{"id":2105,"date":"2021-08-03T17:46:18","date_gmt":"2021-08-03T21:46:18","guid":{"rendered":"https:\/\/openbooks.macewan.ca\/rcommander\/?post_type=chapter&p=2105"},"modified":"2024-02-08T13:27:26","modified_gmt":"2024-02-08T18:27:26","slug":"4-3-defining-events-using-random-variable-notation","status":"publish","type":"chapter","link":"https:\/\/openbooks.macewan.ca\/introstats\/chapter\/4-3-defining-events-using-random-variable-notation\/","title":{"raw":"4.3 Defining Events Using Random Variable Notation","rendered":"4.3 Defining Events Using Random Variable Notation"},"content":{"raw":"We can define events using the notation of random variables and we can compute probabilities of events based on the probability distributions of the variables. For example, the event of having one sibling can be written as [latex]\\{X=1\\}[\/latex] and its probability is [latex]P(X=1) =0.2[\/latex]. The event of having at least one sibling is [latex]\\{X \\geq 1 \\}[\/latex] and its probability is\r\n[latex]\\begin{align*}\r\nP(X \\geq 1) &= P(X=1 \\text{ or } X=2 \\text{ or } X=3) \\\\\r\n&= P(X=1) + P(X=2) + P(X=3) \\\\\r\n&= 0.2 + 0.4 + 0.2 = 0.8.\r\n\\end{align*}[\/latex]\r\nAlternatively, we can apply the complement rule to find the probability:\r\n[latex]\\begin{align*}\r\nP(X \\geq 1) &= P(X=1 \\text{ or } X=2 \\text{ or } X=3) \\\\\r\n&= 1 - P(X=0) \u00a0\\\\\r\n&= 1 - 0.2 = 0.8.\r\n\\end{align*}[\/latex]\r\n
\r\n\r\n\"\"\r\n\r\n<\/div>\r\n
\r\n

Exercise: Define Events Using Random Variable<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nConsider the probability distribution of X = # of siblings below.\r\n\r\n\r\n\r\n\r\n
[latex]x[\/latex]<\/strong><\/td>\r\n0<\/td>\r\n1<\/td>\r\n2<\/td>\r\n3<\/td>\r\n<\/tr>\r\n
[latex] P(X=x)[\/latex]<\/strong><\/td>\r\n0.2<\/td>\r\n0.2<\/td>\r\n0.4<\/td>\r\n0.2<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nDefine the following events using the variable X and find the probabilities:\r\n
    \r\n \t
  1. Having one and a half siblings.<\/li>\r\n \t
  2. Having zero to two siblings exclusively.<\/li>\r\n \t
  3. Having zero to two siblings inclusively.<\/li>\r\n<\/ol>\r\n
    Show\/Hide Answer<\/summary>\r\n
      \r\n \t
    1. Event: [latex]\\{X=1.5 \\}[\/latex]. Since the possible values of [latex]X[\/latex] are [latex]x=0, 1, 2, 3[\/latex], it is impossible for [latex]X[\/latex] to be 1.5; therefore, [latex]P(X=1.5) =0[\/latex].<\/li>\r\n \t
    2. Event: [latex]\\{ 0 < X <2 \\}[\/latex]. Since 1 is the only possible value that is greater than 0 and smaller than 2, [latex]\\{ 0 < X < 2 \\} = \\{ X=1 \\} [\/latex]. This implies [latex]P(0 < X <2) = P(X=1) = 0.2[\/latex].<\/li>\r\n \t
    3. Event: [latex]\\{ 0 \\leq X \\leq 2 \\} = \\{ X=0 \\text{ or } X=1 \\text{ or } X=2 \\}[\/latex]. Since the events are mutually exclusive, i.e., they don't overlap, the special addition rule gives:\r\n[latex]\\begin{align*} P(0 \\leq X \\leq 2) &= P(X=0 \\text{ or } X=1 \\text{ or } X=2) \\\\\r\n&= P(X=0) + P(X=1) + P(X=2) \\\\\r\n&= 0.2 + 0.2 + 0.4 \\\\\r\n&= 0.8.\r\n\\end{align*}[\/latex]<\/li>\r\n<\/ol>\r\n<\/details><\/div>\r\n<\/div>","rendered":"

