Fundamental Counting Principle - Math Reviewer Grade 10 - 3rd Quarter PDF

MATHEMATICS REVIEWER GRADE *Using the FCP, we can simply multiply 3 10 – THIRD QUARTER and 2. We will get 6. But let us use other By: Rv Sahagun & Danielle Roberto methods. THE FUNDAMENTAL COUNTING...

MATHEMATICS REVIEWER GRADE *Using the FCP, we can simply multiply 3 10 – THIRD QUARTER and 2. We will get 6. But let us use other By: Rv Sahagun & Danielle Roberto methods. THE FUNDAMENTAL COUNTING 1. Tree Diagram - A tree diagram is a PRINCIPLE / FUNDAMENTAL visual way to represent the total outcomes PRINCIPLE OF COUNTING you could have. Fundamental Counting Principle - Let Event 1 and Event 2 be two events. If the first event can occur in m different ways and after first event has occurred, second event can occur in n different ways, then together the events can occur in (m)(n) different ways. (Assuming that event 2 is not influenced by event 1) *This is simply multiplying the events stated. Example: How many different pairs of letters from the English alphabet are possible? Solution: The alphabet contains 26 letters. So, the number of two letter words that can be formed is (26) (26). - There are 676 different pairs of letters. Another example: Tarryn has five different 2. Sample Space - Sample space can be skirts, four different tops and three pairs of listed. The sample space for the outfits is shoes. Assuming that all the colours {Black Shirt Black Pants, Black Shirt Jeans, complement each other, how many different Brown Shirt Black Pants, Brown Shirt Jeans, outfits can she put together? White Shirt Black Pants, White Shirt Jeans}. As you can see, there are 6 combinations Solution: Using the FCP, multiply 5, 4, and 3 listed because there are 6 total possible together: (5)(4)(3) outcomes. - She can put 60 different outfits 3. Table of outcomes - A table of outcomes is a table where the first row and There are other ways to get the number of first column represent the possible outcomes possible outcomes if there are only few in each event. For instance, the first column events. These are tree diagram, sample would be each shirt, and the first row would space and table of outcomes. be each pair of pants. Then we simply fill in the table with the possible outcomes Note: These are applicable if and only if there are few events. Given Problem: Suppose you have a black shirt, a brown shirt, and a white shirt, and you also have a pair of jeans and a pair of black pants. How many possible outfits can you make?  PERMUTATION 34& ! 3. 3)& ! Permutation is a way, especially one of several possible variations, in which a set or Solution: number of things can be ordered or 𝑛+1 𝑛 𝑛−1 ! arranged. 𝑛−1 ! *(n-1)! will be CANCELLEDT Let S be a set of n elements and let 1 ≤ 𝑟 ≤ (n+1) (n) 𝑛. A permutation of r elements of S is an arrangement, without repetition of r = 𝒏𝟐 + 𝒏 elements. nPr Notation Example: How many permutations are there Formula for Permutation WITHOUT of 4 children selected from a group of 8? REPETITION: 3! Solution: The first child can be selected in 8 nPr = 3)9 ! ways, then the next can be selected in 7 ways, Example: How many three – letter words can the third in 6 ways and the fourth in 5 ways. be formed from the letters of the word Using FCP, the total number of permutation SUNDAY? is (8) (7) (6) (5) or 1,680 ways. Solution: n=6;r=3 Factorial Notation Using the formula, For each positive integer n, 6! 