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Mathematik » Stochastik und Statistik » Bernoulli, Bayes-Theorie
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Universität/Hochschule Bernoulli, Bayes-Theorie
sharkk
Aktiv Letzter Besuch: im letzten Quartal
Dabei seit: 11.11.2013
Mitteilungen: 72
Aus: Hannover
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag  Themenstart: 2020-05-28


Guten Tag,

Ich habe hier ein paar Aufgaben bekommen, die benotet werden und die ich gerne einmal im Vorhinein überprüfen möchte bevor ich sie abgebe.
Nicht wundern, das Englisch des Professors ist sehr schlecht, dementsprechend sind die Aufgaben etwas komisch formuliert.



LÖSUNG:
1)        The probability that plant 1 produced the part is: 1/5 and the probability that  the part is defective is: 0,04.
The probability that plant 2 produced the part is: 3/10 and the probability that the part is defective is 0,02.
The probability that plant 3 produced the part is: ½ and the probability that the part is defective is 0,01

Now I am going to calculate the total probability for the indices of a defective item, which is the addition of the probability of a defective item from all items:
Total probability: (1/5 * 0,04)+(3/10 * 0,02)+(1/2 * 0,01) = 0,019
Now I am going to calculate the a posteriori for each plant by dividing the probability of receiving a defective item from each one and dividing it by the total probability:

A posteriori plant 1: (1/5 * 0,04)/0,019 = 0,421

 A posteriori plant 2: (3/10 * 0,02)/0,019 = 0,3157

A posteriori plant 3: (1/2 * 0,01)/0,019 = 0,2631

Conclusion: It is more likely that we received the defective item from plant 1


2)        Total probability: (3/5 * 0,1) + (2/5 *0,2) = 0,14
A posteriori: (2/5 *0,2)/0.14 = 0,57




5.4. In order to find the probability that at least one arrow hits the target I calculate the probability that no arrow hits the target and subtract one from it.

1 -  ( 0,2*0,1) = 0,98

The probability of at least one arrow hitting the target is 0,98.


1.5.1. There are 6 messages of which EACH (?) can be distorted with a probability of p = 0,2.
            The probability that one message is not distorted: p = 0,8.
-        to find the combinations of which 4 out of 6 messages can be not distorted we   calculate: 6C4: 6!/[4!(6!-4!)] = 15
-        now we calculate the probability that exactly 4 out of 6 to be not distorted: (0,2)2*(0,8)4=0,016384
-        now we take the different combinations into account: 15*0,0163=0,245




 In order to find the probability that both robots find the image I calculate the probability that they don’t find the pattern and substract 1 from it

1 - ( 0,5*0,5) = 0,75

The probability is 0,75

Due to the ratio the probability that the transmitted signal is a 1 is P(trans1)= 5/8

Due to the ratio the probability that the transmitted signal is a 0 is P(trans0)= 3/8

The probability that the transmitted signal is a 1 and the signal is not distorted is: 3/5

The probability that the transmitted signal is a 0 and the signal is not distorted is: 2/3

The probability that the transmitted signal is a 1 and the signal is distorted is: 2/5

The probability that the transmitted signal is a 0 and the signal is distorted is: 1/3

We calculate the total probability that the received signal is a 1: 5/8 * 3/5 + 3/8 * 1/3 = 1/2

We calculate the total probability that the received signal is a 0: 5/8 * 2/5 + 3/8 * 2/3 = ½

We calculate posteriori for transm and receiv. 1: (5/8 * 3/5)/0,5 = ¾

We calculate posteriori for transm and receiv. 0: (5/8 * 3/5)/0,5 = ½



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sharkk
Aktiv Letzter Besuch: im letzten Quartal
Dabei seit: 11.11.2013
Mitteilungen: 72
Aus: Hannover
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag  Beitrag No.1, vom Themenstarter, eingetragen 2020-05-29


niemand der mal kurz rübergucken könnte?

oder sollte ich es übersetzen?



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