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Protokoll vom 2.3.2016, Themen: Difference between conditional propability and intersection

Protokoll von --Tws98OB (Diskussion) 01:31, 6. Mär. 2016 (CET) (Schuljahr 2016)
Lehrer C.-J. Schmitt (5 Unterrichtsstunden)
verbessert von --Tws98OB (Diskussion) 20:31, 10. Mär. 2016 (CET)

Exercise sheet 4

Task #1

Initial Situation

We have a box with 100 balls, 70 of which are wooden while the other 30 are made of plastic.

25 of the wooden balls are red and 45 green.

10 of the plastic balls are red and 20 green.

Ü4 #1

This is a 2x2 Table. It represents the intersection and amount of different events. For example: the first column and first row show how many of the balls in the pit are wooden and red. It does not give you the chance of getting one but rather the amount of this kind of ball. The second row shows the amount of balls that are wooden and green while the outer row shows the total amount of wooden balls.

I am feeling a plastic ball. What is the chance of it being green?

Kugel - Baum 2

\overline{H}:Plastic ball

\overline{R}:Green colour

Bayes' Theorem gives us the ability to calculate the chance of an outcome which is depending on a previous event.

P(\overline{H} \cap \overline{R})=P(\overline{H}) \cdot P_{\overline{H}}(\overline{R})

Now we have to adjust this term to find the chance of a plastic ball, being green.

P_{\overline{H}}(\overline{R})=\frac{P(\overline{H} \cap \overline{R})}{P(\overline{H})}=\frac{0,2}{0,3}

=\frac{2}{3} for a green coloured plastic ball

The outcome of R (the colour) depends on what we drew first (H). In this case, we drew a plastic ball and want to know what the chance of it being a green and wooden ball is.

The chance of it being red?

P_{\overline{H}}(R)=\frac{P(\overline{H} \cap R)}{P(\overline{H})}=\frac{0,1}{0,3}

=\frac{1}{3} for a red coloured plastic ball

I am feeling a wooden ball. What is the chance of it being red?

P_{H}(R)=\frac{P(H \cap R)}{P(H)}=\frac{0,25}{0,7}



P_{H}(R)\neq P(R) => Conditional

To find out if one outcome depends on another one you need to calculate in independent chance of the event and the dependent one and check if they are equal or not.

If they do not math then the event is conditional.

The chance of it being green?

P(H \cap R)=0,25

P_{H}(\overline{R})=\frac{P(H \cap \overline{R})}{P(H)}=\frac{0,45}{0,7}


I drew a red ball. What is the chance of it being made out of wood?

This works by the same priciple. The only difference is that the first event is now the colour and not the material.

Kugel - Baum

P(H \cap R)=0,25

P(H \cap R)=P(R) \cdot P_{R}(H)

P_{R}(H)=\frac{P(R \cap H)}{P(R)}=\frac{0,25}{0,35}



P(H) \neq P_{R}(H) => Conditional

The chance of it being made out of plastic?

P_{R}(\overline{H})=\frac{P(R \cap \overline{H})}{P(R)}=\frac{0,1}{0,35}


I drew a green ball. What is the chance of it being made out of wood?

P_{\overline{R}}(H)=\frac{P(\overline{R} \cap H)}{P(\overline{R})}=\frac{0,45}{0,65}



P(H) \neq P_{\overline{R}}(H) => Conditional

The chance of it being made out of plastic?

P_{\overline{R}}(\overline{H})=\frac{P(\overline{R} \cap \overline{H})}{P(\overline{R})}=\frac{0,2}{0,65}


Task #2

This 2x2 table shows the distribution of mobile phones in a business college.

Ü4 #2



H:Mobile phone

\overline{H}:No mobile phone

If you meet a woman, what is the probability of her having a phone?

Ü4 #2

We are looking for P_{W}(H)

Intersection between "female" and "having a phone": P(W \cap H)=0,41

P(W \cap H)=P(W) \cdot P_W(H)

P_W(H)=\frac{P(W \cap H)}{P(W)}=\frac{0,41}{0,524}



P(H) \neq P_W(H)

The property to be female and to have a phone depend on each other. This means that males have a different percentage to own a phone than females.

