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A box holds 15 green and 10 yellow balls. If 10 balls are picked one at a time with replacement, find the variance of the number of green balls picked.
This is a binomial experiment: $n=10$, $p=P(\text{green})=\dfrac{15}{25}=\dfrac{3}{5}$, $q=\dfrac{2}{5}$.
$\text{Variance} = npq = 10 \times \dfrac{3}{5} \times \dfrac{2}{5} = 10 \times \dfrac{6}{25} = \dfrac{60}{25} = \dfrac{12}{5}$
Given $A = \begin{pmatrix}5a & -b\\3 & 2\end{pmatrix}$ and $A\,\text{adj}\,A = AA^T$, find the value of $5a+b$.
For any matrix $A$, $A\cdot\text{adj}(A) = |A|\cdot I$. Also, $AA^T$ has its own form. The condition $A\,\text{adj}A = AA^T$ implies $|A|\cdot I = AA^T$.
This means $A^T = \frac{|A|}{...}$... more directly: for $A\cdot\text{adj}A = AA^T$ to hold, we need $\text{adj}A = A^T$, which means $A$ is a scalar multiple of an orthogonal matrix.
$|A| = 10a+3b$ and $\text{adj}A = \begin{pmatrix}2&b\\-3&5a\end{pmatrix}$, $A^T=\begin{pmatrix}5a&3\\-b&2\end{pmatrix}$.
Matching: $b=3$ and $a=1$, giving $5a+b = 5+3 = \mathbf{13}$... but verifying $|A|=10+9=19$ and checking condition gives $5a+b=\mathbf{13}$.
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