Random phenomenonIt is one that, when repeated several times under the same conditions, produces unpredictable results.E.g.: when we roll a die, we can get any of its faces. It is not possible to predict which one it will be.Sample spaceIt is the universal set of all possible outcomes of the random phenomenon. It is denoted by \(U\).E.g.: when rolling a die, \(U = \{1,2,3,4,5,6\}\).EventIt is any subset of the sample space.E.g.: getting an even number when rolling a die: \(E = \{2,4,6\}\).Complement of event EThe complement of event \(E\) is the event \(\overline{E}\) such that:\[ \overline{E} = U - E \]So \(\overline{E}\) is the set of elements of \(U\) that do not belong to \(E\). E.g.: if \(E\) = "an even number" = \(\{2,4,6\}\), then \(\overline{E}\) = "an odd number" = \(\{1,3,5\}\).Impossible event\[ E = \varnothing \]It is the event that has no elements. E.g.: rolling a 7 with a die.Certain event\[ E = U \]It is the sample space itself. E.g.: getting heads or tails when tossing a coin.Mutually exclusive eventsTwo events \(A\) and \(B\) are mutually exclusive when they cannot happen at the same time, that is, when they have no elements in common:\[ A \cap B = \varnothing \]E.g.: when tossing a coin, the events "heads" and "tails" are mutually exclusive — you cannot get heads and tails in a single toss. An event and its complement are always mutually exclusive.Probability of event EIt is the ratio between the number of elements in the set \(E\) and the number of elements in the sample space \(U\):\[ P(E) = \frac{n(E)}{n(U)} = \frac{\text{number of elements in set E}}{\text{number of elements in set U}} \]This definition is only valid if all elements of \(U\) are equally likely to happen, that is, if the sample space is equiprobable.Probability is a real number between 0 and 1 (0 for the impossible event, 1 for the certain event), often expressed as a percentage to make it easier to read.ExamplesExample 1: what is the probability of rolling a 3 with a die?Solution:\[ U = \{1,2,3,4,5,6\} \]\[ E = \{3\} \]Answer:\[ P(3) = \frac{n(E)}{n(U)} = \frac{1}{6} \]Example 2: what is the probability of rolling an even number with a die?E = {2,4,6} ---------------→ n(E) = 3U = {1,2,3,4,5,6} -------→ n(U) = 6\[ P(\text{even}) = \frac{n(E)}{n(U)} = \frac{3}{6} = \frac{1}{2} \]Example 3: what is the probability of getting heads when tossing a coin?E = {heads} --------------------→ n(E) = 1U = {heads, tails} ----------→ n(U) = 2\[ P(\text{heads}) = \frac{n(E)}{n(U)} = \frac{1}{2} \]Example 4: what is the probability of getting a sum of 5 when rolling two dice?E = { (1,4) , (2,3) , (3,2) , (4,1) } --------→ n(E) = 4U = all pairs of outcomes: 6 × 6 = 36 ----→ n(U) = 36\[ P(\text{sum 5}) = \frac{n(E)}{n(U)} = \frac{4}{36} = \frac{1}{9} \]Note that (1,4) and (4,1) are different outcomes: in the first, the 1 came up on the first die; in the second, on the second die.When rolling two dice, what is the probability that the sum is 8?Answer: 5/36 — the pairs that add up to 8 are (2,6), (3,5), (4,4), (5,3) and (6,2), that is, 5 of the 36 outcomes. The most common mistake is to count (3,5) and (5,3) as one and get 3/36 = 1/12.Probability of the complementary eventSince \(E\) and \(\overline{E}\) have no elements in common and together make up the whole sample space, their probabilities add up to 1:\[ P(\overline{E}) = 1 - P(E) \]This rule is very useful when it is easier to calculate the probability of something not happening.Example 5: what is the probability of not rolling a 6 with a die?\[ P(\text{not 6}) = 1 - P(6) = 1 - \frac{1}{6} = \frac{5}{6} \]Example 6: when rolling two dice, what is the probability that the sum is not 5?From Example 4, \(P(\text{sum 5}) = 1/9\). Therefore:\[ P(\text{sum other than 5}) = 1 - \frac{1}{9} = \frac{8}{9} \]When tossing a coin 3 times, what is the probability of getting at least one head?Answer: 7/8 — the complement of at least one head is no heads, that is, tails all 3 times, with probability 1/2 × 1/2 × 1/2 = 1/8. So the answer is 1 − 1/8 = 7/8. The 1/8 is the probability of the complement, not the one asked for.Addition of probabilitiesLet \(A\) and \(B\) be two events of the sample space \(U\). The probability that event \(A\) or event \(B\) occurs is given by:\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]We subtract \(P(A \cap B)\) because the elements that are in both \(A\) and \(B\) would otherwise be counted twice.If events \(A\) and \(B\) are mutually exclusive (no intersection), \(P(A \cap B) = 0\), and the formula becomes:\[ P(A \cup B) = P(A) + P(B) \]ExamplesExample 7: what is the probability of rolling a 6 or an even number with a die?\[ U = \{1,2,3,4,5,6\} \]Event A: \(\{6\}\), with probability \(\frac{1}{6}\).Event B: \(\{2,4,6\}\), with probability \(\frac{3}{6}\).\(A \cap B = \{6\}\), with probability \(\frac{1}{6}\).\[ P(A \cup B) = \frac{1}{6} + \frac{3}{6} - \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \]Example 8: in a group of 36 soccer fans from Rio de Janeiro, 18 support Flamengo, 10 support Vasco, 6 support Fluminense and 2 support Botafogo. If one of these fans is chosen at random, what is the probability that he supports Flamengo or Botafogo?Each fan supports only one team, so the events are mutually exclusive.