Respuesta :
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.
[tex]P(A\cap B)=P(A)\times P(B)[/tex]
(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)
[tex]=P(A)+P(B)-P(A\cap B)+P(A\cap B)\\=P(A)+P(B)\\=0.06+0.03\\=0.09[/tex]
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:
[tex]P(A\cap B)=P(A)\times P(B)\\=0.06\times0.03\\=0.0018[/tex]
Thus, the probability that both defect are found is 0.0018.
(d)
The probability that none of the defect is found is:
[tex]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[/tex]
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.
