{"id":1757,"date":"2025-07-28T18:54:22","date_gmt":"2025-07-28T18:54:22","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1757"},"modified":"2026-03-26T20:52:49","modified_gmt":"2026-03-26T20:52:49","slug":"probability-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/probability-learn-it-3\/","title":{"raw":"Probability: Learn It 3","rendered":"Probability: Learn It 3"},"content":{"raw":"

Computing the Probability of the Union of Two Events<\/h2>\r\nWe are often interested in finding the probability that one of multiple events occurs. Suppose we are playing a card game, and we will win if the next card drawn is either a heart or a king. We would be interested in finding the probability of the next card being a heart or a king. The union of two events<\/strong> [latex]E\\text{ and }F,\\text{written }E\\cup F[\/latex], is the event that occurs if either or both events occur.\r\n

[latex]P\\left(E\\cup F\\right)=P\\left(E\\right)+P\\left(F\\right)-P\\left(E\\cap F\\right)[\/latex]<\/p>\r\n\r\n

\r\n

probability of the union of two events<\/h3>\r\nThe union of two events<\/strong> [latex]E\\text{ and }F,\\text{written }E\\cup F[\/latex], is the event that occurs if either or both events occur.\r\n\r\n \r\n\r\nThe probability of the union of two events [latex]E[\/latex] or<\/strong> [latex]F[\/latex] (written [latex]E\\cup F[\/latex] ) equals the sum of the probability of [latex]E[\/latex] and the probability of [latex]F[\/latex] minus the probability of [latex]E[\/latex] and [latex]F[\/latex] occurring together [latex]\\text{(}[\/latex] which is called the intersection<\/strong> of [latex]E[\/latex] and<\/strong> [latex]F[\/latex] and is written as [latex]E\\cap F[\/latex] ).\r\n\r\n \r\n

[latex]P(E \\text{ or }F) = P(E \\cup F)=P\\left(E\\right)+P\\left(F\\right)-P\\left(E\\cap F\\right)[\/latex]<\/p>\r\n\r\n<\/section>

\"ASuppose the spinner below is spun. We want to find the probability of spinning orange or spinning a [latex]b[\/latex].That is, find [latex]P(\\text{orange} \\cup b)[\/latex].\r\n
    \r\n \t
  • There are a total of [latex]6[\/latex] sections, and [latex]3[\/latex] of them are orange. So, the probability of spinning orange is [latex]P(\\text{orange}) = \\frac{3}{6}[\/latex].<\/li>\r\n \t
  • There are a total of [latex]6[\/latex] sections, and [latex]2[\/latex] of them have a [latex]b[\/latex]. So the probability of spinning a [latex]b[\/latex] is [latex]P(b) = \\frac{2}{6}[\/latex].<\/li>\r\n<\/ul>\r\nIf we added these two probabilities, we would be counting the sector that is both orange and a [latex]b[\/latex] twice. To find the probability of spinning an orange or a [latex]b[\/latex], we need to subtract the probability that the sector is both orange and has a [latex]b[\/latex].\r\n

    [latex]P(\\text{orange} \\cup b) = P(\\text{orange})+ P(b) - P(\\text{orange} \\cap b) = \\dfrac{3}{6}+\\dfrac{2}{6}-\\dfrac{1}{6}=\\dfrac{4}{6}[\/latex]<\/p>\r\nThe probability of spinning orange or a [latex]b[\/latex] is [latex]\\dfrac{4}{6}[\/latex] or [latex]\\dfrac{2}{3}[\/latex].\r\n\r\n<\/section>

    The\u00a0union symbol<\/em> given above [latex]\\cup[\/latex] is the same symbol you used in the past to express the union of two intervals. Here we use it to represent the union of two events. Mathematically, it represents the word\u00a0or.<\/em>\r\n

    The union of\u00a0[latex]E[\/latex]and\u00a0[latex]F[\/latex] includes all the elements that could be present in\u00a0[latex]E[\/latex] or in\u00a0<\/em>[latex]F[\/latex], one or the other.<\/em><\/p>\r\nThe\u00a0intersection symbol<\/em> given above [latex]\\cap[\/latex] may be new notation for you. It is used to express the intersection of events. Mathematically, it represents the word\u00a0and<\/em>.\r\n

    The intersection of\u00a0[latex]E[\/latex] and\u00a0[latex]F[\/latex] includes all the elements present in both <\/em>[latex]E[\/latex] and <\/em>[latex]F[\/latex], and not in just one or the other.<\/em><\/p>\r\nTry the example and practice problem below on paper to get familiar with the formula.\r\n\r\n<\/section>

    A card is drawn from a standard deck. Find the probability of drawing a heart or a [latex]7[\/latex].[reveal-answer q=\"870976\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"870976\"]A standard deck of [latex]52[\/latex] cards contains an equal number of hearts, diamonds, clubs, and spades.\r\n\"\"\r\n
      \r\n \t
    • The probability of drawing a heart is [latex]\\frac{13}{52} = \\frac{1}{4}[\/latex].<\/li>\r\n \t
    • There are four [latex]7[\/latex]s in a standard deck, and there are a total of [latex]52[\/latex] cards. So, the probability of drawing a [latex]7[\/latex] is [latex]\\frac{4}{52} = \\frac{1}{13}[\/latex].<\/li>\r\n \t
    • The only card in the deck that is both a heart and a [latex]7[\/latex]<\/span> is the <\/span>[latex]7[\/latex]<\/span> of hearts, so the probability of drawing both a heart and<\/strong> a [latex]7[\/latex] is [latex]\\frac{1}{52}[\/latex].<\/span><\/li>\r\n<\/ul>\r\nSubstitute [latex]P\\left(H\\right)=\\dfrac{13}{52}, P\\left(7\\right)=\\dfrac{4}{52}, \\text{and} P\\left(H\\cap 7\\right)=\\dfrac{1}{52}[\/latex] into the formula.\r\n

