{"id":1758,"date":"2025-07-28T18:53:55","date_gmt":"2025-07-28T18:53:55","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1758"},"modified":"2026-03-26T19:35:24","modified_gmt":"2026-03-26T19:35:24","slug":"probability-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/probability-learn-it-2\/","title":{"raw":"Probability: Learn It 2","rendered":"Probability: Learn It 2"},"content":{"raw":"

Computing Probabilities of Equally Likely Outcomes<\/h2>\r\nLet [latex]S[\/latex] be a sample space for an experiment. When investigating probability, an event is any subset of [latex]S[\/latex]. When the outcomes of an experiment are all equally likely, we can find the probability of an event by dividing the number of outcomes in the event by the total number of outcomes in [latex]S[\/latex].\r\n\r\n
\r\n

probability of an event with equally likely outcomes<\/h3>\r\nThe probability of an event [latex]E[\/latex] in an experiment with sample space [latex]S[\/latex] with equally likely outcomes is given by\r\n\r\n \r\n

[latex]P(E)=\\dfrac{\\text{number of elements in }E }{\\text{number of elements in }S } = \\dfrac{n(E)}{n(S)}[\/latex]<\/p>\r\n \r\n\r\n[latex]E[\/latex] is a subset of [latex]S[\/latex], so it is always true that [latex]0\\le P\\left(E\\right)\\le 1[\/latex].\r\n\r\n<\/section>

A number cube (a fair six-sided die) is rolled.\r\n
    \r\n \t
  1. Find the probability of rolling an odd number.[reveal-answer q=\"320469\"]Show Solution[\/reveal-answer][hidden-answer a=\"320469\"]The event \"rolling an odd number\" contains three outcomes. [latex]E = \\{1,3,5\\}[\/latex]. There are 6 equally likely outcomes in the sample space, [latex]S = \\{1,2,3,4,5,6\\}[\/latex]. Divide to find the probability of the event.\r\n[latex]P\\left(E\\right)=\\dfrac{3}{6}=\\dfrac{1}{2}[\/latex][\/hidden-answer]<\/li>\r\n \t
  2. Find the probability of the event \"rolling a number less than or equal to [latex]4[\/latex].\"\r\n[reveal-answer q=\"728391\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"728391\"]There are [latex]4[\/latex] possible outcomes [latex]\\{1,2,3,4\\}[\/latex] in the event and [latex]6[\/latex] possible outcomes in [latex]S[\/latex], so the probability of the event is [latex]\\frac{4}{6}=\\frac{2}{3}[\/latex].[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/section>
    [ohm_question hide_question_numbers=1]322243[\/ohm_question]<\/section>
    [ohm_question hide_question_numbers=1]322244[\/ohm_question]<\/section>","rendered":"

    Computing Probabilities of Equally Likely Outcomes<\/h2>\n

    Let [latex]S[\/latex] be a sample space for an experiment. When investigating probability, an event is any subset of [latex]S[\/latex]. When the outcomes of an experiment are all equally likely, we can find the probability of an event by dividing the number of outcomes in the event by the total number of outcomes in [latex]S[\/latex].<\/p>\n

    \n

    probability of an event with equally likely outcomes<\/h3>\n

    The probability of an event [latex]E[\/latex] in an experiment with sample space [latex]S[\/latex] with equally likely outcomes is given by<\/p>\n

     <\/p>\n

    [latex]P(E)=\\dfrac{\\text{number of elements in }E }{\\text{number of elements in }S } = \\dfrac{n(E)}{n(S)}[\/latex]<\/p>\n

     <\/p>\n

    [latex]E[\/latex] is a subset of [latex]S[\/latex], so it is always true that [latex]0\\le P\\left(E\\right)\\le 1[\/latex].<\/p>\n<\/section>\n

    A number cube (a fair six-sided die) is rolled.<\/p>\n
      \n
    1. Find the probability of rolling an odd number.\n