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- https://stats.libretexts.org/Courses/Fresno_City_College/Book%3A_Business_Statistics_Customized_(OpenStax)/04%3A_Discrete_Random_Variables/4.01%3A_IntroductionTo use the combinatorial formula we would solve the formula as follows: \[\left(\begin{array}{l}{4} \\ {2}\end{array}\right)=\frac{4 !}{(4-2) ! 2 !}=\frac{4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1 \cdot 2 ...To use the combinatorial formula we would solve the formula as follows: \[\left(\begin{array}{l}{4} \\ {2}\end{array}\right)=\frac{4 !}{(4-2) ! 2 !}=\frac{4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1 \cdot 2 \cdot 1}=6\nonumber\] If we wanted to know the number of unique 5 card poker hands that could be created from a 52 card deck we simply compute: \[\left(\begin{array}{c}{52} \\ {5}\end{array}\right)\nonumber\] where 52 is the total number of unique elements from which we are drawing and 5 is the siz…
- https://stats.libretexts.org/Courses/Saint_Mary's_College_Notre_Dame/BFE_1201_Statistical_Methods_for_Finance_(Kuter)/04%3A_Random_Variables/4.01%3A_Introduction_to_Discrete_Random_VariablesAn alternative to listing the complete sample space and counting the number of elements we are interested in, is to skip the step of listing the sample space, and simply figuring out the number of ele...An alternative to listing the complete sample space and counting the number of elements we are interested in, is to skip the step of listing the sample space, and simply figuring out the number of elements in it and doing the appropriate division.
- https://stats.libretexts.org/Courses/Fresno_City_College/Introduction_to_Business_Statistics_-_OER_-_Spring_2023/04%3A_Discrete_Random_Variables/4.01%3A_Introduction_to_Discrete_Random_VariablesTo use the combinatorial formula we would solve the formula as follows: \[\left(\begin{array}{l}{4} \\ {2}\end{array}\right)=\frac{4 !}{(4-2) ! 2 !}=\frac{4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1 \cdot 2 ...To use the combinatorial formula we would solve the formula as follows: \[\left(\begin{array}{l}{4} \\ {2}\end{array}\right)=\frac{4 !}{(4-2) ! 2 !}=\frac{4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1 \cdot 2 \cdot 1}=6\nonumber\] If we wanted to know the number of unique 5 card poker hands that could be created from a 52 card deck we simply compute: \[\left(\begin{array}{c}{52} \\ {5}\end{array}\right)\nonumber\] where 52 is the total number of unique elements from which we are drawing and 5 is the siz…
