{"id":2121,"date":"2015-11-12T18:30:41","date_gmt":"2015-11-12T18:30:41","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=2121"},"modified":"2015-11-12T18:30:41","modified_gmt":"2015-11-12T18:30:41","slug":"computing-the-probability-of-the-union-of-two-events","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/computing-the-probability-of-the-union-of-two-events\/","title":{"raw":"Computing the Probability of the Union of Two Events","rendered":"Computing the Probability of the Union of Two Events"},"content":{"raw":"

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.\n<\/p>

[latex]P\\left(E\\cup F\\right)=P\\left(E\\right)+P\\left(F\\right)-P\\left(E\\cap F\\right)[\/latex]<\/div>\nSuppose the spinner in Figure 2\u00a0is spun. We want to find the probability of spinning orange or spinning a [latex]b[\/latex].\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"AFigure 2<\/b>[\/caption]\n\nThere are a total of 6 sections, and 3 of them are orange. So the probability of spinning orange is [latex]\\frac{3}{6}=\\frac{1}{2}[\/latex]. There are a total of 6 sections, and 2 of them have a [latex]b[\/latex]. So the probability of spinning a [latex]b[\/latex] is [latex]\\frac{2}{6}=\\frac{1}{3}[\/latex]. 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].\n
[latex]\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{6}=\\frac{2}{3}[\/latex]<\/div>\nThe probability of spinning orange or a [latex]b[\/latex] is [latex]\\frac{2}{3}[\/latex].\n
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A General Note: Probability of the Union of Two Events<\/h3>\nThe probability of the union of two events [latex]E[\/latex] and [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 [latex]F[\/latex] and is written as [latex]E\\cap F[\/latex] ).\n
[latex]P\\left(E\\cup F\\right)=P\\left(E\\right)+P\\left(F\\right)-P\\left(E\\cap F\\right)[\/latex]<\/div>\n<\/div>\n
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Example 3: Computing the Probability of the Union of Two Events<\/h3>\nA card is drawn from a standard deck. Find the probability of drawing a heart or a 7.\n\n<\/div>\n
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Solution<\/h3>\nA standard deck contains an equal number of hearts, diamonds, clubs, and spades. So the probability of drawing a heart is [latex]\\frac{1}{4}[\/latex]. There are four 7s in a standard deck, and there are a total of 52 cards. So the probability of drawing a 7 is [latex]\\frac{1}{13}[\/latex].\n\nThe only card in the deck that is both a heart and a 7 is the 7 of hearts, so the probability of drawing both a heart and a 7 is [latex]\\frac{1}{52}[\/latex]. Substitute [latex]P\\left(H\\right)=\\frac{1}{4}, P\\left(7\\right)=\\frac{1}{13}, \\text{and} P\\left(H\\cap 7\\right)=\\frac{1}{52}[\/latex] into the formula.\n
[latex]\\begin{array}{l}P\\left(E{\\cup }^{\\text{ }}F\\right)=P\\left(E\\right)+P\\left(F\\right)-P\\left(E{\\cap }^{\\text{ }}F\\right)\\hfill \\\\ \\text{ }=\\frac{1}{4}+\\frac{1}{13}-\\frac{1}{52}\\hfill \\\\ \\text{ }=\\frac{4}{13}\\hfill \\end{array}[\/latex]<\/div>\nThe probability of drawing a heart or a 7 is [latex]\\frac{4}{13}[\/latex].\n\n<\/div>\n
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Try It 3<\/h3>\nA card is drawn from a standard deck. Find the probability of drawing a red card or an ace.\n\nSolution<\/a>\n\n<\/div>","rendered":"

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.\n<\/p>\n

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

Suppose the spinner in Figure 2\u00a0is spun. We want to find the probability of spinning orange or spinning a [latex]b[\/latex].<\/p>\n

\"A<\/p>\n

Figure 2<\/b><\/p>\n<\/div>\n

There are a total of 6 sections, and 3 of them are orange. So the probability of spinning orange is [latex]\\frac{3}{6}=\\frac{1}{2}[\/latex]. There are a total of 6 sections, and 2 of them have a [latex]b[\/latex]. So the probability of spinning a [latex]b[\/latex] is [latex]\\frac{2}{6}=\\frac{1}{3}[\/latex]. 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]\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{6}=\\frac{2}{3}[\/latex]<\/div>\n

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

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A General Note: Probability of the Union of Two Events<\/h3>\n

The probability of the union of two events [latex]E[\/latex] and [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 [latex]F[\/latex] and is written as [latex]E\\cap F[\/latex] ).<\/p>\n

[latex]P\\left(E\\cup F\\right)=P\\left(E\\right)+P\\left(F\\right)-P\\left(E\\cap F\\right)[\/latex]<\/div>\n<\/div>\n
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Example 3: Computing the Probability of the Union of Two Events<\/h3>\n

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

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Solution<\/h3>\n

A standard deck contains an equal number of hearts, diamonds, clubs, and spades. So the probability of drawing a heart is [latex]\\frac{1}{4}[\/latex]. There are four 7s in a standard deck, and there are a total of 52 cards. So the probability of drawing a 7 is [latex]\\frac{1}{13}[\/latex].<\/p>\n

The only card in the deck that is both a heart and a 7 is the 7 of hearts, so the probability of drawing both a heart and a 7 is [latex]\\frac{1}{52}[\/latex]. Substitute [latex]P\\left(H\\right)=\\frac{1}{4}, P\\left(7\\right)=\\frac{1}{13}, \\text{and} P\\left(H\\cap 7\\right)=\\frac{1}{52}[\/latex] into the formula.<\/p>\n

[latex]\\begin{array}{l}P\\left(E{\\cup }^{\\text{ }}F\\right)=P\\left(E\\right)+P\\left(F\\right)-P\\left(E{\\cap }^{\\text{ }}F\\right)\\hfill \\\\ \\text{ }=\\frac{1}{4}+\\frac{1}{13}-\\frac{1}{52}\\hfill \\\\ \\text{ }=\\frac{4}{13}\\hfill \\end{array}[\/latex]<\/div>\n

The probability of drawing a heart or a 7 is [latex]\\frac{4}{13}[\/latex].<\/p>\n<\/div>\n

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Try It 3<\/h3>\n

A card is drawn from a standard deck. Find the probability of drawing a red card or an ace.<\/p>\n

Solution<\/a><\/p>\n<\/div>\n\n\t\t\t

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Candela Citations<\/h3>\n\t\t\t\t\t
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