SHS-MAT201 Week 11 Study Guide PDF

MAT201-SHS-SG-WK11 FUTURE READY FUTURE SMART PROGRAM STUDY GUIDE Subject STATISTIC...

MAT201-SHS-SG-WK11 FUTURE READY FUTURE SMART PROGRAM STUDY GUIDE Subject STATISTICS AND PROBABILITY Topic Probability and Random Variable Level Grade 11 Week Week 11 School Year 3rd Quarter/SY 2021-2022 I. Objectives − Recall solving simple probability problems. − Illustrate a random variable (discrete and continuous). − Distinguish between a discrete and a continuous random variable. − Find the possible values of a random variable. − Illustrate a probability distribution for a discrete random variable and its properties. II. Discussion Probability An experiment is any procedure that can be infinitely repeated and has a well-defined set of possible outcomes. “A sample space, also called an outcome space, is simply the set of all possible outcomes in a given selection or in a combination of selections (Albay 16)”. Determining the number of outcomes is important in determining the probability of an event. In addition, determining the sample spaces and events is also important. A sample space or outcome space is the set of all possible outcomes of an experiment. An event refers to a part or portion of the sample space. Remember that sample space and event are sets so they should be written in a set way (roster form). Example 1: Imagine a coin is tossed. The possible outcomes are tail and head (T and H). Then the sample space is 𝑆 = {𝑇, 𝐻}. There are two possible events and those are 𝑋 = {𝑇, } and 𝑌 = {𝐻}. Example 2: Mrs. Reyes has 5 applicants for the positions of his secretaries. The five applicants are Khryzhelle, Diana, Elaine, Trisha, and Lovelyn. Mr. Reyes is about to choose 1 secretary ©2020 Good Samaritan Colleges Page 1 of All rights reserved. No part of this publication may be reproduced, distributed, or transmitted in any form or by any means, including photocopying, recording, or other electronic or mechanical methods, without the prior written permission of the Good Samaritan Colleges. 14  MAT201-SHS-SG-WK11 from five applicants (Khryzhelle, Diana, Elaine, Trisha, and Lovelyn). The sample space for this is 𝑆 = {𝐾ℎ𝑟𝑦𝑧ℎ𝑒𝑙𝑙𝑒, 𝐷𝑖𝑎𝑛𝑎, 𝐸𝑙𝑎𝑖𝑛𝑒, 𝑇𝑟𝑖𝑠ℎ𝑎, 𝐿𝑜𝑣𝑒𝑙𝑦𝑛} and some of the possible events are 𝑀 = {𝐸𝑙𝑎𝑖𝑛𝑒} and 𝐷 = {𝐿𝑜𝑣𝑒𝑙𝑦𝑛}. Simple Probability Since you know how to determine the sample space and events, then you can now compute the probability of an event. The probability of an event is a numerical value that describes the possibility that an event will happen or not. It is simply calculated using the ratio of the number of events and the number of the sample space (Albay 17). Probability of an Event The probability of an event, with the symbol P(E), is the numerical measure of the possibility that an event will happen. It is calculated by determining the quotient of the number of favorable outcomes and the total number of possible outcomes. 𝑛(𝐸) 𝑃(𝐸) = 𝑛(𝑆) 𝑛(𝐸) = the number of the elements in the event 𝑛(𝑆) = the number of the elements in the sample space Example 1: Imagine a coin is tossed. The possible outcomes are tail and head. Then the sample space is 𝑆 = {𝑡𝑎𝑖𝑙, ℎ𝑒𝑎𝑑}. There are two possible events and those are 𝑋 = {𝑡𝑎𝑖𝑙} and 𝑌 = {ℎ𝑒𝑎𝑑}. What is the probability of getting a tail from the coin? ©2020 Good Samaritan Colleges Page 2 of All rights reserved. No part of this publication may be reproduced, distributed, or transmitted in any form or by any means, including photocopying, recording, or other electronic or mechanical methods, without the prior written permission of the Good Samaritan Colleges. 14  MAT201-SHS-SG-WK11 Our event pertains to getting a tail. In tossing a coin, you can only get a tail once. Therefore, the number of elements in the event is only 1. Therefore, 𝑛(𝐸) = 1. The sample space contains 2 possible outcomes which are 1 head and 1 tail. Therefore, 𝑛(𝑆) =2 𝑛(𝐸) = 1 𝑛(𝑆) = 2 Computing for the probability: 𝑛(𝐸) 𝑃(𝐸) = 𝑛(𝑆) 𝟏 𝑷(𝑬) = 𝟐 𝑷(𝑬) = 𝟎. 𝟓𝟎 𝑷(𝑬) = 𝟓𝟎% Therefore, there is a 50% possibility that you will get a tail in tossing a coin. Example 2: David has a bag with 6 red, 4 blue, and 8 green marbles. What is the probability that a marble chosen at random is not red? There are 12 marbles that are not red (4 blue + 8 green = 12 marbles). Therefore, 𝑛(𝐸) = 12. Since the total of marbles is 18, then 𝑛(𝑆) = 18. 