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Math  /  Data & Statistics

QuestionArbeitsblatt 2 zu Baumdiagrammen Aufgabe 1 Bei einer Produktionskontrolle werden in 3 Prüfungsgängen Länge, Breite und Höhe eines Metallstückes geprüft (in jedem Prüfungsgang nur eine Größe!). Diese sind erfahrungsgemäß mit den Wahrscheinlichkeiten 0,2 (Länge) bzw. 0,25 (Breite) bzw. 0,15 (Höhe) außerhalb vorgegebener Toleranzgrenzen. Ein Metallstück wird nicht ausgeliefert, wenn mindestens zwei der Kontrollen negativ ausgehen. Mit welcher Wahrscheinlichkeit wird ein kontrolliertes Werkstück ausgeliefert? In welcher Reihenfolge würden Sie die Überprüfungen durchführen?

Studdy Solution

STEP 1

1. The probability that the length is outside the tolerance limits is P(L)=0.2P(L) = 0.2.
2. The probability that the width is outside the tolerance limits is P(B)=0.25P(B) = 0.25.
3. The probability that the height is outside the tolerance limits is P(H)=0.15P(H) = 0.15.
4. The events of the length, width, and height being outside the tolerance limits are independent.
5. A metal piece is not delivered if at least two out of three tests fail.

STEP 2

1. Calculate the probability of each possible combination of test results (pass/fail).
2. Determine the probability that a metal piece will be delivered.
3. Analyze the order of the tests that would be optimal based on probabilities.

STEP 3

List all possible combinations of the test results for length, width, and height.
The combinations are:
1. All pass: P(L‾∩B‾∩H‾)P(\overline{L} \cap \overline{B} \cap \overline{H})
2. One fails: P(L∩B‾∩H‾),P(L‾∩B∩H‾),P(L‾∩B‾∩H)P(L \cap \overline{B} \cap \overline{H}), P(\overline{L} \cap B \cap \overline{H}), P(\overline{L} \cap \overline{B} \cap H)
3. Two fail: P(L∩B∩H‾),P(L∩B‾∩H),P(L‾∩B∩H)P(L \cap B \cap \overline{H}), P(L \cap \overline{B} \cap H), P(\overline{L} \cap B \cap H)
4. All fail: P(L∩B∩H)P(L \cap B \cap H)

STEP 4

Calculate the probability of each combination using the independence of events.
P(L‾∩B‾∩H‾)=(1−P(L))(1−P(B))(1−P(H))=0.8⋅0.75⋅0.85 P(\overline{L} \cap \overline{B} \cap \overline{H}) = (1 - P(L))(1 - P(B))(1 - P(H)) = 0.8 \cdot 0.75 \cdot 0.85

STEP 5

Continue calculating probabilities for the scenarios where exactly one test fails.
P(L∩B‾∩H‾)=P(L)⋅(1−P(B))⋅(1−P(H))=0.2⋅0.75⋅0.85 P(L \cap \overline{B} \cap \overline{H}) = P(L) \cdot (1 - P(B)) \cdot (1 - P(H)) = 0.2 \cdot 0.75 \cdot 0.85
P(L‾∩B∩H‾)=(1−P(L))⋅P(B)⋅(1−P(H))=0.8⋅0.25⋅0.85 P(\overline{L} \cap B \cap \overline{H}) = (1 - P(L)) \cdot P(B) \cdot (1 - P(H)) = 0.8 \cdot 0.25 \cdot 0.85
P(L‾∩B‾∩H)=(1−P(L))⋅(1−P(B))⋅P(H)=0.8⋅0.75⋅0.15 P(\overline{L} \cap \overline{B} \cap H) = (1 - P(L)) \cdot (1 - P(B)) \cdot P(H) = 0.8 \cdot 0.75 \cdot 0.15

STEP 6

Calculate probabilities for the scenarios where exactly two tests fail.
P(L∩B∩H‾)=P(L)⋅P(B)⋅(1−P(H))=0.2⋅0.25⋅0.85 P(L \cap B \cap \overline{H}) = P(L) \cdot P(B) \cdot (1 - P(H)) = 0.2 \cdot 0.25 \cdot 0.85
P(L∩B‾∩H)=P(L)⋅(1−P(B))⋅P(H)=0.2⋅0.75⋅0.15 P(L \cap \overline{B} \cap H) = P(L) \cdot (1 - P(B)) \cdot P(H) = 0.2 \cdot 0.75 \cdot 0.15
P(L‾∩B∩H)=(1−P(L))⋅P(B)⋅P(H)=0.8⋅0.25⋅0.15 P(\overline{L} \cap B \cap H) = (1 - P(L)) \cdot P(B) \cdot P(H) = 0.8 \cdot 0.25 \cdot 0.15

STEP 7

Calculate the probability for the scenario where all three tests fail.
P(L∩B∩H)=P(L)⋅P(B)⋅P(H)=0.2⋅0.25⋅0.15 P(L \cap B \cap H) = P(L) \cdot P(B) \cdot P(H) = 0.2 \cdot 0.25 \cdot 0.15

STEP 8

Combine the probabilities of scenarios where at least two tests fail to determine the probability that a metal piece is not delivered.
P(not delivered)=P(L∩B∩H‾)+P(L∩B‾∩H)+P(L‾∩B∩H)+P(L∩B∩H) P(\text{not delivered}) = P(L \cap B \cap \overline{H}) + P(L \cap \overline{B} \cap H) + P(\overline{L} \cap B \cap H) + P(L \cap B \cap H)
=0.2⋅0.25⋅0.85+0.2⋅0.75⋅0.15+0.8⋅0.25⋅0.15+0.2⋅0.25⋅0.15 = 0.2 \cdot 0.25 \cdot 0.85 + 0.2 \cdot 0.75 \cdot 0.15 + 0.8 \cdot 0.25 \cdot 0.15 + 0.2 \cdot 0.25 \cdot 0.15

STEP 9

Calculate the combined probability from STEP_6.
P(not delivered)=0.0425+0.0225+0.03+0.0075=0.1025 P(\text{not delivered}) = 0.0425 + 0.0225 + 0.03 + 0.0075 = 0.1025

STEP 10

Calculate the probability that a metal piece is delivered.
P(delivered)=1−P(not delivered)=1−0.1025=0.8975 P(\text{delivered}) = 1 - P(\text{not delivered}) = 1 - 0.1025 = 0.8975

STEP 11

Determine the order of tests based on the probability of failing each test. It is optimal to start with the test with the highest failure probability to possibly save on further testing costs and efforts.
The order should be:
1. Width (P(B)=0.25P(B) = 0.25)
2. Length (P(L)=0.2P(L) = 0.2)
3. Height (P(H)=0.15P(H) = 0.15)
Solution: The probability that a controlled metal piece will be delivered is 0.89750.8975. The optimal order of testing is width, length, and then height.

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