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+# Noter til probability m2
+
+## Random variables
+
+Her mapper man fra et sample space S til en variabel.
+Her kalder man variablen et stort tal R eller sådan noget.
+
+Derfor er et random variabel egentlig en transformation mellem S og real tal.
+
+*X er en descrete random variable hvis dens range er countable.*
+
+For continues random variables the following is true:
+
+$$
+P(X = x) = 0
+$$
+
+## Functions beskriver ens random variable
+
+### Cumulative Distribution function
+
+Her måler man prob for at ens random er mindre end et bestemt tal.
+
+$$
+F(x) = P(X \leq x)
+$$
+
+Man kan også finde det for en range:
+
+$$
+P(a < X \leq b) = F(b) - F(a)
+$$
+
+Ved discrete random variables vil denne være en slags trappe.
+
+Kan sige at den er *continues from the right* eftersom man har $\leq$ i definition.
+
+### Probability Mass Function
+
+Works only for discrete random variables.
+Is defines as the probability that $X = a$:
+
+$$
+p(a) = P(X = a)
+$$
+
+From here CDF can be found:
+
+$$
+    F(a) = \sum_{all x \leq a} p(a)
+$$
+
+
+
+### Probability Density Function
+
+Her finder man P i et evigt lille interval:
+Is the derivative of the CDF.
+
+$$
+    F(a) = P(X \in (-\infty,a]) = \int_{-\infty}^a f(x) dx \\
+    f(a) = \frac{d}{da} F(a)
+$$
+
+The following must be true:
+
+$$
+    \int_{-\infty}^{\infty} f(x) dx = 1
+$$
+
+## Multiple random variables
+
+Have multiple random variables, which can be or is not correlated.
+Can define the joined CDF:
+
+$$
+    F_{XY}(x,y) = P(X \leq x, Y \leq y)
+$$
+
+One can also find the probability of one of the variables. (The *marginal*)
+
+$$
+    F_X(x) = P(X \leq x) = P(X \leq, Y < \intfy) = F(x, \infty)
+$$
+
+One can not go from marginal to the joined, as they do not contain enough information.
+This is only possible if X and Y are **independent**.
+
+$$
+    F_XY(x,y) = F_X(x) \cdot F_Y(x) \\
+    p(x,y) = p_X(x) \cdot p_Y(y) \\
+    f(x,y) = f_X(x) \cdot f_Y(y)
+$$
+
+### Joined PMF
+
+$$
+    P_{XY}(x,y) = P(X = x, Y = y)
+$$
+
+    vim: spell spelllang=da,en
+
diff --git a/sem6/prob/m2/opgaver.md b/sem6/prob/m2/opgaver.md
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+# Opgaver til prob m2
+
+## Opgave 1
+
+Man kan sige at chances for at en kvinde kommer først er 50%.
+
+$$
+P(1) = \frac 1 2
+$$
+
+Herefter kræver det at en mand for først og en kvinde får næste.
+
+$$
+P(2) = \frac 5 10 \cdot \frac 5 9 \\
+P(3) = \frac 5 10 \cdot \frac 4 9 \frac 5 8 \\
+P(4) = \frac 5 10 \cdot \frac 4 9 \frac 3 8 \cdot \frac 5 8
+$$
+
+osv.
+
+## Opgave 2
+
+### A)
+
+Dette har jeg gjort på papir.
+
+### B)
+
+$$
+P\left(X > \frac 1 2\right) = 1 - F\left(\frac 1 2\right) = 1 - \frac 1 4 = \frac 3 4
+$$
+
+### C)
+
+$$
+P(2 < X \leq 4) = F(4) - F(2) = 1 - \frac {11} {12} = \frac 1 {12}
+$$
+
+### D)
+
+$$
+P(X < 3) = \frac {11}{12}
+$$
+
+### E)
+
+$$
+P(X = 1) = \frac 2 3 - \frac 1 2 = \frac 1 6
+$$
+
+## Opgave 3
+
+