6−3 ! n! = n (n-1) (n-2) (n-3) (n-4) … (3) (2) (1). 6! (By definition 0! Is equal to 1) 3! = 120 three – letter words Examples: Simplify the following: (or just simply type to your calculator 6P3 and then you will get 120 myghadit ang hirap na nga ng buhay wag na natin dagdagan 1. 6! + 2! paghihirap naten) Solution: Another example: In how many ways can 7 (6) (5) (4) (3) (2) (1) + (2) (1) students be arranged in a straight line? 720 + 2 Solution: n = 7 ; r = 7 6! + 2! = 722 7P7 = 5,040 ways &'! 2. Formula for permutation WITH &')* ! REPETITION Solution: 𝑛! 10! 𝑝! 𝑞! 7! 362880 Note: This rule can be extended for any 5040 number of objects that are repeated. = 720 Example: How many diff ways can letter of each word be arranged? 1. PARALLEL Another example: In how many ways can you - n(A) = 2; n(L) = 3 select a committee of 3 students out of 10 8! / (2!) (3!) = 3,360 ways students? 2. MAMMALS Solution: 10C3 = 120 ways. - n(M) = 3 ; n(A) = 2 7! / (3!) (2!) = 420 ways PERMUTATION vs. COMBINATION 3. ADDITION PERMUTATION – ORDER is - n(D) = 2 ; n(I) = 2 IMPORTANT 8! / (2!) (2!) = 10,080 ways COMBINATION – ORDER is NOT IMPORTANT CIRCULAR PERMUTATION The formula for circular permutation is: (n-1)! Example: Five people are seated in a round table. In how many ways can they be arranged? Solution: (5-1) ! = 4! = 24 ways Another example: In how many ways can 8 keys be arranged on a key ring? Solution: (8-1)! = 7! = 5,040 ways COMBINATION Let S be a set of n elements and let 1 ≤ 𝑟 ≤ 𝑛. A combination of r elements of S is a subset of S that contains r distinct elements. The number of combinations of r elements that can be obtained from a set of n elements is : 3! nCr = wherein 1 ≤ 𝑟 ≤ 𝑛 3)9 !9! Example: In how many ways you can choose 8 of 32 playing cards not considering their order? Solution: Or simply type 32C8 on your calculator. You will get 10,518,300 ways.  PROBABILITY TERMS: c. Complement ( ’ ) 1.) Probability -The complement of an event A is the event - is the likelihood of something happening. that A does not occur, that is the event When someone tells you the probability of consisting of all sample points that are not in something happening, they are telling you event A. The complement of A is A’. how likely that something is. U = {1,2,3….10} 2.) Experiment A = {1,2,3} B={4,5} - is the process of making an observation that leads to a single outcome which cannot be A’ = {4,5,6,7,8,9,10} B’ = {1,2,3,6,7,8,9,10} predicted with certainty. 3.) Outcome/s Formula of Probability: -is a possible result of an experiment or trial. 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒐𝒖𝒕𝒄𝒐𝒎𝒆𝒔 𝒊𝒏 𝒂𝒏 𝒆𝒗𝒆𝒏𝒕 3.) Sample point 𝑷 𝑬 = 𝒕𝒐𝒕𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒐𝒖𝒕𝒄𝒐𝒎𝒆𝒔 -the most basic outcome of an experiment. 𝒊𝒏 𝒕𝒉𝒆 𝒔𝒂𝒎𝒑𝒍𝒆 𝒔𝒑𝒂𝒄𝒆 4.) Sample space -is the collection of all its sample points. TYPES OF PROBABILITY 5.) Event 1. Probability of Union of Mutually Exclusive and Inclusive Events -is a specific collection of sample points. 