And the probability of a male having a phone?

P_{\overline{W}}(H)=\frac{P(\overline{W} \cap H)}{P(\overline{W})}=\frac{0,397}{0,476}


If you see someone with a phone, what is the percentage of the person being female or male?

Ü4 #2


P_H(W)=\frac{P(W \cap H)}{P(H)}=\frac{0,41}{0,807}



P(W) \neq P_H(W) => Conditional


P_H(\overline{W})=\frac{P(\overline{W} \cap H)}{P(\overline{H})}=\frac{0,397}{0,807}


What if the person does not have a phone?

P_{\overline{H}}(W)=\frac{P(W \cap \overline{H})}{P(\overline{H})}=\frac{0,114}{0,193}

=0,5907=59,07% The percentage for the person being female.

P_{\overline{H}}(\overline{W})=40,93% The percentage for the person being male.

Task #3

40% of people, who alre older than 15 smoke. 0,5% of all people have lung cancer and 80% of the people who have lung cancer smoke.

Ü2 #3


\overline{R}:Non-smoker (=60%)

L:Lung cancer patient (=0,5%)

\overline{L}:Healthy (=99,5%)

This task requires precise thinking to distinguish between conditional propability and the intersection between two events.


"80% of people who have lung cancer smoke" is conditional because it says, that out of the 0,5% of people who are sick, 80% smoke. This means that you can not use the 80% in the 2x2 Table for P(L \cap R). You can use Bayes' theorem to get the union of smoking and having lung cancer.

Ü2 #3

P(L \cap R)=P(L) \cdot P_{L}(R)

=0,005 \cdot 0,8

=0,004=0,4% This can be used to fill in the rest of the 2x2 table.

What is the chance of a smoker to get lung cancer?

Ü2 #3

P(L \cap R)=0,004

P_{R}(L)=\frac{P(R \cap L}{P(R)}=\frac{0,004}{0,4}



P(L) \neq P_R(L) => Conditional

The chance of a smoker to get lung cancer is 1%.

What is the chance of a non-smoker to get lung cancer?

P_{\overline{R}}(L)=\frac{P(\overline{R} \cap L}{P(\overline{R})}=\frac{0,001}{0,6}


The chance of a non-smoker to get lung cancer is \frac{1}{6}%.

If you compare the chances of non-smokers and smokers to get lung cancer, you can see that a smoker is six times more likely to get sick than a non-smoker.

P_R(L)=6 \cdot P_{\overline{R}}(L)


This task can be solved in a different way, using a graphic representation.

The chance of a lung cancer patient to be a smoker

Lung cancer and smoking #1

P(L \cap R)= \frac{|L \cap R|}{|S|}=(P_L(R) \cdot \frac{|L|}{|S|})=P_L(R) \cdot P(L)

The percentage of the interesection of L and R can be calculated by dividing the amount of R and L by the total amount S.

\frac{|L \cap R|}{|L|}=P_L(R) => We can use this to find the percentage

=\frac{0,004}{0,005}=0,8=80% => 80% of lung cancer patients smoke

The chance of a non-smoker to get lung cancer

Lung cancer and smoking #2

P(\overline{R} \cap L) = \frac{|L \cap \overline{R}|}{|S|} = \frac{|L| - |R \cap L|}{|S|}

The intersection between lung cancer and non-smoker is the same as all the people that are sick minus the people that smoke and have cancer. The pink parking in this picture are the non-smokers while the orange lines represent the people that don't smoke and have cancer.

=\frac{|L|}{|S|} - \frac{|R \cap L|}{|S|}= P(L) - P(R \cap L)

The chance of someone who does not have lung cancer to not be smoking

P_{\overline{L}}(\overline{R})=\frac{P(\overline{L} \cap \overline{R})}{\overline{L}}



Recap of the data

P_L(R)=80% chance to be a smoker if you have lung cancer.

