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damaskus [11]
3 years ago
11

Two different suppliers, A and B, provide a manufacturer with the same part. All supplies of this part are kept in a large bin.

in the past, 5% of the parts supplied by A and 9% of the parts supplied by B have been defective. A supplies four times as many parts as B. Suppose you reach into the bin and select a part, and find it is nondetective. What is the probability that it was supplied by A
Mathematics
1 answer:
koban [17]3 years ago
5 0

Answer:

The probability of selecting a non-defective part provided by supplier A is 0.807.

Step-by-step explanation:

Let <em>A</em> = a part is supplied by supplier A, <em>B</em> = a part is supplied by supplier B and <em>D</em> = a part is defective.

<u>Given</u>:

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 <em>A</em> and <em>B</em> is:

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

Compute the probability of selecting a defective product:

P(D)=P(D|A)P(A)+P(D|B)P(B)\\=(0.05\times0.80)+(0.09\times0.20)\\=0.058

The probability of selecting a non-defective part provided by supplier A is:

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

Thus, the probability of selecting a non-defective part provided by supplier A is 0.807.

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0 =  - 49t {}^{2}  + 270t + 24
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49t {}^{2}  - 270t - 24 = 0Solve the quadratic equation.

t =  \frac{ - ( - 270) +  -  \sqrt{( - 270) {}^{2} - 4 \times 49 \times ( - 24) } }{2 \times 49}
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t =  \frac{270 +  -  \sqrt{72900  + 4704} }{98}
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