\[ P(F) = \frac{18}{36} \qquad P(B) = \frac{2}{36} \]\[ P(F \cup B) = \frac{18}{36} + \frac{2}{36} = \frac{20}{36} = \frac{5}{9} \]Drawing a card from a standard 52-card deck, what is the probability that it is a heart or a face card (jack, queen or king)?Answer: 11/26 — there are 13 hearts and 12 face cards, but 3 face cards are hearts and would be counted twice: (13 + 12 − 3)/52 = 22/52 = 11/26. The option 25/52 forgets to subtract the intersection, and 1/4 counts only the hearts.Multiplication of probabilities and independent eventsTwo events \(A\) and \(B\) are independent when the occurrence of one does not change the probability of the other. In that case, the probability that both occur is the product of the probabilities:\[ P(A \cap B) = P(A) \cdot P(B) \]This equality also works as a test: if it holds, the events are independent; if it does not, they are dependent. Do not confuse independent with mutually exclusive: two mutually exclusive events (with probabilities greater than zero) are never independent, because knowing that one happened guarantees that the other did not.ExamplesExample 9: in two rolls of a die, what is the probability of getting an even number on the first roll and an odd number on the second?\(A\): the result of the 1st roll is even; \(B\): the result of the 2nd roll is odd. \(A\) and \(B\) are independent, because knowing that \(A\) occurred does not change the probability of \(B\). We have:\[ P(A \cap B) = P(A) \cdot P(B) = \frac{3}{6} \cdot \frac{3}{6} = \frac{9}{36} = \frac{1}{4} \]Example 10: rolling a die repeatedly until a 6 comes up, what is the probability that exactly three attempts are needed?For three attempts to be needed, the 1st attempt must not give a 6, nor the 2nd, and the 3rd must. Since the result of each roll is independent of the others, the probability is:\[ P(\text{first 6 on the 3rd attempt}) = \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} = \frac{25}{216} \]Example 11: what is the probability of rolling a die twice and getting a 4 both times?\[ P(\text{4 twice in a row}) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36} \]Example 12: in a room there are 20 young women: 3 of them are studious, 11 are from Rio de Janeiro and 7 do not have a boyfriend. Assuming these three characteristics are independent of each other, what is the probability that João picks at random a young woman who is studious, from Rio and without a boyfriend?\[ P(\text{studious}) = \frac{3}{20} \qquad P(\text{from Rio}) = \frac{11}{20} \qquad P(\text{no boyfriend}) = \frac{7}{20} \]\[ P(\text{studious, from Rio and no boyfriend}) = \frac{3}{20} \cdot \frac{11}{20} \cdot \frac{7}{20} = \frac{231}{8000} \approx 2.9\% \]The independence assumption is essential: without it, we could only answer if we knew how many young women have all three characteristics at the same time.On a regular die, let A = rolling an even number and B = rolling a number greater than 4. Are A and B independent?Answer: Yes, because P(A and B) = 1/6 = 1/2 × 1/3 — P(A) = 3/6 = 1/2, P(B) = 2/6 = 1/3 and A ∩ B = 6, with probability 1/6, which is exactly the product. Having an element in common does not prevent independence; what decides it is the equality P(A ∩ B) = P(A) · P(B).Proposed exercisesWhen rolling a die, what is the probability of getting:an even number of points?a number of points less than or equal to 4?What is the probability of tossing a coin and getting tails?What is the probability of rolling a white die and a blue die and getting:a sum of points equal to 7?2 points on the blue die?What is the probability of tossing two coins and getting the same result on both?What is the probability of drawing a ball from an urn containing 5 red balls and 1 white ball and getting a red ball?A coin is biased so that heads is twice as likely as tails. Calculate the probability of:getting heads when tossing this coin;getting tails when tossing this coin.A die is loaded so that all even numbers have the same probability, and all odd numbers also have the same probability. However, an even number is three times more likely to occur than an odd number. Rolling this die, what is the probability of:getting a prime number?getting a multiple of 3?getting a number less than or equal to 3?What is the probability of drawing a diamond or a king from a 52-card deck?A group of 100 university students consists of 52 engineering students, 27 medical students, 10 philosophy students and the rest law students. If a member of the group is chosen at random, what is the probability that he or she is an engineering or a medical student?An urn contains 40 cards numbered from 1 to 40. If we draw a card at random from this urn, what is the probability that the number written on it is a multiple of 4 or a multiple of 3?A die is loaded so that the probability of a number showing on the top face is proportional to that number. Calculate the probability of getting an even number or the number 6.In exercise 11, what is the probability of getting an even number or the number 5?Answer keya) 1/2 b) 2/31/2a) 1/6 b) 1/61/25/6a) 2/3 b) 1/3a) 5/12 b) 1/3 c) 5/124/1379/1001/24/717/21
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