P_L(\overline{R})=20%chance to not smoke when you have lung cancer.

P_R(L)=1%chance to have lung cancer when you are a smoker.

P_R(\overline{L})=99%chance to not have lung cancer when you smoke.

P_{\overline{R}}(L)=\frac{1}{6}% => P_R(L)=6 \cdot P_{\overline{R}}(L) Having lung cancer as a smoker is six times more likely than having it as a non smoker.

P_{\overline{L}}(\overline{R})=39,79%chance to not be a smoker if you don't have lung cancer.

P_{\overline{L}}(R)=60,21%chance to be a smoker if you don't have lung cancer.

Wheel of fortune


We have a wheel of fortune with 2 panels.

The red field has the angle \alpha



E: The same colour twice

P(E)=p^2 + q^2

The chance to get red is p so the possibility to get red twice is p sqared. The same goes for white.

=p^2 + (1-p)^2

=p^2 + 1-2p + p^2


Now we need to find p if the bet is 5€ and the win if you get the same colour twice is 8€.

P(E) \cdot 8 =5




\frac{1}{2} ^+_-\sqrt{(-\frac{1}{2})^2-\frac{3}{16}}

p_1=0,75 -> \alpha_{r_1}=270^o

p_2=0,25 -> \alpha_{r_2}=90^o



Both of these angles make this a fair game.

Protokoll vom 9.3. und 11.3.2015, Themen: Additionssatz und Kombinatorik

Protokoll von --Nico98OB (Diskussion) 15:24, 12. Mär. 2016 (CET)-- (Schuljahr 2016)
Lehrer C.-J. Schmitt (5 Unterrichtsstunden)
verbessert von--Nico98OB (Diskussion) 14:03, 19. Mär. 2016 (CET)

Der Additionssatz

Bei einem ganz normalen Würfel gilt

Ereignis 1 sei {6} mit der Wahrscheinlichkeit P({ E }_{ 1 })=\frac { 1 }{ 6 }
Ereignis 2 sei {1,3,5} mit der Wahrscheinlichkeit P({ E }_{ 2 })=\frac { 3 }{ 6 }
{ E }_{ 1 }\cap { E }_{ 2 }={}
Mengen die keine gemeinsamen Elemente haben nennt man disjunkt (dass heißt es gibt kein Element , was in beiden Mengen ist.)

{ E }_{ 1 }\cup { E }_{ 2 }=\left\{1,3,5,6\right\}
P({ E }_{ 1 }\cup { E }_{ 2 })=\frac { 4 }{ 6 } =P({ E }_{ 1 })+P({ E }_{ 2 })

{ E }_{ 1 }=\left\{ 1,2 \right\}
{ E }_{ 2 }=\left\{ 5,6 \right\}
{ E }_{ 1 }\cap { E }_{ 2 }=\left\{  \right\}
{ E }_{ 1 }\cup { E }_{ 2 }=\frac { 4 }{ 6 }=P({ E }_{ 1 })+P({ E }_{ 2 })

{ E }_{ 1 }=\left\{ 6 \right\}
{ E }_{ 2 }=\left\{ 2,4,6 \right\}
{ E }_{ 1 }\cap { E }_{ 2 }=\left\{ 6 \right\}
Mengen nennt man nicht disjunkt, dass heist es gibt ein oder mehrere Elemente ,welche in beiden Mengen sind. { E }_{ 1 }\cup { E }_{ 2 }=\left\{ 4,2,6\right\}
P({ E }_{ 1 }\cup { E }_{ 2 })=\frac { 3 }{ 6 }
P({ E }_{ 1 }\cup { E }_{ 2 })=P({ E }_{ 1 })+P({ E }_{ 2 })-P({ E }_{ 1 }\cap { E }_{ 2 })=\frac { 1 }{ 6 } +\frac { 3 }{ 6 } -\frac { 1 }{ 6 }

P({ E }_{ 1 }\cup { E }_{ 2 })=P({ E }_{ 1 })+P({ E }_{ 2 })-P({ E }_{ 1 }\cap { E }_{ 2 })

