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## Probability, Part II

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**Learning Objectives**By the end of this lecture, you should be able to: • Describe the difference between events that are disjoint, and those that are non-disjoint. • Apply the addition rule for disjoint events. • Describe what is meant by ‘complement’ and how to use the complement rule.**New Term: Disjoint**• Important for our next probability rule… • Disjoint essentially means ‘Mutually Exclusive’ • Disjoint is used to discuss the events that you are interested in. • Definition: If two events are disjoint, this means that if one event takes place, the other event can NOT have taken place. Examples: • Event 1: You flip a coin and get a heads. • Event 2: You flip a coin and get a tails. • Event 1: You roll a die and get a 5. • Event 2: You roll a die and get a 6.**Non-Disjoint**Two events are NOT disjoint if they can BOTH be true. Examples: What is the probability of rolling a 4 or an even number? • Event 1: You roll a 4. • Event 2: Your roll an even number. What is the probability of picking a Queen or picking a heart from a deck of cards? • Event 1: You pick a card from the deck and it is a Queen. • Event 2: You pick a card from the deck and it is a Heart.**Examples: Disjoint or Non-Disjoint?**• You flip two coins at the same time. What is the probability of getting 2 heads or two tails? • Event 1: Heads and Heads appears • Event 2: Tails and Tails appears • If the result is event #1, then event #2 is impossible. And vice-versa. In other words, if one of the events is true, the other event can NOT be true. Therefore, these two events are disjoint. • Note how important it is to be sure that you identify things such as the events accurately! • You flip two coins at the same time. What is the probability of getting 1 heads on either coin or 1 tails on either coin? • Event 1: A heads appears • Event 2: A tails appears • When you flip the two coins, it is possible for both of these events to be true. Therefore, these two events are not disjoint.**Examples: Disjoint or Non-Disjoint?**You roll a die. What is the probability that you will roll a number greater than 5, or that you will roll an even number or that the die you rolled is colored red? • Event 1: A number greater than 5 appears • Event 2: An even number is rolled. • Event 3: The die is red • Because ALL of these events can be true, it is a non-disjoint situation. You roll a die twice. What is the probability that on the first roll you get a 5 and on the second roll you will get an even number? • Event 1: You roll a 5 on the first roll. • Event 2: You roll an even number on the second roll. • Even though 5 and even appear to be disjoint events, in this case, we are talking about two separate rolls. Therefore, both events can be true, so these events are non-disjoint.**Venn diagrams:**A and B disjoint Probability rules (cont’d ) 3) Two events A and B are disjoint if they have no outcomes in common and can never happen together. If two events A and B are disjoint, then the probability that A or B occurs is the sum of their individual probabilities. P(A or B) aka P(A U B) = P(A) + P(B) This rule is called the addition rule for disjoint events. A and B not disjoint (Later we will learn a different formula to use for NON-disjoint events) When you find yourself using the word “OR” in a probability question, you are probably going to need to use some form of the addition rule. You will need to decide whether to use addition rule for disjoint events, or the addition rule for non-disjoint events.**“OR” Questions**• Very often in the ‘real world’ we find ourselves trying to come up with probabilities that involve an “or” situation: • I need to decide what to wear to work: What is the probability that it will be snowing or it will be raining? • I need to win a hand at poker: What is the probability that I will draw either a 7 or a Queen? • My friend, my spouse, and I all bought a ticket to win an all-expenses paid trip at a fundraiser: What is the probability that either I will win or my friend will win or my spouse will win? We will now learn a formula that we can use to answer many such “or” probability questions.**The Addition Rule for Disjoint Events**• This probability rule for disjoint events says that the probability of A happening OR the probability of B happening is simply equal to the sum of the two probabilities. • However, this is true only if the events are disjoint. • Put as a fomula: P(A or B) = P(A) + P(B) • Again: It must be noted that this rule only applies for “disjoint” events. If the events are not disjoint, this formula will give you the wrong answer!! • If you are dealing with non-disjoint events, there is a modified version of this formula that we can use. We will learn about it soon.