Question

Two types of defects are observed in the production of integrated circuit boards. It is estimated that 6% of the boards have solder defects and 3% have some surface-finish defects. The occurrences of the two types of defects are assumed to be independent of each other. If a circuit board is randomly selected from the output, find the probabilities for the following situations:(a) Either a solder defect or a surface-finish defect or both are found.(b) Only a solder defect is found.(c) Both types of defects are found.(d) The board is free of defects.(e) If a surface-finish

Answer 1

Answer:

(a) 0.09

(b) 0.06

(c) 0.0018

(d) 0.9118

(e) 0.03

Step-by-step explanation:

Let A = boards have solder defects and B = boards have surface defects.

The proportion of boards having solder defects is, P (A) = 0.06.

The proportion of boards having surface-finish defects is, P (B) = 0.03.

It is provided that the events A and B are independent, i.e.

$P(A\cap B)=P(A)\times P(B)$

(a)

Compute the probability that either a solder defect or a surface-finish defect or both are found as follows:

= P (A or B) + P (A and B)

$=P(A)+P(B)-P(A\cap B)+P(A\cap B)\\=P(A)+P(B)\\=0.06+0.03\\=0.09$

Thus, the probability that either a solder defect or a surface-finish defect or both are found is 0.09.

(b)

The probability that a solder defect is found is 0.06.

(c)

The probability that both defect are found is:

$P(A\cap B)=P(A)\times P(B)\\=0.06\times0.03\\=0.0018$

Thus, the probability that both defect are found is 0.0018.

(d)

The probability that none of the defect is found is:

$P(A^{c}\cup B^{c})=1-P(A\cup B)\\=1-P(A)-P(B)+P(A\cap B)\\=1-P(A)-P(B)+[P(A)\times P(B)]\\=1-0.06-0.03+(0.06\times0.03)\\=0.9118$

Thus, the probability that none of the defect is found is 0.9118.

(e)

The probability that the defect found is a surface finish is 0.03.