Ein Beispiel:


\left| { E }_{ 1 } \right| =4
    \left| { E }_{ 2 } \right| =8

\left| { E }_{ 1 }{ \cap E }_{ 2 } \right| =1\quad (KaroBube)
P({ E }_{ 1 }\cup { E }_{ 2 })=\frac { 4 }{ 32 } +\frac { 8 }{ 32 } -\frac { 1 }{ 32 } =\frac { 11 }{ 32 }

Bei disjunkten Mengen ist die Erechnung der Vereinigungsmenge Einfacher, da die Schnittmenge eine leere Menge ist, da sich die Mengen ja nicht schneiden also: P({ E }_{ 1 }\cup { E }_{ 2 })=P({ E }_{ 1 })+P({ E }_{ 2 })
So lässt sich auch die Erechnung der Wahrscheinlichkeit, des Gegenereignises, durch Abzug der Wahrscheinlichkeit, des Ereignis von 1 beweisen.

{ E }_{ 1 }\cup { E }_{ 2 }=S

{ E }_{ 1 }\cap { E }_{ 2 }=\left\{  \right\}
P({ E }_{ 1 }\cup { E }_{ 2 })=P(E)+P(\overline { E } )-P(E\cap \overline { E } )
1=P(E)+P(\overline { E } )
1-P(E)=P(\overline { E } )


Um in die Kombinatorik einzusteigen beschäftigen wir uns mit einem Problem des Fürsten von Toskana. In einem Würfelspiel in dem man 3mal würfelt käme die Augensumme 10 nach Beobachtung öfter vor, als die Augensumme 9 obwohl es jeweils 6 Wurfmöglichkeiten gibt.
Sollte eine Kombination 3 verschiedene Elemente haben so lässt sie sich 6 mal verschieden darstellen (Ein Beispiel mit (1+2+6):{(1,2,6),(2,1,6),(2,6,1),(1,6,2),(6,1,2),(6,2,1)} Wenn es 2 verschiedene Elemente gibt lassen sich 3 verschiedene Kombinationen erstellen und wenn alle Elemente gleich sind, logischer Weise nur eine.

Wir unterteilen nun beide Augensummen.

Tabelle 9999.PNG
(von 3+3+3 gibt es nur eine Variation da keine Zahlen vertauscht werden können ohne das wieder 3+3+3 entsteht.)


Unbenannt dfdfhh.PNG
Die ANzahl aller Möglichkeiten beträgt S=216=6^3 da wir 6 Augen haben und 3 mal gewürfelt wird. Also 6\cdot 6\cdot 6

Nun sagen wir Augensumme 9 ={ E }_{ 1 }
und Augensumme 10 ={ E }_{ 2 }
\left| { E }_{ 1 } \right| =6 \cdot 6-3-3-5=25
\left| { E }_{ 2 } \right| =6 \cdot 6-3-3-3=27
P(E_1)=\frac { 25 }{ 216 } =11,5%

P(E_2)=\frac {27}{216}=12,5%
Dieses Problem wurde übrigens von Galileo Galilei (1564-1624) gelöst.

Galileo war ein Wissenschaftler der in der Astronomie ,in der Mathematik und in der Physik tätig war.

Durch seine Theorie die Erde drehe sich um die Sonne wurde er vor die Heilige Inquisition gerufen und sollte

wegen seinen ketzerischen Ideen verbrannt werden wenn er nicht davon abschwören würde.
Jetzt noch ein paar Zitate aus der Wissenschaft: Alle Wahrheiten sind leicht verständlich von dem Zeitpunkt an, wo sie aufgedeckt werden. Die Frage ist, ob sie aufgedeckt werden.
Im großen Buch der Natur kann nur der lesen, der die Sprache kennt, in welcher dieses Buch geschrieben ist, und diese Sprache ist die Mathematik.
Ich glaube nicht, daß derselbe Gott, der uns Sinne, Vernunft und Verstand gab, uns ihren Gebrauch verbieten wollte.
Und sie bewegt sich doch ( zugesprochenen Worte Galileis nach der Verhandlung mit der Heiligen Inquisition)

