Counting techniques and more complex experiments (pp ) Counting techniques determining the number of outcomes for an experiment

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1 Counting techniques and more complex experiments (pp ) In our introduction to probability, we looked at examples of simple experiments. These examples had small sample spaces and were easy to evaluate. But what happens when the experiments get complex, have large sample spaces, or multiple stages. In this section we will look at techniques and methods that allow us to investigate the probabilities of more complex situations. Counting techniques determining the number of outcomes for an experiment To determine the probability of a situation, you don t actually need to know the details of every possible outcome, just how many there are. That is where counting techniques can help us out. This first technique uses a visual model to help determine the number of outcomes for a multi-stage situation. Tree diagrams Example 1: Suppose you can get a one-scoop ice cream cone with 2 choices of cone (regular or sugar cone), 4 choices of ice cream flavor (vanilla, chocolate, strawberry or butter pecan), and 2 choices of sprinkles (nuts or candy). How many different ice cream treats is it possible to make given these choices? For this problem, we will draw a tree diagram to visually solve the problem. Let s think of the choices you would need to make at each stage of putting the ice cream cone together. Each choice (stage) is represented in our diagram with branches for each choice. So for this problem the picture might look like this: regular sugar vanilla chocolate strawberry Butter pecan vanilla chocolate strawberry Butter pecan nuts candy nuts candy nuts candy nuts candy nuts candy nuts candy nuts candy nuts candy If you count all of the boxes at the bottom of the last row of branches you find there are 16 different outcomes. Is there a faster way to get this answer? Yes. Just multiply how many choices we have at each stage: 2 x 4 x 2 = 16. This is called the Fundamental Counting Principle and it is a useful tool for determining how many different outcomes a multi-stage situation may have.

2 Fundamental Counting Principle If event A can occur in R ways, and for each of these R ways, an event B can occur in S ways, then events A and B, in succession, can happen in R x S ways. This works for any number of events, as long as they happen in succession and as long as they are independent meaning one event has no bearing on the next event. Let s look at another example of the Fundamental Counting Principle. Example 2. At the Costume Design Shop you can create your unique clown costume for $20. You are given several choices to make sure your costume is unique. The shop offers a costume consisting of various choices for wig, makeup colors, nose, shirt, pants, shoes, and gloves. They offer 4 styles of wig, 6 makeup styles, 3 types of nose, 8 types of shirt, 7 types of pants, 4 styles of shoes, and 3 types of gloves. How many differ clown costumes can be created? Answer: 4 x 6 x 3 x 8 x 7 x 4 x 3 = 48,384. So the probability that someone would have exactly the same costume as you selected would be 1/48384! This doesn t tell us exactly what each costume would look like, but we don t need to know that to determine the size of the sample space. Probability tree diagrams. The tree diagram we looked at previously only gave us the various outcomes for a specific situation, but we can also use tree diagrams to answer questions about the probabilities of certain events in the given situation. When we insert the probabilities into the tree diagram, we call it a probability tree. I do my probability trees a little differently than the book (see page 622) because I think they make it too complicated! Let s set up a tree diagram for a simple two stage experiment where I draw two marbles from a jar with replacement. With replacement means that after I draw my first marble and note the color, I put it back in the jar. This means that what I do on the first draw has no impact on my second draw, so the two draws are independent events. Without replacement means I don t put the marble back after the first draw, so the probabilities for the draw of the second marble are altered. The two draws are dependent events. Example 3 (with replacement): A jar contains 4 red (R) and 2 black (B) marbles. Suppose 2 marbles are draw with replacement. What is the probability that both marbles are black or P(BB)?

3 Draw1 Draw 2 So we end up with 4 two-stage outcomes: RR, RB, BR, and BB. The probabilities for these outcomes can be found by multiplying down each branch in the probability tree. So we get this: P(RR) = x = 4/9 P(RB) = x = 2/9 P(BR) = x = 2/9 P(BB) = x = 1/9. So the probability of getting two black marbles in a row is 1/9. Note: if I add P(RR) + P(RB) + P(BR) + P(BB) = 4/9 + 2/9 + 2/9 + 1/9 = 1. If I wanted to know what the probability would be that I d bet two marbles of the same color, I d add P(RR) + P(BB) = 4/9 + 1/9 = 5/9. If I wanted to know the probability that I d draw two marbles of different colors, then I d add P(RB) + P(BR) = 2/9 + 2/9 = 4/9. Rules for probability trees: Multiply down branches to get the probability of sequenced events, add across branches to get probabilities for a combination of more than one outcome. What if we did the experiment without replacement? Would the probabilities be different? Let s find out. Example 3 (without replacement): A jar contains 4 Red and 2 Black marbles. Suppose you draw one marble out of the jar, note its color, set the marble aside (don t put it back in the jar), then draw a second marble. What is the probability that both marbles are Black (P(BB))?

4 Draw 1 Draw 2 P(R)=3/5 P(B) = 2/5 P(R)=4/5 P(B) =1/5 Notice that the probabilities in the second row change. That is because what happens on the first draw impacts what is left in the jar and so the probabilities for the second draw are different. For example if we draw a Red marble on the first draw, then we only have 3 Red marbles and two Black marbles left in the jar, so the probability of getting a Red on the second draw is 3/5. Now our 4 two-stage outcomes are: P(RR) = x 3/5 = 6/15 = 2/5 P(RB) = x 2/5 = 4/15 P(BR) = x 4/5 = 4/15 P(BB) = x 1/5= 1/15. So the probability of getting two black marbles in a row is 1/15. So if you do the experiment without replacement, the probability of getting two black marbles is less than if you do the experiment with replacement. Is this game fair? As you are working with middle school students, they are very focused on situations being fair. This allows us to explore probability with them from the perspective of determining if certain game situations are fair or unfair. Let s consider this next example. Example 4. Game ONE. There are two players in the game: Player A and Player B. Player A goes first. In a bag are three chips: two are green and one is yellow. Each player draws one chip (without replacement so Player A goes first, keeps her chip, then Player B draws.) Now the players compare chips. If the chips are the same color, then Player A wins. If the chips are different colors, then Player B wins. Is this a fair game? Find a partner (or play against yourself) and try playing the game a few times get some experimental data and see what you think.

5 Now let s draw a probability tree diagram for the situation. P(G)= P(Y) = P(G)=1/2 P(Y) = 1/2 P(G)=2/2 P(Y) =0/2 So our probabilities for the various two-stage outcomes are: P(GG) = x 1/2 = 2/6 = P(GY) = x 1/2 = 2/6 = P(YG) = x 2/2 = 2/6 = P(YY) = x 0/2 = 0/6 = 0 Player A wins if the two chips are the same color, so P(Player A wins) = P(GG) + P(YY) = + 0 =. Player B wins if the two chips are different colors, so P(Player B wins) = P(GY) + P(YG) = + =. So the game is not fair! Player B has a chance of winning the game. Game TWO. Play the game the same way as Game ONE, but now the bag contains three green chips and one yellow chip. Draw the probability tree for this, analyze the probabilities and determine whether adding an additional green chip to the bag will make the game fair.

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