𝑛(𝐸) 𝑃(𝐸) = 𝑛(𝑆) 𝟏𝟐 𝑷(𝑬) = 𝟏𝟖 Reduce the fraction to its lowest term. 𝟏𝟐 𝟔 𝑷(𝑬) = ÷ 𝟏𝟖 𝟔 𝟐 𝑷(𝑬) = 𝟑 𝑷(𝑬) = 𝟎. 𝟔𝟕 ©2020 Good Samaritan Colleges Page 3 of All rights reserved. No part of this publication may be reproduced, distributed, or transmitted in any form or by any means, including photocopying, recording, or other electronic or mechanical methods, without the prior written permission of the Good Samaritan Colleges. 14  MAT201-SHS-SG-WK11 𝑷(𝑬) = 𝟔𝟕% Random Variable “A random variable, also called a stochastic variable, is a rule that assigns a numerical value or characteristic to an outcome of an experiment. It is essentially a variable, usually denoted as X or any capital letter of the alphabet., because its value is not constant” (Albay 34). For example, a die is rolled three times and a random variable X is assigned as the number of times a “3” appears. The random variable X can take on the values 0, 1, 2, and 3 as the outcome may vary from trial to trial. “There are two categories of random variable: discrete and continuous random variables. A discrete random variable takes on the countable number of distinct values, which are whole numbers such as 0, 1, 2, 3, 4, 5 … while a continuous random variable assumes an infinite number of possible values including the decimals between two counting numbers” (Albay 34). The values of a discrete random variable are “counts” and those of a continuous random variable are “measurements”. Example 1: Discrete vs. Continuous Variables a. A fair coin tossed 8 times and the number of times X that a tail appears is a discrete random variable since its possible values may be determined by counting, such as 0, 1, 2, 3, 4, 5, 6, 7, 8. b. A machine ran and the recorded time it starts to experience a glitch Y illustrates a continuous random variable since the value of the variable may be assigned using measurement. The set of all values possible for a given random variable is called the range space. For example, when two fair coins are tossed and the random variable X is defined as the number of tails that appear, the range space is the set {0, 1, 2}. The “0” in the range space stands for the event that there is no tail that will appear. The “1” in the range space stands for the event that there is 1 tail that will appear. The “2” in the range space stands for the event that there are 2 tails that will appear. ©2020 Good Samaritan Colleges Page 4 of All rights reserved. No part of this publication may be reproduced, distributed, or transmitted in any form or by any means, including photocopying, recording, or other electronic or mechanical methods, without the prior written permission of the Good Samaritan Colleges. 14  MAT201-SHS-SG-WK11 Discrete Probability Distribution “We don’t only determine the value of a random variable but also the probability that the variable will assume that particular value. If you have a random variable and you know all the possible values that the variable can take on, you can easily determine the probability that the random variable will have a particular value” (Albay 36). For example, consider the random event of tossing four coins and the variable X gives the number of heads that appear. Your range space, then, will be the set {0, 1, 2, 3, 4}. You know that there are 16 possible outcomes as follows: 2 possible outcomes – head and tail 4 coins 24 = 16 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 HHHH HTHH THHH TTHH HHHT HTHT THHT TTHT HHTH HTTH THTH TTTH HHTT HTTT THTT TTTT Random Variable: 𝑿 = {𝟎, 𝟏, 𝟐, 𝟑, 𝟒} Probability of random variable X when it assumes a value of 0: There is only one possible outcome of not getting head and that is the “TTTT”. ©2020 Good Samaritan Colleges Page 5 of All rights reserved. No part of this publication may be reproduced, distributed, or transmitted in any form or by any means, including photocopying, recording, or other electronic or mechanical methods, without the prior written permission of the Good Samaritan Colleges. 14  MAT201-SHS-SG-WK11 𝑃(𝑋 = 0) = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑋 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑎𝑠𝑠𝑢𝑚𝑒𝑠 𝑎 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 0 𝑛(𝐸) 𝑃(𝑋 = 0) = 𝑛(𝑆) 𝑛(𝐸) = 1 𝑛(𝑆) = 16 1 𝑃(𝑋 = 0) = 16 𝑷(𝑿 = 𝟎) = 𝟎. 