Question

fter production, a computer component is given a quality score of A, B, and C. On the average U of the components were given a quality score A, ( of the components were given a quality score B, and of the components were given a quality score C. Furthermore, it was found that actually of the components given a quality score A failed, of the components given a quality score B failed, and of the components given a quality score C failed. 11-A. What is the probability that a randomly selected component is NOT failed and received a quality score B

Answer 1

Answer:

Follows are the solution to this question:

Step-by-step explanation:

In the given question some of the data is missing so, its correct question is defined in the attached file please find it.

Let

A is quality score of A

B is quality score of B

C is quality score of C

$\to P[A] =0.55\\\\\to P[B] =0.28\\\\\to P[C] =0.17$

Let F is a value of the content so, the value is:

$\to P[\frac{F}{A}] =0.15\\\\\to P[\frac{F}{B}] =0.12\\\\\to P[\frac{F}{C}] =0.14$

Now, we calculate the tooling value:

$\to p[\frac{C}{F}]$

using the baues therom:

$\to p[\frac{C}{F}] = \frac{p[C] \times p[\frac{F}{C} ]}{p[A] \times p[\frac{F}{A}] + p[B] \times p[\frac{F}{B}]+p[C] \times p[\frac{F}{C}] }$  

            $= \frac{ 0.17 \times 0.14 }{0.55 \times 0.15 + 0.28 \times 0.12 + 0.17 \times 0.14 } \\\\= \frac{0.0238}{0.0825 + 0.0336 + 0.0238} \\\\= \frac{0.0238}{0.1399} \\\\=0.1701$