Question
Answer
$$(1-p)^2+2rp\cdot(1-p)+r^2p^2=p^2r^2-2p^2r+p^2+2pr-2p+1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1-p)^2+2rp\cdot(1-p)+r^2p^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}1-2p+p^2+2rp\cdot(1-p)+r^2p^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}1-2p+p^2+2pr-2p^2r+r^2p^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-2p^2r+p^2+2pr-2p+1+r^2p^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}p^2r^2-2p^2r+p^2+2pr-2p+1\end{aligned} $$ | |
| ① | Find $ \left(1-p\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 1 } $ and $ B = \color{red}{ p }$. $$ \begin{aligned}\left(1-p\right)^2 = \color{blue}{1^2} -2 \cdot 1 \cdot p + \color{red}{p^2} = 1-2p+p^2\end{aligned} $$ |
| ② | Multiply $ \color{blue}{2pr} $ by $ \left( 1-p\right) $ $$ \color{blue}{2pr} \cdot \left( 1-p\right) = 2pr-2p^2r $$ |
| ③ | Combine like terms: $$ 1-2p+p^2+2pr-2p^2r = -2p^2r+p^2+2pr-2p+1 $$ |
| ④ | Combine like terms: $$ p^2r^2-2p^2r+p^2+2pr-2p+1 = p^2r^2-2p^2r+p^2+2pr-2p+1 $$ |
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