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Question: An article manufactured by a company consists of two parts X and Y. In the process of manufacture of...

An article manufactured by a company consists of two parts X and Y. In the process of manufacture of part X, 9 out of 104 parts may be defective. Similarly, 5 out of 100 are likely to be defective in the manufacture of part Y. Calculate the probability that the assembled product will not be defective.

Explanation

Solution

The probability is just a number that lies between 0 and 1 of an event. It’s represented by “P(A)”.
The formula for the probability
⇒P(A)=N(A)N(S)\Rightarrow P\left( A \right)=\dfrac{N(A)}{N(S)}
Where, P (A) is the probability of an event ′A′'A'
N(A) is the total number of favorable outcomes and
N(S) is the total number of an event in the sample space.
The range of probability of an event is 0≤P(A)≤10\le P(A)\le 1
There are some rule which will help us
Rules of complementary events P(A′)+P(A)=1P\left( A' \right)+P\left( A \right)=1
Independent event P(A∩B)=P(A).P(B)P(A\cap B)=P(A).P(B)

Complete step by step solution:
Total number of manufactured part X=104=104
Total number of manufactured part Y=100=100
Let’s assume that event A ==Total number of part X that are defective
⇒A=9\Rightarrow A=9
The probability of A is (Parts are defective)
⇒P(A)=9104\Rightarrow P(A)=\dfrac{9}{104}
Then the probability of A’==Part X is not defective
⇒P(A′)=1−P(A)\Rightarrow P(A')=1-P(A)
⇒P(A′)=1−9104\Rightarrow P(A')=1-\dfrac{9}{104}
⇒P(A′)=104−9104=95104\Rightarrow P(A')=\dfrac{104-9}{104}=\dfrac{95}{104}
⇒P(A′)=95104\Rightarrow P(A')=\dfrac{95}{104}
Now let’s event B==Total number of part X that are defective
⇒B=5\Rightarrow B=5
The probability of B (Part are defective)
⇒P(B)=5100\Rightarrow P(B)=\dfrac{5}{100}
Then the probability of B’==Part X is not defective
⇒P(B′)=1−P(B)\Rightarrow P(B')=1-P(B)
⇒P(B′)=1−5100\Rightarrow P(B')=1-\dfrac{5}{100}
⇒P(B′)=100−5100=95100\Rightarrow P(B')=\dfrac{100-5}{100}=\dfrac{95}{100}
⇒P(B′)=95100\Rightarrow P(B')=\dfrac{95}{100}
Now we can say that P(A′)=95104P(A')=\dfrac{95}{104}and P(B′)=95100P(B')=\dfrac{95}{100}
Therefore the probability of an assembled product that is not defective mean both Part X and Part Y are not defective
⇒P(A′).P(B′)=95104×95100\Rightarrow P(A').P(B')=\dfrac{95}{104}\times \dfrac{95}{100}
⇒95104×1920\Rightarrow \dfrac{95}{104}\times \dfrac{19}{20}
⇒95104×1920=19104×194\Rightarrow \dfrac{95}{104}\times \dfrac{19}{20}=\dfrac{19}{104}\times \dfrac{19}{4}
⇒19104×194=361416\Rightarrow \dfrac{19}{104}\times \dfrac{19}{4}=\dfrac{361}{416}
⇒361416\Rightarrow \dfrac{361}{416}

The probability of an assembled product that is not defective will be361416\dfrac{361}{416}.

Note:
The probability is a ratio where we compare how many times an outcome can occur compared to all possible sample space.
Two events are independent when the outcome of the first event does not affect the outcome of the second event is called the independent event.