1(A). Probability of Union of Mutually 2 Types of Events: Exclusive Events -If two events A and B are mutually a.) Simple events- single outcome only exclusive, the probability of the union of A b.) Compound events- two or more outcomes and B equals the sum of the probabilities of A and B. SETS OPERATIONS: P(AUB) = P(A) + P(B) a.) Union of Sets “U” Example 1: -The union of two events A and B is the event A die is rolled once what is the probability of that occurs if either A or B or both occur on a getting an even number or a five? single performance of the experiment. It is 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒐𝒖𝒕𝒄𝒐𝒎𝒆𝒔 𝒊𝒏 𝒂𝒏 𝒆𝒗𝒆𝒏𝒕 denoted by AUB which contains all the 𝑷 𝑬 = 𝒕𝒐𝒕𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒐𝒖𝒕𝒄𝒐𝒎𝒆𝒔 sample points of A or B or both. 𝒊𝒏 𝒕𝒉𝒆 𝒔𝒂𝒎𝒑𝒍𝒆 𝒔𝒑𝒂𝒄𝒆 A = {1,2,3} B = {4,5} Even numbers: P(A) = {2,4,6} = 3/6 AUB = {1,2,3,4,5} P(B)={5} = 1/6 b.) Intersection of Sets “Ç” 3 1 4 2 + = 𝑜𝑟 -The intersection if two events A and B is the 6 6 6 3 event that occurs if both A and B occur in a 𝟐 𝒊𝒔 𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒐𝒇 𝒈𝒆𝒕𝒕𝒊𝒏𝒈 𝒂𝒏 𝒆𝒗𝒆𝒏 𝒏𝒖𝒎𝒃𝒆𝒓 single performance if the experiment. It is 𝟑 denoted by AÇB which contains the sample 𝒐𝒓 𝒂 𝒇𝒊𝒗𝒆 points in both A and B. A = {1,2,3,4,5,6} B = {0,2,4,6,7} AÇB = {2,4,6} Example 2: Divisible by 4: P(B) = {4,8,12} = 3/15 A coin is tossed three times. Find the P(AÇB) = {12} = 1/15 probability of getting either all heads or exactly 1 head. (List all the probabilities). P(AUB) = P(A) + P(B) – P(AÇB) 5 3 1 7 HHH, HHT, HTH, HTT, THH, THT, TTH, + − = TTT 15 15 15 15 𝟕 All heads: P(A) = {HHH} = 1/8 𝒊𝒔 𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒕𝒉𝒂𝒕 𝒂 𝒏𝒖𝒎𝒃𝒆𝒓 𝒊𝒔 𝟏𝟓 One head: P(B) = {HTT,THT,TTH} = 3/8 𝒅𝒊𝒗𝒊𝒔𝒊𝒃𝒍𝒆 by 3 and divisible by 4 P(A)+P(B) 1 3 4 1 2. Probability of Independent and + = 𝑜𝑟 Dependent Events 8 8 8 2 𝟏 2(A). Probability of Two Independent 𝒊𝒔 𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒐𝒇 𝒈𝒆𝒕𝒕𝒊𝒏𝒈 𝒆𝒊𝒕𝒉𝒆𝒓 𝒂𝒍𝒍 Events 𝟐 𝒉𝒆𝒂𝒅𝒔 𝒐𝒓 𝒆𝒙𝒂𝒄𝒕𝒍𝒚 𝟏 𝒉𝒆𝒂𝒅 -To find the probability if two independent events that occur in a sequence, find the probability of each event occurring 1(B). Probability of Inclusive/Non- separately, and then multiply the Mutually Exclusive Events probabilities. -The probability of the union of events A and -If two events, A and B are independent, B is the sum of the probability of events A then the probability of two events occurring and B minus the probability of the is found as follows. intersection of events A and B. P(AÇB) = P(A) x P(B) or P(AUB) = P(A) + P(B) – P(AÇB) P(A and B) = P(A) x P(B) Example 1: Example 1: A die is rolled once. What is the probability of getting an even number or a number A coin is tossed and a die is rolled. Find the greater than 2? probability of landing on the head side of the coin and rolling a 3 on the die. Even numbers: P(A) = {2,4,6} = 3/6 Number greater than 2: P(B) = {3,4,5,6}= Probabilities: 4/6 Let P(A) be the probability of landing on the P(AÇB) = {4,6} = 2/6 head P(AUB) = P(A) + P(B) – P(AÇB) Let P(B) be the probability of rolling a 3 on the die 3 4 2 5 + − = & 6 6 6 6 P(A) = P(head) = Z 𝟓 𝒊𝒔 𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒐𝒇 