action game choices Total possible choices = 2/6 = 1/3

Example 4 S olve a multi-step problem There is 1 red bead, 1 blue bead, and 1 yellow bead in a bag. You randomly select a bead from the bag and then put it back. Your two friends each do the same. What is the probability that you each choose the same color bead? Step 1: List the favorable outcomes. There are 3. R-R-RB-B-BY-Y-Y Step 2: Find the total number of outcomes using the multiplication counting principle. Total outcomes = = 27 Step 3: Find the probablity. P(all the same) = Favorable outcomes Total outcomes = 3/27 = 1/9