We can generalize the idea of combinations: the number of arrangements of n units into r groups of sizes \(n_1,n_2,n_3...n_r\) is \[ n=n_1+n_2+n_3...+n_r\\ \begin{pmatrix}n\\n_1,n_2..n_r\end{pmatrix}={n!\over{n_1!n_2!n_3!..n_r!}} \]
Example, ways to build group of 2,3,3 from 8 students:
\[ \begin{pmatrix}8\\2,3,3\end{pmatrix}={8!\over{2!3!3!}}=560 \]
Example, a communication system consist of 13 antennas work as long as no two of them arrange next to each other. suppose 5 of them stop functioning.
8 functioning, 5 non functioning 9 possible locations for non functioning antennas
\[ \begin{pmatrix}9\\5\end{pmatrix}={9!\over{5!4!}}=126 \]
(b)probability for the system still functional
total arrangement for the antenna \[ \begin{pmatrix}13\\5\end{pmatrix}={13!\over{5!8!}}=1287\\ P(funcitonal)=126/1287 \]
The probability mass function of a discrete random variable X is a list of the probabilities p(x) for each value x in the sample pace \(S_x\) of X.
Rolling two dies. \[ P(x=2)={1\over{36}}...P(x=12)={1\over{36}} \]
Example: random sample of size n =3 from 10 sample, where 4 of them are defective.
p(x)|
for x=0, we select 3 non-defective from 6 non-defective sample, divided by the general chance of selecting 3 item from 10.
\[ P(x=0)={\begin{pmatrix}6\\3\end{pmatrix}\over{\begin{pmatrix}10\\3\end{pmatrix}}}=0.167 \]
for x=1, we select 2 non-defective from 6 non-defective sample, 1 defective from 4 defective, then divided by the general chance of selecting 3 item from 10
\[ P(x=1)={\begin{pmatrix}6\\2\end{pmatrix}\begin{pmatrix}4\\1\end{pmatrix}\over{\begin{pmatrix}10\\3\end{pmatrix}}}=0.5 \]
for x=2,… \[ P(x=2)={\begin{pmatrix}6\\1\end{pmatrix}\begin{pmatrix}4\\2\end{pmatrix}\over{\begin{pmatrix}10\\3\end{pmatrix}}}=0.3 \]
for x=3,… \[ P(x=3)={\begin{pmatrix}4\\3\end{pmatrix}\over{\begin{pmatrix}10\\3\end{pmatrix}}}=0.033 \]
\[ P(x=0)+P(x=1)+P(x=2)+P(x=3)=1 \]
Simulation of experiment in R
# replace = TRUE means we take sample with replacement (no put back operation)
x = sample(0:3, size = 10000, replace = TRUE, prob = c(0.167,0.5,0.3,0.033))
table(x)/10000## x
## 0 1 2 3
## 0.1648 0.5012 0.3015 0.0325Conditional probability: the relative frequency we can expect an event to occur under the condition that additional, preexisting information is known about some other event. Denoted by \(P(A|B)\) (probability of A given B)
During the process, Event B becomes the new sample space.
\[ P(A|B)=P(A\cap B)/P(B) \]