      [latex]\\begin{align}P\\left(H\\cup 7\\right)&=P\\left(H\\right)+P\\left(7\\right)-P\\left(H\\cap 7\\right) \\\\ &=\\dfrac{13}{52}+\\dfrac{4}{52}-\\dfrac{1}{52} \\\\ &=\\dfrac{16}{52} = \\dfrac{4}{13}\\end{align}[\/latex]<\/p>\r\nThe probability of drawing a heart or a [latex]7[\/latex] is [latex]\\dfrac{4}{13}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

      [ohm_question hide_question_numbers=1]322249[\/ohm_question]<\/section>","rendered":"

      Computing the Probability of the Union of Two Events<\/h2>\n

      We are often interested in finding the probability that one of multiple events occurs. Suppose we are playing a card game, and we will win if the next card drawn is either a heart or a king. We would be interested in finding the probability of the next card being a heart or a king. The union of two events<\/strong> [latex]E\\text{ and }F,\\text{written }E\\cup F[\/latex], is the event that occurs if either or both events occur.<\/p>\n

      [latex]P\\left(E\\cup F\\right)=P\\left(E\\right)+P\\left(F\\right)-P\\left(E\\cap F\\right)[\/latex]<\/p>\n

      \n

      probability of the union of two events<\/h3>\n

      The union of two events<\/strong> [latex]E\\text{ and }F,\\text{written }E\\cup F[\/latex], is the event that occurs if either or both events occur.<\/p>\n

       <\/p>\n

      The probability of the union of two events [latex]E[\/latex] or<\/strong> [latex]F[\/latex] (written [latex]E\\cup F[\/latex] ) equals the sum of the probability of [latex]E[\/latex] and the probability of [latex]F[\/latex] minus the probability of [latex]E[\/latex] and [latex]F[\/latex] occurring together [latex]\\text{(}[\/latex] which is called the intersection<\/strong> of [latex]E[\/latex] and<\/strong> [latex]F[\/latex] and is written as [latex]E\\cap F[\/latex] ).<\/p>\n

       <\/p>\n

      [latex]P(E \\text{ or }F) = P(E \\cup F)=P\\left(E\\right)+P\\left(F\\right)-P\\left(E\\cap F\\right)[\/latex]<\/p>\n<\/section>\n

      \"ASuppose the spinner below is spun. We want to find the probability of spinning orange or spinning a [latex]b[\/latex].That is, find [latex]P(\\text{orange} \\cup b)[\/latex].<\/p>\n
        \n
      • There are a total of [latex]6[\/latex] sections, and [latex]3[\/latex] of them are orange. So, the probability of spinning orange is [latex]P(\\text{orange}) = \\frac{3}{6}[\/latex].<\/li>\n
      • There are a total of [latex]6[\/latex] sections, and [latex]2[\/latex] of them have a [latex]b[\/latex]. So the probability of spinning a [latex]b[\/latex] is [latex]P(b) = \\frac{2}{6}[\/latex].<\/li>\n<\/ul>\n

        If we added these two probabilities, we would be counting the sector that is both orange and a [latex]b[\/latex] twice. To find the probability of spinning an orange or a [latex]b[\/latex], we need to subtract the probability that the sector is both orange and has a [latex]b[\/latex].<\/p>\n

        [latex]P(\\text{orange} \\cup b) = P(\\text{orange})+ P(b) - P(\\text{orange} \\cap b) = \\dfrac{3}{6}+\\dfrac{2}{6}-\\dfrac{1}{6}=\\dfrac{4}{6}[\/latex]<\/p>\n

        The probability of spinning orange or a [latex]b[\/latex] is [latex]\\dfrac{4}{6}[\/latex] or [latex]\\dfrac{2}{3}[\/latex].<\/p>\n<\/section>\n

        The\u00a0union symbol<\/em> given above [latex]\\cup[\/latex] is the same symbol you used in the past to express the union of two intervals. Here we use it to represent the union of two events. Mathematically, it represents the word\u00a0or.<\/em><\/p>\n

        The union of\u00a0[latex]E[\/latex]and\u00a0[latex]F[\/latex] includes all the elements that could be present in\u00a0[latex]E[\/latex] or in\u00a0<\/em>[latex]F[\/latex], one or the other.<\/em><\/p>\n

        The\u00a0intersection symbol<\/em> given above [latex]\\cap[\/latex] may be new notation for you. It is used to express the intersection of events. Mathematically, it represents the word\u00a0and<\/em>.<\/p>\n

        The intersection of\u00a0[latex]E[\/latex] and\u00a0[latex]F[\/latex] includes all the elements present in both <\/em>[latex]E[\/latex] and <\/em>[latex]F[\/latex], and not in just one or the other.<\/em><\/p>\n

        Try the example and practice problem below on paper to get familiar with the formula.<\/p>\n<\/section>\n

        A card is drawn from a standard deck. Find the probability of drawing a heart or a [latex]7[\/latex].<\/p>\n