𝒈𝒆𝒕𝒕𝒊𝒏𝒈 𝒂𝒏 & 𝟔 P(B) = P(3) = [ 𝒆𝒗𝒆𝒏 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒓 𝒂 𝒕𝒘𝒐 P(AÇB) = P(head and 3) = P(head) x P(3) & & & Example 2: P(A) x P(B) = x = Z [ &Z 𝟏 A bowl contains 15 chips numbered 1-15, if 𝒊𝒔 𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒐𝒇 𝒈𝒆𝒕𝒕𝒊𝒏𝒈 𝒂 𝒉𝒆𝒂𝒅 𝒂𝒏𝒅 𝟑 𝟏𝟐 a chip is drawn randomly from the bowl what is the probability that a number is divisible by 3 and divisible 4? Divisible by 3: P(A) = { 3,6,9,12, 15} = 5/15 Example 2: Defective: 3 computers A jar contains 3 red, 5 green, 2 blue and 6 We have three variables here: yellow marbles. A marble is chosen at Let P(A) be the first detective chosen. random from the jar. After replacing it, a Let P(B) be the second defective chosen. second marble is chosen. What is the probability of choosing a green and then a Let P(C) be the third defective chosen. yellow marble? P(All three defectives) Let P(A) be the probability of choosing a = P(A) x P(B, after A) x P(C, after B) green marble. * Z & [ & Let P(B) be the probability of choosing a x x = = Z' &_ &] []a' &&a' yellow marble. 𝟏 Total no. of marbles: 16 marbles 𝒊𝒔 𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒕𝒉𝒂𝒕 𝒂𝒍𝒍 𝒕𝒉𝒓𝒆𝒆 𝒂𝒓𝒆 𝟏𝟏𝟒𝟎 \ 𝒅𝒆𝒇𝒆𝒄𝒕𝒊𝒗𝒆 𝒊𝒇 𝒕𝒉𝒆 𝒇𝒊𝒓𝒔𝒕 𝒂𝒏𝒅 𝒔𝒆𝒄𝒐𝒏𝒅 P(A) = P(green) = &[ 𝒂𝒓𝒆 𝒏𝒐𝒕 𝒓𝒆𝒑𝒍𝒂𝒄𝒆𝒅. [ P(B) = P(yellow) = &[ Explanation: P(AÇB) = P(green and yellow) = P(green) x P(yellow) Dahil nga naapektuhan yung mga naunang probability, mababawasan na yung chances \ [ *' &\ P(A) x P(B) = x = = sa mga susunod na probabilities. Nung una, &[ &[ Z\[ &Z] may 3 defectives and 20 computers kaya 𝟏𝟓 3/20, ngayon dahil sinabi sa problem na 𝒊𝒔 𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒐𝒇 𝒈𝒆𝒕𝒕𝒊𝒏𝒈 𝒂 𝒈𝒓𝒆𝒆𝒏 𝟏𝟐𝟖 walang replacement dahil kinuha mo na 𝒂𝒏𝒅 𝒚𝒆𝒍𝒍𝒐𝒘 𝒎𝒂𝒓𝒃𝒍𝒆 yung isang defective, kaya magiging 2 defectives nalang at yung total computer ay 19 nalang dahil computer parin ung natanggal 2(B). Probability of Dependent Events na defective, kaya 2/19 nalang ung second probability. Ngayon sa third na pagpili, - Two events are dependent if the outcome nakuha na ung dalawang defectives kaya isa or occurrence of the first affects the outcome nalang yung natira and yung dalawa na iyon or occurrence of the second so that the ay imiminus sa total computers kaya 18 probability is changed. nalang. Kaya ang probability sa third -If two events, A and B, are dependent, then defective computer chosen ay 1/18. Para lang the probability of both events occurring is siyang permutation without replacement. found as follows: P(A and B) = P(A) x P(B following A) Example 2: Note: Maapektuhan ng unang probability Four cards are chosen at random from a deck yung pangalawang probability. of 52 cards without replacement. What is the Example 1: probability of choosing a ten, a nine, an eight and a seven in order? In a shipment of 20 computers, 3 are defective. Three computers are randomly We have four variables here: selected and tested. What is the probability Let P(A) be the probability of choosing a that all three are defective if the first and ten. second ones are not replaced after being Let P(B) be the probability of choosing a tested? nine. Total number of computers: 20 computers Let