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

Follows are the solution to this question:

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[(F)/(A)] =0.15\\\\\to P[(F)/(B)] =0.12\\\\\to P[(F)/(C)] =0.14\\`

Now, we calculate the tooling value:

`\to p[(C)/(F)]`

using the baues therom:

`\to p[(C)/(F)] = (p[C] * p[(F)/(C) ])/(p[A] * p[(F)/(A)] + p[B] * p[(F)/(B)]+p[C] * p[(F)/(C)] )`

`= ( 0.17 * 0.14 )/(0.55 * 0.15 + 0.28 * 0.12 + 0.17 * 0.14 ) \\\\= (0.0238)/(0.0825 + 0.0336 + 0.0238) \\\\= (0.0238)/(0.1399) \\\\=0.1701`