𝟎𝟔𝟐𝟓 Probability of random variable X when it assumes a value of 1: There are 4 possible outcomes of getting 1 head and those are the “HTTT, THTT, TTHT, TTTH”. 𝑃(𝑋 = 1) = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑋 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑎𝑠𝑠𝑢𝑚𝑒𝑠 𝑎 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 1 𝑛(𝐸) 𝑃(𝑋 = 1) = 𝑛(𝑆) 𝑛(𝐸) = 4 𝑛(𝑆) = 16 4 𝑃(𝑋 = 1) = 16 4 4 𝑃(𝑋 = 1) = ÷ 16 4 1 𝑃(𝑋 = 1) = 4 𝑷(𝑿 = 𝟏) = 𝟎. 𝟐𝟓 Probability of random variable X when it assumes a value of 2: There are 6 possible outcomes of getting 2 heads and those are the “HHTT, HTHT, HTTH, THHT, THTH, TTHH”. 𝑃(𝑋 = 2) = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑋 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑎𝑠𝑠𝑢𝑚𝑒𝑠 𝑎 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 2 ©2020 Good Samaritan Colleges Page 6 of All rights reserved. No part of this publication may be reproduced, distributed, or transmitted in any form or by any means, including photocopying, recording, or other electronic or mechanical methods, without the prior written permission of the Good Samaritan Colleges. 14  MAT201-SHS-SG-WK11 𝑛(𝐸) 𝑃(𝑋 = 2) = 𝑛(𝑆) 𝑛(𝐸) = 6 𝑛(𝑆) = 16 6 𝑃(𝑋 = 2) = 16 6 2 𝑃(𝑋 = 2) = ÷ 16 2 3 𝑃(𝑋 = 2) = 8 𝑷(𝑿 = 𝟐) = 𝟎. 𝟑𝟕𝟓 Probability of random variable X when it assumes a value of 3: There are 4 possible outcomes of getting 3 heads and those are the “HHHT, HHTH, HTHH, THHH”. 𝑃(𝑋 = 3) = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑋 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑎𝑠𝑠𝑢𝑚𝑒𝑠 𝑎 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 3 𝑛(𝐸) 𝑃(𝑋 = 3) = 𝑛(𝑆) 𝑛(𝐸) = 4 𝑛(𝑆) = 16 4 𝑃(𝑋 = 3) = 16 4 4 𝑃(𝑋 = 3) = ÷ 16 4 1 𝑃(𝑋 = 3) = 4 𝑷(𝑿 = 𝟑) = 𝟎. 𝟐𝟓 Probability of random variable X when it assumes a value of 4: There is only one possible outcome of getting 4 heads and that is the “HHHH”. ©2020 Good Samaritan Colleges Page 7 of All rights reserved. No part of this publication may be reproduced, distributed, or transmitted in any form or by any means, including photocopying, recording, or other electronic or mechanical methods, without the prior written permission of the Good Samaritan Colleges. 14  MAT201-SHS-SG-WK11 𝑃(𝑋 = 4) = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑋 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑎𝑠𝑠𝑢𝑚𝑒𝑠 𝑎 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 4 𝑛(𝐸) 𝑃(𝑋 = 4) = 𝑛(𝑆) 𝑛(𝐸) = 1 𝑛(𝑆) = 16 1 𝑃(𝑋 = 4) = 16 𝑷(𝑿 = 𝟒) = 𝟎. 𝟎𝟔𝟐𝟓 Number of Heads (X) 0 1 2 3 4 Probability (P(X)) 0.0625 0.25 0.375 0.25 0.0625 This table is called a probability distribution which is also known as probability mass function. Probability Distribution “A probability distribution, also known as probability mass function, is a table that gives a list of probability values along with their associated value in the range of a discrete random variable” (Albay 36). Note from the previous probability mass function that the following properties are observed given that pi is the individual probabilities for each value in the range of a discrete random variable. 1. Each probability value ranges from 0 to 1, in symbols, 0 ≤ 𝑝i ≤ 1. 2. The sum of all the individual probabilities in the distribution is equal to 1; thus, n 𝑝1 + 𝑝2 + 𝑝3 +... +𝑝n = Σ 𝑝i = 1 i=1 Also, note that like any other statistical distribution, a probability mass function may be graphed using a histogram in which the horizontal axis represents the values of the random variable X and the vertical axis gives the corresponding probabilities, P(X). For example, shown below is the related histogram of the random variable X (showing a head after tossing four coins). ©2020 Good Samaritan Colleges Page 8 of All rights reserved. No part of this publication may be reproduced, distributed, or transmitted in any form or by any means, including photocopying, recording, or other electronic or mechanical methods, without the prior written permission of the Good Samaritan Colleges. 14  MAT201-SHS-SG-WK11 ©2020 Good Samaritan Colleges Page 9 of All rights reserved. No part of this publication may be reproduced, distributed, or transmitted in any form or by any means, including photocopying, recording, or other electronic or mechanical methods, without the prior written permission of the Good Samaritan Colleges. 14