P(C) be the probability of choosing an k lÇm P BA = OR eight. k l Let P(D) be the probability of choosing a P(A Ç B) P AB = seven. P(B) = P(A) x P(B, after A) x P(C, after B) Example 1: x P(D, after C) A math teacher gave her class two tests. Total number of cards: 52 25% of the class passed both tests and 42% of the class passed the first test. What Number of ten: 4 (heart, diamond, clubs, percent of those who passed the first test spade) also passed the second test? Number of nine: 4 (heart, diamond, clubs, -This problem describes a conditional spade) probability since it asks us to find the Number of eight: 4 (heart, diamond, clubs, probability that the second test was passed spade) given that the first test was passed. (may condition) Number of seven: 4 (heart, diamond, clubs, spade) Let P(A) be the probability of passing the first test = 42% 4 4 4 4 256 𝑥 𝑥 𝑥 = Let P(B) be the probability of passing the 52 51 50 49 6,497,400 second test 32 = P(A and B) =those who passed both tests = 812,175 25% 𝟑𝟐 𝒊𝒔 𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒐𝒇 𝒄𝒉𝒐𝒐𝒔𝒊𝒏𝒈 Let P(B|A) be the probability of passing the 𝟖𝟏𝟐, 𝟏𝟕𝟓 second test after passing the first test 𝒂 𝒕𝒆𝒏, a nine, an eight and seven in order. P A and B Explanation: P BA = P A Bakit yung denominator lang yung P First and Second nababawasan? Dahil yung hindi naman pare- P second first = P First parehas na number yung gusto mong kunin kaya 4 parin ung chances na makuha mo ung P(0.25) P BA = = 0.60 = 60% ibang mga numbers. Ngayon pag sinabing P(0.42) with no replacement ibig sabihin hindi Explanation: nababalik yung card kaya pabawas ng pabawas yung 52 cards. P(B|A)- ang ibig sabihin nito ano ang probability na mangyari ang event B kung nangyari or after mangyari ang event A. Dito 3. Conditional Probability sa problem given na yung intersection ng dalawa which is 25% kaya hindi na -Conditional probabilities can be recognized kailangang isolve pa, then idivide siyasa by words like ‘given’, ‘if’, or ‘among’ probability ng event A which is given na rin. -The conditional probability of an event B in relationship to an event A is the probability that event B occurs given that event A has Example 2: already occurred. The notation for In a group of 20 males and 15 females, 12 conditional probability is P(B|A), read as the males and 8 females are service holders. probability of B given A. The formula for What is the probability that a person selected conditional probability is: at random from the group is a service holder given that the selected person is a male? Condition: The condition is that the selected * 𝟏 P(A) = { (4, 6), (5, 5), (6, 4) } = = person is a male. *[ 𝟏𝟐 Let P(A) be the probability of selecting a Let P(B) be the probability of getting the 4 on male. one of the dice. 