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Answer:
The probability of selecting a non-defective part provided by supplier A is 0.807.
Step-by-step explanation:
Let A = a part is supplied by supplier A, B = a part is supplied by supplier B and D = a part is defective.
Given:
P (D|A) = 0.05, P(D|B) = 0.09
A supplies four times as many parts as B, i.e. n (A) = 4 and n (B) = 1.
Then the probability of event A and B is:
[tex]P(A)=\frac{n(A)}{n(A)+N(B)}= \frac{4}{4+1}=0.80\\P(B)\frac{n(B)}{n(A)+N(B)}= \frac{1}{4+1}=0.20[/tex]
Compute the probability of selecting a defective product:
[tex]P(D)=P(D|A)P(A)+P(D|B)P(B)\\=(0.05\times0.80)+(0.09\times0.20)\\=0.058[/tex]
The probability of selecting a non-defective part provided by supplier A is:
[tex]P(A|D')=\frac{P(D'|A)P(A)}{P(D')} = \frac{(1-P(D|A))P(A)}{1-P(D)}\\=\frac{(1-0.05)\times0.80}{(1-0.058)}\\ =0.80679\\\approx0.807[/tex]
Thus, the probability of selecting a non-defective part provided by supplier A is 0.807.
The required probability of selecting a non-defective part provided by supplier A is 0.807.
Let,
A part is supplied by supplier A,
B part is supplied by supplier B,
And D = a part is defective.
Given:
P ([tex]\frac{D}{A}[/tex]) =5% = 0.05,
P([tex]\frac{D}{B}[/tex]) = 9% = 0.09
A supplies four times as many parts as B, .
Then, n (A) = 4 and n (B) = 1.
The probability of event A and B is:
Probability of event P(A) = [tex]\frac{n (A)}{n (A) + n(B)}[/tex] = [tex]\frac{4}{4+1}[/tex]
P(A) = [tex]\frac{4}{5}[/tex]
And Probability of Event P(B) = [tex]\frac{n (B )}{n (A) + n(B)} = \frac{1}{4+1}[/tex]
P(B) = [tex]\frac{1}{5}[/tex]
Then , the probability of selecting a defective product:
P(D) = [tex]P(\frac{D}{A}) P(A) + P(\frac{D}{B}) P(B)[/tex]
P(D) = 0.50×0.80 + 0.09×0.20
P(D) = 0.058
The probability of selecting a non-defective part provided by supplier A is
[tex]P(\frac{A}{D'} )= \frac{P(D'(A)) . P(A)}{P(D')} \\\\[/tex]
= [tex]\frac{1- P(D(A)). P(A)}{1-P(A)} \\\\\frac{(1-0.05) . 0.80}{1- 0.05} \\\\[/tex]
= 0.807
Hence, the probability of selecting a non-defective part provided by supplier A is 0.807.
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