K(1,0), L(4,1), M(1,4), N(-2, 3) 90% counterclockwise rotation
1 answer:
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Answer:
(a) 0.09
(b) 0.06
(c) 0.0018
(d) 0.9118
(e) 0.03
Step-by-step explanation:
Let <em>A</em> = boards have solder defects and <em>B</em> = 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.
(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)
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:
Thus, the probability that both defect are found is 0.0018.
(d)
The probability that none of the defect is found is:
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.