# Probability Distribution

• Apr 26th 2006, 03:37 PM
Arbitur
Probability Distribution
I have no idea if this is right....
for a i get .021%
for b i get 3
for c i get 24.38%
• Apr 26th 2006, 06:28 PM
ThePerfectHacker
Let me remind the binomial probability formula, an event $m$ out of $n$ times exactly with probability $p$ is,
$\left( \begin{array}{c} n\\m \end{array} \right) p^m (1-p)^{n-m}$.

Over here
$\left\{ \begin{array}{c} n=20\\m=3\\p=.15 \end{array}$
Thus,
$\left( \begin{array}{c} 20\\3 \end{array} \right) (.15)^3(.85)^{17}\approx .2428$
• Apr 26th 2006, 06:32 PM
ThePerfectHacker
For the second you need to find $m$ such as,
$\left( \begin{array}{c} 20\\m \end{array} \right) (.15)^m(.85)^{20-m}$ is maximized.
Basic trail and error, you get,
$m=3$
• Apr 26th 2006, 06:44 PM
ThePerfectHacker
For the third to first.
You need to find, (m=3,4,5,....,20) because "at least"
$\left( \begin{array}{c} 20\\3 \end{array} \right)(.15)^3(.85)^{17}+\left( \begin{array}{c} 20\\4 \end{array} \right)(.15)^4(.85)^{16}+$... $+\left( \begin{array}{c} 20\\20 \end{array} \right)(.15)^{20}(.85)^0$
You can add them up but that is the n00bish way of doing this problem. It is easier to find,
$\left( \begin{array}{c}20 \\0\end{array} \right)(.15)^0(.85)^{20}+\left( \begin{array}{c}20 \\1 \end{array} \right)(.15)^1(.85)^{19}+$ $\left( \begin{array}{c} 20\\2 \end{array} \right) (.15)^2 (.85)^{18}\approx .4049$
Now from one subtract that to get,
$\approx .5951$