      We can define events using the notation of random variables and we can compute probabilities of events based on the probability distributions of the variables. For example, the event of having one sibling can be written as [latex]\\{X=1\\}[\/latex] and its probability is [latex]P(X=1) =0.2[\/latex]. The event of having at least one sibling is [latex]\\{X \\geq 1 \\}[\/latex] and its probability is
      \n[latex]\\begin{align*} P(X \\geq 1) &= P(X=1 \\text{ or } X=2 \\text{ or } X=3) \\\\ &= P(X=1) + P(X=2) + P(X=3) \\\\ &= 0.2 + 0.4 + 0.2 = 0.8. \\end{align*}[\/latex]
      \nAlternatively, we can apply the complement rule to find the probability:
      \n[latex]\\begin{align*} P(X \\geq 1) &= P(X=1 \\text{ or } X=2 \\text{ or } X=3) \\\\ &= 1 - P(X=0) \u00a0\\\\ &= 1 - 0.2 = 0.8. \\end{align*}[\/latex]<\/p>\n

      \n

      \"\"<\/p>\n<\/div>\n

      \n
      \n

      Exercise: Define Events Using Random Variable<\/p>\n<\/header>\n

      \n

      Consider the probability distribution of X = # of siblings below.<\/p>\n\n\n\n\n
      [latex]x[\/latex]<\/strong><\/td>\n0<\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n<\/tr>\n
      [latex]P(X=x)[\/latex]<\/strong><\/td>\n0.2<\/td>\n0.2<\/td>\n0.4<\/td>\n0.2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

      Define the following events using the variable X and find the probabilities:<\/p>\n

        \n
      1. Having one and a half siblings.<\/li>\n
      2. Having zero to two siblings exclusively.<\/li>\n
      3. Having zero to two siblings inclusively.<\/li>\n<\/ol>\n
        \nShow\/Hide Answer<\/summary>\n
          \n
        1. Event: [latex]\\{X=1.5 \\}[\/latex]. Since the possible values of [latex]X[\/latex] are [latex]x=0, 1, 2, 3[\/latex], it is impossible for [latex]X[\/latex] to be 1.5; therefore, [latex]P(X=1.5) =0[\/latex].<\/li>\n
        2. Event: [latex]\\{ 0 < X <2 \\}[\/latex]. Since 1 is the only possible value that is greater than 0 and smaller than 2, [latex]\\{ 0 < X < 2 \\} = \\{ X=1 \\}[\/latex]. This implies [latex]P(0 < X <2) = P(X=1) = 0.2[\/latex].<\/li>\n
        3. Event: [latex]\\{ 0 \\leq X \\leq 2 \\} = \\{ X=0 \\text{ or } X=1 \\text{ or } X=2 \\}[\/latex]. Since the events are mutually exclusive, i.e., they don’t overlap, the special addition rule gives:
          \n[latex]\\begin{align*} P(0 \\leq X \\leq 2) &= P(X=0 \\text{ or } X=1 \\text{ or } X=2) \\\\ &= P(X=0) + P(X=1) + P(X=2) \\\\ &= 0.2 + 0.2 + 0.4 \\\\ &= 0.8. \\end{align*}[\/latex]<\/li>\n<\/ol>\n<\/details>\n<\/div>\n<\/div>\n","protected":false},"author":19,"menu_order":3,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2105","chapter","type-chapter","status-publish","hentry"],"part":535,"_links":{"self":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/2105","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/users\/19"}],"version-history":[{"count":17,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/2105\/revisions"}],"predecessor-version":[{"id":5235,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/2105\/revisions\/5235"}],"part":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/parts\/535"}],"metadata":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/2105\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/media?parent=2105"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapter-type?post=2105"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/contributor?post=2105"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/license?post=2105"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}