**Example: Addition Rule for Disjoint Events**Example:If you flip two coins, what is the probability of getting only heads or ONLY tails? S = {HH, HT, TH, TT}. We first need to determine the probabilities. Let’s create a probabilty model: Because this is an ‘or’ question, it is a good time to think about using the addition rule. Before we do so, we must confirm that th events we are interested in are disjoint with each other. In this case, the events {HH} and {TT} are indeed disjoint. Therefore, we can use our addition rule for disjoint events. Answer: So, the probability that of getting only heads (HH) or only tails (HH):P(HH or TT) = P(HH) + P(TT) = 0.25 + 0.25 = 0.50**More Examples**• Example: P(draw 6 of Spades or draw an even numbered card) • Yes, we see the word ‘OR’ here, which makes it tempting to use the addition rule for disjoint events. However, note that it IS possible for both of these events to be true. Therefore, the rule can not be used. • Example: Suppose you have a group of 100 athletes. Of those 100 athletes, 30% are basketball players. It is also noted that within the group of 100 athletes, 25% are over 6-feet tall. What is the probability that an athlete is a basketball player OR is 6 feet tall? • As with the previous example, you may be tempted to use our addition rule for disjoint events: P(Basketball Player) + P(6-feet tall) • However, because these two events are not disjoint, doing so would give an incorrect result! • It is possible for any member of the population to be BOTH a basketball player and also 6 feet tall. The addition rule that we have discussed will NOT give a valid answer in this case. We will soon modify our addition formula to deal with non-disjoint events.**Example:**On a single roll of a die, what is the probability of rolling a 1 or a 3 or a 5? • In this case we have 3 events we are interested in: • Event 1: Roll a 1 • Event 2: Roll a 3 • Event 3: Roll a 5 You see the word ‘OR’, so you should start thinking about the addition rule. No we need to decide if the events disjoint. These 3 events are indeed disjoint. That is, if one of these events is true, the other events by definition can not be true. Therefore, we can now proceed with our addition rule: P(1 or 3 or 5) = P(1) + P(3) + P(5) = (1/6) + (1/6) + (1/6) = 3/6 (i.e. 0.5)**Coming up with the probabilities in our probability model:**Example: A couple wants three children. What are the numbers of girls they could end up with? What are the probabilities for each outcome? Create a probability table. To calculate the probabilities for each possible value, we can use the addition rule. Sample space: Let X = possible number of girls: {0, 1, 2, 3}P(X = 0) = P(BBB) = 1/8P(X = 1) = P(BBG or BGB or GBB) = P(BBG) + P(BGB) + P(GBB) = 3/8 • P(X = 2) = P(BGG or GBG or GGB) = P(BBG) + P(BGB) + P(GBB) = 3/8 • P(X = 3) = P(GGG) = 1/8 Important: Note that the sample space in this question is NOT {BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG} Remember that it is vitally important to properly identify your sample space!!!**Examples using a Probability Table**X = # of girls in a household with 3 children Example: What percentage of households has fewer than 2 girls? Answer: P(X<2) = 3/8 + 1/8 = 0.5 Example: What is the probability that a randomly selected household has at least one girl? Answer: P(X>=1) = 3/8 + 3/8 + 1/8 = 7/8**Coin Toss Example:**S ={Head, Tail} Probability of heads = 0.5 Probability of tails = 0.5 Probability rules (cont’d) 4) The complement of any event A refers to an event in which A does not occur. It is written as Ac. Put as a formula: The complement rule states that the probability of an event not occurring is 1 minus the probability that is does occur. P(not A) = P(Ac) = 1 − P(A) Tailc = not Tail = Head P(Tailc) = 1 − P(Head) = 0.5 Venn diagram: Sample space made up of an event A and its complementary Ac, i.e., everything that is not A.**The Complement Rule:**• Sometimes, the complement rule is a powerful tool where it would be essentially impossible to solve the problem without it. At other times, it can just be a useful shortcut. • As a simple example, suppose you want to calculate the probability of rolling a 1 or a 2 or a 3 or a 4 or a 5 on a single roll of a die. This could be determined by calcuating P(1) + P(2) + P(3) + P(4) + P(5) • However, an easier way would be to calculate the complement of rolling a 6: • P(6C)= 1 – P(6) = 1-1/6 = 5/6**M&M candies**If you draw an M&M candy at random from a bag, the candy will have one of six colors. The probability of drawing each color depends on the proportions manufactured, as described here: What is the probability that a single M&M chosen at random is blue? S = {brown, red, yellow, green, orange, blue} P(S) = P(brown) + P(red) + P(yellow) + P(green) + P(orange) + P(blue) = 1 P(blue) = 1 – [P(brown) + P(red) + P(yellow) + P(green) + P(orange)] = 1 – [0.3 + 0.2 + 0.2 + 0.1 + 0.1] = 0.1 What is the probability that a single M&M is either red, yellow, or orange? P(red or yellow or orange) = P(red) + P(yellow) + P(orange) = 0.2 + 0.2 + 0.1 = 0.5**M&M candies**Suppose I told you there was an SEVENTH color, ‘fuscia’. Now what is the probability that an M&M chosen at random is blue? • Answer: You can’t tell. The probabilities must add up to 1. All colors except blue and fuscia add up to 0.9. Blue and fuscia must make up the remaining 0.1. We have no way of knowing the exact amounts. • eg: Blue = 0.06, Fuscia = 0.04? • Blue = 0.05, Fuscia= 0.05? • Blue = 0.03, Fuscia = 0.07? • etc**Don’t take the complement rule for granted!**• You will see down the road that some problems that appear to be extremely difficult can, in fact, be solved fairly easily if you remember this rule. We will go through some examples shortly.