A probability experiment is conducted in which the sample space of the experiment is \( \mathrm{S}=\{9,10,11,12,13,14,15,16,17,18,19,20\} \), event \( \mathrm{F}=\{11,12,13,14,15\} \), and \( \operatorname{svent} \mathrm{G}=\{15,16,17,16\} \) Assume that each outcome is equally likely List the outcomes in \( F \) or \( G \). Find \( P(F \) or \( G) \) by counting the number of outcomes in \( F \) or \( G \). Determine \( P(F \) or \( G) \) using the general addition ruie. List the outcomes in \( F \) or \( G \). Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( F \) or \( G=\{11,12,13,14,15,16,17,18\} \) (Use a comma to separate answers as needed ) B. For \( \mathrm{G}=\{ \} \) Find \( P(F \) or \( G) \) by counting the number of outcomes in \( F \) or \( G \). \[ P(F \text { or } G)=0.667 \] (Type an integer or a decimal rounded to three decimal places as needed.) Determine P(F or G) using the general addition rute. Select the correct choice below and fill in any answer boxes within your choice. (Type the terms of your expression in the same order as they appear in the original expression. Round to three decimal places as needed.) A. \( \mathrm{P}(\mathrm{F} \) or G\( )= \) \( \square \) \( + \) - \( \square \) \( - \) \( \square \) B. \( P(F \) or \( G)=\square+\square=\square \)
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The outcomes in \( F \) or \( G \) combine the elements of both sets. Since \( F = \{11, 12, 13, 14, 15\} \) and \( G = \{15, 16, 17, 16\} \) (note \( 16 \) is listed twice), the unique outcomes are \( \{11, 12, 13, 14, 15, 16, 17\} \). So, the correct choice is: A. \( F \) or \( G=\{11,12,13,14,15,16,17\} \) To find \( P(F \) or \( G) \) by counting, count the unique outcomes from \( F \) or \( G \). There are 7 outcomes in total, and the sample space \( S \) has 12 outcomes. Therefore, \[ P(F \text{ or } G) = \frac{7}{12} \approx 0.583\] For the general addition rule, we have: A. \( P(F \text{ or } G) = P(F) + P(G) - P(F \text{ and } G) \) Calculating: - \( P(F) = \frac{5}{12} \) (5 outcomes in \( F \)) - \( P(G) = \frac{3}{12} \) (3 unique outcomes in \( G \), counting 15 only once) - \( P(F \text{ and } G) = \frac{1}{12} \) (only 15 is in both) Putting it together: A. \( P(F \text{ or } G) = \frac{5}{12} + \frac{3}{12} - \frac{1}{12} = \frac{7}{12} \approx 0.583 \) So fill in: \( P(F) = \frac{5}{12} \), \( P(G) = \frac{3}{12} \), and \( P(F \text{ and } G) = \frac{1}{12} \).