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒐𝒖𝒕𝒄𝒐𝒎𝒆𝒔 𝒊𝒏 𝒂𝒏 𝒆𝒗𝒆𝒏𝒕 P(B) = { (1, 4), (2, 4), (3, 4), (4, 1), (4, 2), P(A) = 𝒕𝒐𝒕𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒐𝒖𝒕𝒄𝒐𝒎𝒆𝒔 (4, 3), (4, 5), (4, 6), (5, 4), (6, 4) } 𝒊𝒏 𝒕𝒉𝒆 𝒔𝒂𝒎𝒑𝒍𝒆 𝒔𝒑𝒂𝒄𝒆 𝟐𝟎 = Get their intersection: 𝟑𝟓 P(A) = { (4, 6), (5, 5), (6, 4) } Let P(B) be the probability of selecting a service holder. P(B) = { (1, 4), (2, 4), (3, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (5, 4), (6, 4) } THEN: 𝟐 𝟏 P(AÇB) = 𝟏𝟐 P(AÇ𝐁) = { (4, 6), (6, 4) } = = 𝟑𝟔 𝟏𝟖 𝟑𝟓 Then solve: P(A Ç B) P BA = P(A) P(A Ç B) BA = &Z Z' P(A) P BA = ÷ *\ *\ 1 1 3 P BA = ÷ 18 12 P BA = 5 Z P BA = 𝟑 * 𝒊𝒔 𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒕𝒉𝒂𝒕 𝒂 𝒑𝒆𝒓𝒔𝒐𝒏 𝒔𝒆𝒍𝒆𝒄𝒕𝒆𝒅 𝟓 𝟐 𝒊𝒔 𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒕𝒉𝒂𝒕 𝒐𝒏𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒕𝒘𝒐 𝒅𝒊𝒄𝒆 𝒂𝒕 𝒓𝒂𝒏𝒅𝒐𝒎 𝒊𝒔 𝒂 𝒔𝒆𝒓𝒗𝒊𝒄𝒆 𝒉𝒐𝒍𝒅𝒆𝒓 𝒈𝒊𝒗𝒆𝒏 𝒕𝒉𝒂𝒕 𝟑 𝒔𝒆𝒍𝒆𝒄𝒕𝒆𝒅 𝒑𝒆𝒓𝒔𝒐𝒏 𝒊𝒔 𝒂 𝒎𝒂𝒍𝒆. 𝒉𝒂𝒔 𝒔𝒉𝒐𝒘𝒏 𝒂 𝟒. Explanation: Explanation: Ang condition daw dito ay kailangan ang Sa problem na ito ang hinihingi ay ang selected service holder ay male. Ang unang probability na lilitaw ang 4 sa isang dice, event ay probability na makakapili ng lalaki, given na yung sum ay 10. Ang unang dapat sa problem may 20 na lalaki at 35 ang total gawin ay, ilista ang numbers satwo dice na sample points so P(A) ay 20/35. Ang P(B) ay kapag niroll ay magsusum ng 10. selecting a service holder, ngayon ang Pagkatapos, ilista yung mga instances na intersection between the two events ay 12 pwedeng lumabas ang 4 sa isang die dahil sa P(A) may 20 males at 12 roon ay regardless of the sum. Ang sunod na ginawa service holder so ang intersection between ay hinanap ang intersection ng 2 events at males and service holders ay 12. Then, idinivide sa probability ng event A. perform the operation. Note: Ang kadalasang event A ay yung condition. Example 3: A pair of dice is thrown together and the sum of points of the two dice is noted to be 10. What is the probability that one of the two dice has shown the point 4? Condition: The sum of points of the two dice is noted to be 10 Let P(A) be the probability of getting the sum of points to be 10.  REMINDER: How to know what kind of probability a problem is? Mutually Exclusive- remember that it happened at the same time, simultaneously and there cannot be any intersection. -Kunyare a die is rolled once probability of getting an even number and a five, syempre hindi magkakaroon ng intersection between sample points dahil hindi even number yung five. -Pwede mong gawin yung listing method. Inclusive events- There is an intersection sa pagitan ng dalawang hinihingi. -Kunyare a die is rolled once, probability of getting a number greater than 1 and an even number. Merong numbers greater than 1 na pwedeng even, may mga even numbers din na more than 1. See. Inclusive. May intersection. Pero remember iisang event ito at sabay nangyayari. Independent Events- Remember na events to na pwedeng magkaiba o magkasunod pero kung sabay kailangan magkaibang event or experiment. -Kunyare a die is rolled and a coin is flipped. Or choosing an ace and then choosing a four. -Hindi rin nito naapektuhan ang isa’t isa. Dependent Events- Ito yung mga problem na naapektuhan nila yung isa’t isa. Meaning yung probability ng isang event ay makakapagpabawas or makakadagdag sa probability ng isang event. -Kadalasang mga problems without replacement. Conditional Probability- lagging may condition, and words like “given that”, “if”. (Note: Hindi pa itinuturo ni Ma’am kaya nagresearch lang ako, dadagdagan ko nalang if ever may tinuro siyang iba)