Galileo Galilei

Nun wollen wir aus einer Urne mit n Kugeln mit und ohne Zurücklegen k tupel bilden

zuerst zur Begriffserklären Tupel : Tupel sind Anordnungen von Objekten bei denen die Reihenfolge eine Rolle spielt (also  (2,4) \neq (4,2) )
Bei einem Würfel kann ein 2-Tupel zum Beispiel (2,4) sein.
Bei Mengen ist die Reihenfolge nicht wichtig (also \left\{ 2,4 \right\} = \left\{ 4,2 \right\}) Nun zum Wesentlichen : Im ersten Versuch zeihen wir n kugeln ohne Zurücklegen aus einer Urne mit n Kugeln.

Als Beispiel werden Kugeln von a bis d genommen.

Zuerst gibt es vier Möglickeiten dan 3 dann 2 dann eine also: 4! Deswegen:

n Objekte lassen sich auf n! verschiedene Weisen einordnen Diese Einordnungen nennt man Permutationen

Nun wiederholen wir das selbe mit zurücklegen
Als Beispiel hierfür verwenden wir die Zahlen 0 und 1
Es gibt insgesamt 8 Möglichkeiten, was 2^3 entspricht

Aus n Ziffern und K Stellen entstehen n^k Tupel

Protokoll vom 16.03.,18.03. und 23.03.2016, Themen: Combinatorics I/II,Binomialcoefficient, Sample Exam

Protokoll von ----Sinan98OB (Diskussion) 09:22, 19. Mär. 2016 (CET)-- (Schuljahr 2016)
Lehrer C.-J. Schmitt (6 Unterrichtsstunden)
verbessert von--Sinan98OB (Diskussion) 18:13, 12. Apr. 2016 (CEST)

Combinatorics I

When we talk about Combinatorics I we talk about Tupel. Tupels are groups of various objects with the attribute that the order is important .Because we already talked about case 1&2 in the previous protocol, i am going to explain case 3:

I /3: Form k_Tupels with n-objects, without putting back



Because we can see how that Experiment works and how the tree diagramm looks like, we can make more examples:


n-objects und 2_Tupel

Amount of Tupel: n\cdot (n-1)= \frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot....\cdot 2\cdot1}{(n-2)\cdot(n-3)\cdot....\cdot 2\cdot1} =  \frac{n!}{(n-2)!}


n-objects und 3_Tupel

Amount of Tupel:n\cdot (n-1)\cdot(n-2)= \frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot....\cdot 2\cdot1}{(n-3)\cdot(n-4)\cdot....\cdot 2\cdot1} =  \frac{n!}{(n-3)!}


Through our examples we are able to create a general formula

n-objects und k_Tupel

Amount of Tupel:

n\cdot (n-1)\cdot(n-2)\cdot.....\cdot(n-(k-1))

n\cdot (n-1)\cdot(n-2)\cdot.....\cdot(n-k+1)

= \frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-k+1)\cdot(n-k)\cdot(n-k-1)\cdot....\cdot 2\cdot1}{(n-k)\cdot(n-k-1)\cdot....\cdot 2\cdot1} =  \frac{n!}{(n-k)!}


Remember: This formula describes the Amount of possibilities how many k_Tupel can be made out of n-objects. Remember: order is important and without putting back. ( German translation: Diese Formel beschreibt die Anzahl der Möglichkeiten wie man aus n-Ziffern k_Tupel bilden kann. Die Reihenfolge ist wichtig und ohne Zurücklegen)

Combinatorics II

In Kombinatorik II we are talking about Sets. A Set is a group of objects where the order is no more important.

II/1 Form Subsets with k-objects out of a Set with n-objects


Make Subsets with 2 objects out of a Set with 3 objects ( without putting back ). In relation with combinatorics I we also calculate the amount of Tupel:


We have 3 Subsets, but we can make 6 Tupel. We can calculate the number of certain Tupels with this formula: Amount of objects in our Subset multiplied with k!


