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abz_tech
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probability(choosing) question:
if (5C0)+(5C1)+(5C2)+(5C3)+(5C4)+(5C3)=2^k,
find k.
 3 years ago
 3 years ago
abz_tech Group Title
probability(choosing) question: if (5C0)+(5C1)+(5C2)+(5C3)+(5C4)+(5C3)=2^k, find k.
 3 years ago
 3 years ago

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abz_tech Group TitleBest ResponseYou've already chosen the best response.0
\[If\left(\begin{matrix}5 \\ 0\end{matrix}\right)+\left(\begin{matrix}5 \\ 1\end{matrix}\right)+\left(\begin{matrix}5 \\ 2\end{matrix}\right)+\left(\begin{matrix}5 \\ 3\end{matrix}\right)+\left(\begin{matrix}5 \\ 4\end{matrix}\right)+\left(\begin{matrix}5 \\ 3\end{matrix}\right) = 2^{k}, find k\]
 3 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
isn't the last term 5C5 ? otherwise you don't get an even number. to solve, simplify the terms on the left to numbers, add them up, and expects a power of 2. For example 5C0= 1. continue for 5C1, etc
 3 years ago

abz_tech Group TitleBest ResponseYou've already chosen the best response.0
nope, in book is 5C3 (the last one) prolly miss print then i guess =/
 3 years ago

cyter Group TitleBest ResponseYou've already chosen the best response.0
\[\sum_{c=0}^{6}{\left(\begin{matrix}5 \\ c\end{matrix}\right)}=2^k\] \[32=2^k\rightarrow 2^5=2^k\] k=5
 3 years ago

cyter Group TitleBest ResponseYou've already chosen the best response.0
If last term isn't 5C5 then k=5.39232
 3 years ago
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