Make Subsets with 3 objects out of a Set with 4 objects ( without putting back ):




n=4 k=2
6 subsets Amount of Tupels: 6\cdot 2! = 12
Amount of subsets: \frac{12}{2!} =6


n=3 k=2
3 subsets Amount of Tupels: 3\cdot 2! = 6
Amount of subsets: \frac{6}{2!} =3

Now we can derive a new formula:

 \frac{k_{-} Tupel}{k!} = Amount of Subsets

The Number of k_Tupels divided by k! factorial gives us the number of the subsets in dependence of the Number of k-Subsets.

Combined with the formula for the Number of k_Tupels we can create a new and very important formula, the so called Binomial coefficient:

\frac{n!}{(n-k)!\cdot k!}=\binom{n}{k} 

If we have a Set with n-objects, we can use this formula to define the amount of the subsets with k-objects. The order is not important and we do not put the object back.



To explain the Binomialcoefficient here some examples:

\binom{4}{3}= \frac{4!}{3!\cdot(4-3)!} =4

\binom{5}{2}= \frac{5!}{2!\cdot(5-2)!} =10

\binom{20}{18} = \frac{20!}{18!\cdot(20-18)!}= 190

Now we have one problem:

Most calculators can not calculate with factorials over 70, so how can we solve this formula:

\binom{75}{73}= \frac{75!}{73!\cdot2!}

We use a Hypothesis:

\binom{n}{k}= \binom{n}{n-k}

With that Hypothesis we can solve this problem:

\binom{75}{73}= \binom{75}{2}= \frac{75!}{73!\cdot 2!} = \frac{75\cdot 74\cdot 73\cdot 72....\cdot3\cdot2\cdot1}{2\cdot 1\cdot73\cdot72....\cdot3\cdot2\cdot1}

We can see how our hypothesis works. We subtract our upper Number with the lower one and can use the result as the lower number. This helps if the upper and the lower Number are similar to each other. For example 75 and 73 or 100 and 98.

\binom{75}{2}= \frac{75\cdot 74}{2\cdot 1} =2775

We write in full what the factorial of the lower number means and write exactly the same number of the Factorial of the upper number, but backwards. If we have the 3 as the lower number and write it out like this:

\frac{66!}{3\cdot\ 2\cdot1}

the next step looks like this:

\frac{66\cdot 65\cdot 64}{3\cdot\ 2\cdot1}

After mental arithmetics of binomialcoefficients, here you can see and check the solutions:

\binom{10}{2}= 45

\binom{20}{2}= 190

\binom{5}{3}= 10

\binom{6}{4}= 15

\binom{100}{98}= 4955

\binom{15}{13}= 105

\binom{12}{2}= 66

\binom{4}{3}= 4

\binom{67}{66}= 67

\binom{10}{7}= 120

\binom{50}{48}= 1225

Example Lottery

We can use our Binomialcoefficient to define the probability of lottery draws:

Winning-Draw: \big\{2,3,4,5,6,7\big\} this draw has a probability of \frac{1}{\binom{49}{6}}

What is the probability of 5 right ones?

5 right ones could look like this:

\big\{ -,3,4,5,6,7\big\} 

or like this:

\big\{2,3,4,5,6, - \big\}

We can see that we create a Subset of 5 numbers out of a Set of 6 numbers: \binom{6}{5}

Now we have to fill the Gap. Each Gap has 43 possibilities to be filled with one wrong number:\binom{43}{1}

To calculate the probability we have to multiply the two Coefficients and divide them with the number of possible draws:

 \frac{\binom{6}{5}\cdot \binom{43}{1}}{\binom{49}{6}}= 0,00184%

We can use this formula for more examples:

4 right ones:  \frac{\binom{6}{4}\cdot \binom{43}{2}}{\binom{49}{6}} = 0,0968%

3 right ones:  \frac{\binom{6}{3}\cdot \binom{43}{3}}{\binom{49}{6}} = 1,77%

Sample Exam

Here you have a link to the sample exam for our exam on the 23.03.2016

Sample exam